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NCERT Exemplar for Class 12 Maths Chapter-4 (Book Solutions)

NCERT Exemplar for Class 12 Maths Chapter-4 (Book Solutions)

NCERT Exemplar for Class 12 Maths - Determinants - Free PDF Download

Free PDF download of NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants solved by expert Maths teachers on Vedantu.com as per NCERT (CBSE) Book guidelines. All Chapter 4 - Determinants Exercise questions with solutions to help you to revise the complete syllabus and score more marks in your Examinations.

If learners want to improve their understanding of ideas while also practising and preparing for tests, the NCERT Exemplar textbooks are all they need. NCERT textbooks are available for all Classes and disciplines. Students can now use the NCERT Exemplar Solutions, which are available subject-by-subject, to clear up any misconceptions they may have about the concepts and issues included in these textbooks. All of the answers are based on the most recent CBSE patterns to assist pupils to improve their grades.

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Access NCERT Exemplar Solutions for Class 12 Mathematics Chapter 4 - Determinants

Solved Examples

Short Answer (S. A)

Example 1: If $| \begin{matrix} 2 x & 5 \\ 8 & x \end{matrix} |$ = $| \begin{matrix} 6 & 5 \\ 8 & 3 \end{matrix} |$ , then find $x .$

Ans: Here, we have $| \begin{matrix} 2 x & 5 \\ 8 & x \end{matrix} |$ = $| \begin{matrix} 6 & 5 \\ 8 & 3 \end{matrix} |$

⇒ $2 x^{2} - 40 = 18 - 40$

⇒ $2 x^{2} = 18$

⇒ $x^{2} = \frac{18}{2}$

⇒ $x^{2} = 9$

⇒ $x = \sqrt{9}$

⇒ $x = \pm 3$

Hence, value of $x$ is $\pm 3 .$

Example 2: If △ $= | \begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix} |$ , △1 = $| \begin{matrix} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{matrix} |$ , then prove that

△+△1=0.

Ans: here, we have △1 $= | \begin{matrix} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{matrix} |$

Now, interchanging rows and columns, we get

⇒ △1 $= | \begin{matrix} 1 & y z & x \\ 1 & z x & y \\ 1 & x y & z \end{matrix} |$

⇒ △1 $= \frac{1}{x y z}$ $| \begin{matrix} x & x y z & x^{2} \\ y & x y z & y^{2} \\ z & x y z & z^{2} \end{matrix} |$

Taking common $x y z$ from $C_{2},$ we get

⇒ △1 $= \frac{1}{x y z} \times x y z$ $| \begin{matrix} x & 1 & x^{2} \\ y & 1 & y^{2} \\ z & 1 & z^{2} \end{matrix} |$

Now, interchanging $C_{1}$ and $C_{2}$ , we get

⇒ △1 $= (- 1)$ $| \begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix} |$

⇒ △1 = -△

⇒ △ + △1 = 0

Hence proved.

Example 3: Without expanding, Show that △ $= | \begin{matrix} c o s e c^{2} θ & c o t^{2} θ & 1 \\ c o t^{2} θ & c o s e c^{2} θ & - 1 \\ 42 & 40 & 2 \end{matrix} | = 0$

Ans: here, we have △ $= | \begin{matrix} c o s e c^{2} θ & c o t^{2} θ & 1 \\ c o t^{2} θ & c o s e c^{2} θ & - 1 \\ 42 & 40 & 2 \end{matrix} |$

Applying $C_{1} \to C_{1} - C_{2}$

⇒ △ $= | \begin{matrix} c o s e c^{2} θ - c o t^{2} θ & c o t^{2} θ & 1 \\ c o t^{2} θ - c o s e c^{2} θ & c o s e c^{2} θ & - 1 \\ 2 & 40 & 2 \end{matrix} |$

⇒ △ $= | \begin{matrix} c o s e c^{2} θ - c o t^{2} θ & c o t^{2} θ & 1 \\ - (c o s e c^{2} θ - c o t^{2} θ) & c o s e c^{2} θ & - 1 \\ 2 & 40 & 2 \end{matrix} |$

⇒ △ $= | \begin{matrix} 1 & c o t^{2} θ & 1 \\ - 1 & c o s e c^{2} θ & - 1 \\ 2 & 40 & 2 \end{matrix} |$

⇒ △ = 0 (when two columns are same, then value of determinants is 0)

Hence proved.

Example 4: Show that △ $= | \begin{matrix} x & p & q \\ p & x & q \\ q & q & x \end{matrix} |$ = $(x - p) (x^{2} + p x - 2 q^{2}) .$

Ans: here, we have △ $= | \begin{matrix} x & p & q \\ p & x & q \\ q & q & x \end{matrix} |$

Applying $C_{1} \to C_{1} - C_{2}$

⇒ △ $= | \begin{matrix} x - p & p & q \\ - (x - p) & x & q \\ 0 & q & x \end{matrix} |$

Taking common $(x - p)$ from $C_{1}$

⇒ △ $= (x - p) | \begin{matrix} 1 & p & q \\ - 1 & x & q \\ 0 & q & x \end{matrix} |$

Now, expanding along $C_{1}$

⇒ △ $= (x - p) [1 (x^{2} - q^{2}) + 1 (p x - q^{2}) + 0]$

⇒ △ $= (x - p) (x^{2} + p x - 2 q^{2})$

Hence proved.

Example 5: If △ $= | \begin{matrix} 0 & b - a & c - a \\ a - b & 0 & c - b \\ a - c & b - c & 0 \end{matrix} |$ , then show that △ is equal to zero.

Ans: here, we have △ $= | \begin{matrix} 0 & b - a & c - a \\ a - b & 0 & c - b \\ a - c & b - c & 0 \end{matrix} |$

Taking “-1” as common from $R_{1}, R_{2} a n d R_{3}$ , we get

⇒ △ $= {(- 1)}^{3} | \begin{matrix} 0 & a - b & a - c \\ b - a & 0 & b - c \\ c - a & c - b & 0 \end{matrix} |$

Now, interchanging rows and columns, we get

⇒ △ $= (- 1) | \begin{matrix} 0 & b - a & c - a \\ a - b & 0 & c - b \\ a - c & b - c & 0 \end{matrix} |$

⇒ △ = - △

⇒ 2△ = 0

⇒ △ = 0

Hence proved.

Example 6: Prove that ${(A^{- 1})}^{'} = {(A^{'})}^{- 1},$ where A is an invertible matrix.

Ans: Here, matrix A is an invertible matrix, so it is non – singular i.e., $| A | \neq 0$ .

And we know that, $| A | = | A^{'} |$ . So, $| A^{'} | \neq 0 .$

Also, we have $A A^{- 1} = A^{- 1} A = I$

Taking transpose on both sides, we get

⇒ ${(A A^{- 1})}^{^{'}} = {(A^{- 1} A)}^{^{'}} = I^{'}$

⇒ ${(A^{- 1})}^{^{'}} A^{'} = A^{'} {(A^{- 1})}^{^{'}} = I$

Thus, ${(A^{- 1})}^{^{'}}$ is inverse of $A^{'}$ ,

Therefore, ${(A^{- 1})}^{^{'}} = {(A^{'})}^{- 1}$

Hence proved.

Long Answer (L.A.)

Example 7: If $x = - 4$ is a root of △ = $| \begin{matrix} x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x \end{matrix} | = 0,$ then find the other two roots.

Ans: here, we have $| \begin{matrix} x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x \end{matrix} | = 0$

Applying $R_{1} \to R_{1} + R_{2} + R_{3}$

⇒ $| \begin{matrix} x + 4 & x + 4 & x + 4 \\ 1 & x & 1 \\ 3 & 2 & x \end{matrix} | = 0$

Taking $(x + 4)$ common from $R_{1}$

⇒ $(x + 4) | \begin{matrix} 1 & 1 & 1 \\ 1 & x & 1 \\ 3 & 2 & x \end{matrix} | = 0$

Applying $C_{1} \to C_{1} - C_{3}$ and $C_{2} \to C_{2} - C_{3}$

⇒ $(x + 4) | \begin{matrix} 0 & 0 & 1 \\ 0 & x - 1 & 1 \\ 3 - x & 2 - x & x \end{matrix} | = 0$

⇒ $(x + 4) [0 - (3 - x) (x - 1)] = 0$

⇒ $(x + 4) (3 - x) (x - 1) = 0$

⇒ $x = - 4, 3, 1$

Hence, the other two roots are 1 and 3.

Example 8: In a triangle ABC, if $| \begin{matrix} 1 & 1 & 1 \\ 1 + \sin A & 1 + \sin B & 1 + \sin C \\ \sin A + s i n^{2} A & \sin B + s i n^{2} B & \sin C + s i n^{2} C \end{matrix} | = 0,$

then prove that $Δ A B C$ is an isosceles triangle.

Ans: Here, we have $| \begin{matrix} 1 & 1 & 1 \\ 1 + s i n A & 1 + s i n B & 1 + s i n C \\ s i n A + s i n^{2} A & s i n B + s i n^{2} B & s i n C + s i n^{2} C \end{matrix} | = 0$

⇒ Applying $C_{1} \to C_{1} - C_{3}$ and $C_{2} \to C_{2} - C_{3}$

⇒ $| \begin{matrix} 0 & 0 & 1 \\ s i n A - s i n B & s i n B - s i n C & 1 + s i n C \\ s i n A - s i n B + s i n^{2} A - s i n^{2} B & s i n B - s i n C + s i n^{2} B - s i n^{2} C & s i n C + s i n^{2} C \end{matrix} | = 0$

Taking $(s i n A - s i n B)$ and $(s i n B - s i n C)$ from $C_{1}$ and $C_{2}$ respectively

⇒ $(s i n A - s i n B) (s i n B - s i n C) | \begin{matrix} 0 & 0 & 1 \\ 1 & 1 & 1 + s i n C \\ 1 + s i n A + s i n B & 1 + s i n B + s i n C & s i n C + s i n^{2} C \end{matrix} | = 0$

Now, expanding along $R_{1}$ , we get

⇒ $(s i n A - s i n B) (s i n B - s i n C) (1 + s i n B + s i n C - 1 - s i n A - s i n B) = 0$

⇒ $(s i n A - s i n B) (s i n B - s i n C) (s i n C - s i n A) = 0$

⇒ $(s i n A - s i n B) = 0 o r (s i n B - s i n C) = 0 o r (s i n C - s i n A) = 0$

⇒ $\sin A = \sin B o r \sin B = \sin C o r \sin C = \sin A$

⇒ $A = B o r B = C o r C = A$

Hence, triangle ABC is an isosceles triangle.

Hence proved.

Example 9: Show that if the determinant △ = $| \begin{matrix} 3 & - 2 & s i n 3 θ \\ - 7 & 8 & c o s 2 θ \\ - 11 & 14 & 2 \end{matrix} | = 0$ , then $s i n θ = 0$ or $\frac{1}{2}$

Ans: Here, we have $| \begin{matrix} 3 & - 2 & s i n 3 θ \\ - 7 & 8 & c o s 2 θ \\ - 11 & 14 & 2 \end{matrix} | = 0$

Applying $R_{2} \to R_{2} + 4 R_{1}$ and $R_{3} \to R_{3} + 7 R_{1}$ , we get

⇒ $| \begin{matrix} 3 & - 2 & s i n 3 θ \\ 5 & 0 & c o s 2 θ + 4 s i n 3 θ \\ 10 & 0 & 2 + 7 s i n 3 θ \end{matrix} | = 0$

Now, expanding along $C_{2}$ , we get

⇒ 2 (5( $2 + 7 s i n 3 θ) - 10 (c o s 2 θ + 4 s i n 3 θ) = 0)$

⇒ $[(10 + 35 s i n 3 θ) - (10 c o s 2 θ + 40 s i n 3 θ)] = 0$

⇒ $10 - 5 s i n 3 θ - 10 c o s 2 θ = 0$

⇒ $2 - s i n 3 θ - 2 c o s 2 θ = 0$

⇒ 2 $- (3 s i n θ - 4 s i n^{3} x) - 2 (1 - 2 s i n^{2} θ) = 0$

⇒ 2 $- 3 s i n θ + 4 s i n^{3} x - 2 + 4 s i n^{2} θ = 0$

⇒ $- 3 s i n θ + 4 s i n^{3} x + 4 s i n^{2} θ = 0$

⇒ $s i n θ (- 3 + 4 s i n^{2} x + 4 s i n θ) = 0$

⇒ $s i n θ (4 s i n^{2} x + 4 s i n θ - 3) = 0$

⇒ $s i n θ (4 s i n^{2} x + 6 s i n θ - 2 s i n θ - 3) = 0$

⇒ $s i n θ [2 s i n θ (2 s i n θ + 3) - 1 (2 s i n θ + 3)] = 0$

⇒ $s i n θ (2 s i n θ + 3) (2 s i n θ - 1) = 0$

⇒ $s i n θ = 0 o r (2 s i n θ + 3) = 0 o r (2 s i n θ - 1) = 0$

⇒ $s i n θ = 0 o r s i n θ = - \frac{3}{2} o r s i n θ = \frac{1}{2}$

As we know that $s i n θ \in [- 1, 1]$

Thus, $s i n θ = 0 o r s i n θ = \frac{1}{2}$

Hence proved.

Objective Type Questions

Choose the correct answer from the given four options in each of Examples 10 and 11.

Example 10: Let △ = $| \begin{matrix} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{matrix} |$ and △1 = $| \begin{matrix} A & B & C \\ x & y & z \\ z y & z x & x y \end{matrix} |$ , then

(A) △1 = △

(B) △ $\neq$ △1

(D) None of these

Ans: The correct answer is option (C).

Here, we have △ $= | \begin{matrix} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{matrix} |$ and △1 = $| \begin{matrix} A & B & C \\ x & y & z \\ z y & z x & x y \end{matrix} |$

⇒ △1 $= | \begin{matrix} A & B & C \\ x & y & z \\ z y & z x & x y \end{matrix} |$

Now, interchanging rows and columns, we get

⇒ △1 $= | \begin{matrix} A & x & z y \\ B & y & z x \\ C & z & x y \end{matrix} |$

⇒ △1 $= \frac{1}{x y z} | \begin{matrix} A x & x^{2} & x y z \\ B y & y^{2} & x y z \\ C z & z^{2} & x y z \end{matrix} |$

Taking $x y z$ common from $C_{3},$ we get

⇒ △1 $= \frac{1}{x y z} \times x y z$ $| \begin{matrix} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{matrix} |$

⇒ △1 = △

⇒ △ - △1 = 0

Hence, option (C) is the correct answer.

Example 11: If $x, y \in R,$ then the determinant

△ = $| \begin{matrix} \cos x & - \sin x & 1 \\ \sin x & \cos x & 1 \\ \cos (x + y) & - \sin (x + y) & 0 \end{matrix} |$ lies in the interval

(A) $[- \sqrt{2}, \sqrt{2}]$

(B) $[- 1, 1]$

(D) $[- 1, - \sqrt{2}]$

Ans: The correct answer is option (A).

Here, we have △ $= | \begin{matrix} c o s x & - s i n x & 1 \\ s i n x & c o s x & 1 \\ c o s (x + y) & - s i n (x + y) & 0 \end{matrix} |$

Applying $R_{3} \to R_{3} - R_{1} \cos y + R_{2} \sin y$

⇒ △ $= | \begin{matrix} c o s x & - s i n x & 1 \\ s i n x & c o s x & 1 \\ 0 & 0 & s i n y - c o s y \end{matrix} |$

Now, expanding along $R_{3}$ , we get

⇒ △ $= (\sin y - \cos y) (c o s^{2} x + s i n^{2} x)$

⇒ △ $= (\sin y - \cos y)$

⇒ △ $= \sqrt{2} (\frac{1}{\sqrt{2}} s i n y - \frac{1}{\sqrt{2}} c o s y)$

⇒ △ $= \sqrt{2} (c o s \frac{π}{4} s i n y - s i n \frac{π}{4} c o s y)$

⇒ △ $= \sqrt{2} s i n (y - \frac{π}{4})$

As we know that, $- 1 ⩽ s i n (y - \frac{π}{4}) ⩽ 1$

⇒ $- \sqrt{2} ⩽ \sqrt{2} s i n (y - \frac{π}{4}) ⩽ \sqrt{2}$

⇒ $- \sqrt{2} ⩽ Δ ⩽ \sqrt{2}$

Hence, option (A) is the correct answer.

Fill in the blank in each of the examples 12 to 14.

Example 12: If A, B, C are the angles of a triangles, then △ = $| \begin{matrix} s i n^{2} A & c o t A & 1 \\ s i n^{2} B & c o t B & 1 \\ s i n^{2} C & c o t C & 1 \end{matrix} | =$ ………..

Ans: Here, we have △ $= | \begin{matrix} s i n^{2} A & c o t A & 1 \\ s i n^{2} B & c o t B & 1 \\ s i n^{2} C & c o t C & 1 \end{matrix} |$

Applying $R_{2} \to R_{2} - R_{1},$ $R_{3} \to R_{3} - R_{1}$ , we get

⇒ △ $= | \begin{matrix} s i n^{2} A & c o t A & 1 \\ s i n^{2} B - s i n^{2} A & c o t B - c o t A & 0 \\ s i n^{2} C - s i n^{2} A & c o t C - c o t A & 0 \end{matrix} |$

⇒ △ $= | \begin{matrix} s i n^{2} A & c o t A & 1 \\ s i n (B + A) s i n (B - A) & \frac{c o s B}{s i n B} - \frac{c o s A}{s i n A} & 0 \\ s i n (C + A) s i n (C - A) & \frac{c o s C}{s i n C} - \frac{c o s A}{s i n A} & 0 \end{matrix} |$

Here, $A + B + C = π$

⇒ △ $= | \begin{matrix} s i n^{2} A & c o t A & 1 \\ s i n (π - C) s i n (B - A) & \frac{c o s B . s i n A - c o s A . s i n B}{s i n B . s i n A} & 0 \\ s i n (π - B) s i n (C - A) & \frac{c o s C . s i n A - c o s A . s i n C}{s i n C . s i n A} & 0 \end{matrix} |$

⇒ △ $= | \begin{matrix} s i n^{2} A & \frac{c o s A}{s i n A} & 1 \\ s i n C . s i n (B - A) & \frac{s i n (A - B)}{s i n B . s i n A} & 0 \\ s i n B . s i n (C - A) & \frac{s i n (A - C)}{s i n C . s i n A} & 0 \end{matrix} |$

Taking $\sin (B - A)$ and $\sin (C - A)$ common from $R_{2}$ and $R_{3}$ respectively.

⇒ △ = $s i n (B - A) s i n (C - A)$ $| \begin{matrix} {s i n}^{2} A & \frac{c o s A}{s i n A} & 1 \\ s i n C & \frac{- 1}{s i n B . s i n A} & 0 \\ s i n B & \frac{- 1}{s i n C . s i n A} & 0 \end{matrix} |$

Taking $\frac{1}{\sin A}$ common from $C_{2},$ we get

⇒ △ $= s i n (B - A) s i n (C - A)$ $| \begin{matrix} s i n^{2} A & c o s A & 1 \\ s i n C & \frac{- 1}{s i n B} & 0 \\ s i n B & \frac{- 1}{s i n C} & 0 \end{matrix} |$

Now, expanding along $R_{1},$ we get

⇒ △ $= s i n (B - A) s i n (C - A) (- 1 + 1)$

⇒ △ $= s i n (B - A) s i n (C - A) \times 0$

⇒ △ $= 0$

Thus, △ $= | \begin{matrix} s i n^{2} A & c o t A & 1 \\ s i n^{2} B & c o t B & 1 \\ s i n^{2} C & c o t C & 1 \end{matrix} | = 0 .$

Example 13: The determinant △ = $| \begin{matrix} \sqrt{23} + \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{46} & 5 & \sqrt{10} \\ 3 + \sqrt{115} & \sqrt{15} & 5 \end{matrix} |$ is equal to …....

Ans: Here, we have △ $= | \begin{matrix} \sqrt{23} + \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{46} & 5 & \sqrt{10} \\ 3 + \sqrt{115} & \sqrt{15} & 5 \end{matrix} |$

Taking $\sqrt{5}$ common from $C_{2}$ and $C_{3},$ we get

⇒ △ $= \sqrt{5} \times \sqrt{5} | \begin{matrix} \sqrt{23} + \sqrt{3} & 1 & 1 \\ \sqrt{15} + \sqrt{46} & \sqrt{5} & \sqrt{2} \\ 3 + \sqrt{115} & \sqrt{3} & \sqrt{5} \end{matrix} |$

Applying $C_{1} \to C_{1} - \sqrt{3} C_{2} - \sqrt{23} C_{3}$

⇒ △ $= 5 | \begin{matrix} 0 & 1 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{matrix} |$

⇒ △ $= 0$ (All the elements of a row is 0)

Thus, The determinant △ $= | \begin{matrix} \sqrt{23} + \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{46} & 5 & \sqrt{10} \\ 3 + \sqrt{115} & \sqrt{15} & 5 \end{matrix} |$ is equal to 0.

Example 14: The value of the determinant

△ $= | \begin{matrix} s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} & \cos 180^{\circ} \\ - s i n^{2} 67^{\circ} & - s i n^{2} 23^{\circ} & c o s^{2} 180^{\circ} \\ \cos 180^{\circ} & s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} \end{matrix} | =$ …..

Ans: Here, we have △ $= | \begin{matrix} s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} & c o s 180^{\circ} \\ - s i n^{2} 67^{\circ} & - s i n^{2} 23^{\circ} & c o s^{2} 180^{\circ} \\ c o s 180^{\circ} & s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} \end{matrix} |$

⇒ △ $= | \begin{matrix} s i n^{2} 23^{\circ} & s i n^{2} (90 - 23)^{\circ} & - 1 \\ - s i n^{2} 67^{\circ} & - s i n^{2} (90 - 67)^{\circ} & 1 \\ - 1 & s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} \end{matrix} |$

⇒ △ $= | \begin{matrix} s i n^{2} 23^{\circ} & c o s^{2} 23^{\circ} & - 1 \\ - s i n^{2} 67^{\circ} & - c o s^{2} 67^{\circ} & 1 \\ - 1 & s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} \end{matrix} |$

Applying $C_{1} \to C_{1} + C_{2} + C_{3}$

⇒ △ $= | \begin{matrix} s i n^{2} 23^{\circ} + c o s^{2} 23^{\circ} - 1 & c o s^{2} 23^{\circ} & - 1 \\ - (s i n^{2} 67^{\circ} + c o s^{2} 67^{\circ}) + 1 & - c o s^{2} 67^{\circ} & 1 \\ - 1 + s i n^{2} 23^{\circ} + s i n^{2} 67^{\circ} & s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} \end{matrix} |$

⇒ △ $= | \begin{matrix} 1 - 1 & c o s^{2} 23^{\circ} & - 1 \\ - 1 + 1 & - c o s^{2} 67^{\circ} & 1 \\ - 1 + 1 & s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} \end{matrix} |$

⇒ △ $= | \begin{matrix} 0 & c o s^{2} 23^{\circ} & - 1 \\ 0 & - c o s^{2} 67^{\circ} & 1 \\ 0 & s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} \end{matrix} |$

⇒ △ $= 0$ (All the elements of a column is 0)

Hence, △ $= | \begin{matrix} s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} & c o s 180^{\circ} \\ - s i n^{2} 67^{\circ} & - s i n^{2} 23^{\circ} & c o s^{2} 180^{\circ} \\ c o s 180^{\circ} & s i n^{2} 23^{\circ} & s i n^{2} 67^{\circ} \end{matrix} | = 0 .$

State whether the statements in Examples 15 to 18 is True or False.

Example 15: The determinant △ = $| \begin{matrix} c o s (x + y) & - s i n (x + y) & c o s 2 y \\ s i n x & c o s x & s i n y \\ - c o s x & s i n x & c o s y \end{matrix} |$ is independent of $x$ only.

Ans: Here, we have △ = $| \begin{matrix} \cos (x + y) & - \sin (x + y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & \cos y \end{matrix} |$

Applying $R_{1} \to R_{1} + R_{2} \sin y + R_{3} \cos y$ , we get

⇒ △ $= | \begin{matrix} 0 & 0 & 1 + c o s 2 y \\ s i n x & c o s x & s i n y \\ - c o s x & s i n x & c o s y \end{matrix} |$

Now, expanding along $R_{1}$ , we get

⇒ △ $= (1 + c o s 2 y) (s i n^{2} x + c o s^{2} x)$

⇒ △ $= (1 + c o s 2 y)$

Thus, the value of the given determinant is independent of $x$ only.

Hence, the given statement is True.

Example 16: The value of △ = $| \begin{matrix} 1 & 1 & 1 \\ ^{n} C_{1} & ^{n + 2} C_{1} & ^{n + 4} C_{1} \\ ^{n} C_{2} & ^{n + 2} C_{2} & ^{n + 4} C_{2} \end{matrix} |$ is 8.

Ans: Here, we have △ $= | \begin{matrix} 1 & 1 & 1 \\ _{.}^{n} C_{1} & _{.}^{n + 2} C_{1} & _{.}^{n + 4} C_{1} \\ _{.}^{n} C_{2} & _{.}^{n + 2} C_{2} & _{.}^{n + 4} C_{2} \end{matrix} |$

⇒ △ $= | \begin{matrix} 1 & 1 & 1 \\ \frac{n!}{(n - 1)!} & \frac{(n + 2)!}{(n + 1)!} & \frac{(n + 4)!}{(n + 3)!} \\ \frac{n!}{(n - 2)! 2!} & \frac{(n + 2)!}{n! 2!} & \frac{(n + 4)!}{(n + 2)! 2!} \end{matrix} |$

⇒ △ $= | \begin{matrix} 1 & 1 & 1 \\ \frac{n (n - 1)!}{(n - 1)!} & \frac{(n + 2) (n + 1)!}{(n + 1)!} & \frac{(n + 4) (n + 3)!}{(n + 3)!} \\ \frac{n (n - 1) (n - 2)!}{(n - 2)! 2!} & \frac{(n + 2) (n + 1) n!}{n! 2!} & \frac{(n + 4) (n + 3) (n + 2)!}{(n + 2)! 2!} \end{matrix} |$

⇒ △ $= | \begin{matrix} 1 & 1 & 1 \\ n & (n + 2) & (n + 4) \\ \frac{n (n - 1)}{2} & \frac{(n + 2) (n + 1)}{2} & \frac{(n + 4) (n + 3)}{2} \end{matrix} |$

Taking $\frac{1}{2}$ as common from $R_{3}$ ,

⇒ △ $= \frac{1}{2} | \begin{matrix} 1 & 1 & 1 \\ n & (n + 2) & (n + 4) \\ n (n - 1) & (n + 2) (n + 1) & (n + 4) (n + 3) \end{matrix} |$

Applying $C_{1} \to C_{1} - C_{3}$ and $C_{2} \to C_{2} - C_{3}$ , we get

⇒ △ $= \frac{1}{2} | \begin{matrix} 0 & 0 & 1 \\ - 4 & - 2 & (n + 4) \\ - 8 n - 12 & - 4 n - 10 & (n + 4) (n + 3) \end{matrix} |$

Now, expanding along $R_{1},$ we get

⇒ △ $= \frac{1}{2} [- 4 (- 4 n - 10) + 2 (- 8 n - 12)]$

⇒ △ $= \frac{1}{2} [16 n + 40 - 16 n - 24]$

⇒ △ $= \frac{1}{2} \times 16$

⇒ △ = 8

Hence, the given statement is True.

Example 17: If $A = [\begin{matrix} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{matrix}]$ , $x y z = 80$ , $3 x + 2 y + 10 z = 20$ , then

$A a d j . A = [\begin{matrix} 81 & 0 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 81 \end{matrix}]$ .

Ans: here we have $A = | \begin{matrix} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{matrix} |$ , $x y z = 80$ , $3 x + 2 y + 10 z = 20$

As we know that, $A^{- 1} = \frac{1}{| A |} a d j (A)$

⇒ $A A^{- 1} = \frac{1}{| A |} A a d j (A)$

⇒ $I = \frac{1}{| A |} A a d j (A)$

⇒ $A a d j (A) = | A |$ ……… (i)

Here, $| A | = x (y z - 3) - 5 (2 z - 3) + 2 (2 - y)$

⇒ $| A | = x y z - 3 x - 10 z + 15 + 4 - 2 y$

⇒ $| A | = x y z + 19 - (3 x + 2 y + 10 z)$

⇒ $| A | = 80 + 19 - 20$

⇒ $| A | = 79$

Therefore, $A a d j (A) = 79$ , and it is constant value.

Hence, the given statement is False.

Example 18: If A = $[\begin{matrix} 0 & 1 & 3 \\ 1 & 2 & x \\ 2 & 3 & 1 \end{matrix}]$ $A^{- 1} = [\begin{matrix} \frac{1}{2} & - 4 & \frac{5}{2} \\ - \frac{1}{2} & 3 & - \frac{3}{2} \\ \frac{1}{2} & y & \frac{1}{2} \end{matrix}]$ , then $x = 1, y = - 1.$

Ans: Here, we have $A = [\begin{matrix} 0 & 1 & 3 \\ 1 & 2 & x \\ 2 & 3 & 1 \end{matrix}]$

$A^{- 1} = [\begin{matrix} \frac{1}{2} & - 4 & \frac{5}{2} \\ - \frac{1}{2} & 3 & - \frac{3}{2} \\ \frac{1}{2} & y & \frac{1}{2} \end{matrix}]$

As we know that, $A A^{- 1} = I$

⇒ $[\begin{matrix} 0 & 1 & 3 \\ 1 & 2 & x \\ 2 & 3 & 1 \end{matrix}]$ $[\begin{matrix} \frac{1}{2} & - 4 & \frac{5}{2} \\ - \frac{1}{2} & 3 & - \frac{3}{2} \\ \frac{1}{2} & y & \frac{1}{2} \end{matrix}]$ $= [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

⇒ $[\begin{matrix} 0 - \frac{1}{2} + \frac{2}{2} & 0 + 3 + 3 y & 0 - \frac{3}{2} + \frac{3}{2} \\ \frac{1}{2} - 1 + \frac{x}{2} & - 4 + 6 + x y & \frac{5}{2} - 3 + \frac{x}{2} \\ 1 - 1 + 1 & - 8 + 9 + y & 5 - \frac{9}{2} + \frac{1}{2} \end{matrix}]$ = $[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

⇒ $[\begin{matrix} 1 & 3 (1 + y) & 0 \\ - \frac{1}{2} + \frac{x}{2} & 2 + x y & - \frac{1}{2} + \frac{x}{2} \\ 1 & 1 + y & 1 \end{matrix}]$ = $[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

Now, comparing both sides, we get

⇒ $3 (1 + y) = 0$ and $- \frac{1}{2} + \frac{x}{2} = 0$

⇒ $1 + y = 0$ and $- 1 + x = 0$

⇒ $y = - 1$ and x = 1

Hence, the given statement is True.

Exercise 4.3

Using the properties of determinants in Exercise 1 to 6, evaluate

1. $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

Ans: Here, we have $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

Applying $[C_{1} \to C_{1} - C_{2}]$ , we get

$= | \begin{matrix} x^{2} - 2 x + 2 & x - 1 \\ 0 & x + 1 \end{matrix} |$

$= (x^{2} - 2 x + 2) (x + 1) - 0 (x - 1)$

$= x^{3} - 2 x^{2} + 2 x + x^{2} - 2 x + 2$

$= x^{3} - x^{2} + 2$

Hence, $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} | = x^{3} - x^{2} + 2$ .

2. $| \begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix} |$

Ans: Here, we have $| \begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix} |$

Applying $[C_{1} \to C_{1} + C_{2} + C_{3}]$ , we get

$= | \begin{matrix} a + x + y + z & y & z \\ a + x + y + z & a + y & z \\ a + x + y + z & y & a + z \end{matrix} |$

Now, take common $(a + x + y + z)$ from $C_{1}$

$= (a + x + y + z) | \begin{matrix} 1 & y & z \\ 1 & a + y & z \\ 1 & y & a + z \end{matrix} |$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

$= (a + x + y + z) | \begin{matrix} 1 & y & z \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix} |$

Now, expanding along $R_{1}$

$= (a + x + y + z) [(a . a - 0) - y (0 - 0) + z (0 - 0)]$

$= (a + x + y + z) [a^{2} - 0 - 0]$

$= a^{2} (a + x + y + z)$

Hence, $| \begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix} | = a^{2} (a + x + y + z) .$

3. $| \begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{matrix} |$

Ans: Here, we have $| \begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{matrix} |$

Now, taking common $x^{2}, y^{2} a n d z^{2}$ from $C_{1}, C_{2} a n d C_{3}$ respectively.

$= x^{2} y^{2} z^{2} | \begin{matrix} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{matrix} |$

Applying $[C_{2} \to C_{2} - C_{3}]$ , we get

$= x^{2} y^{2} z^{2} | \begin{matrix} 0 & 0 & x \\ y & - y & y \\ z & z & 0 \end{matrix} |$

Now, expanding along $R_{1}$

$= x^{2} y^{2} z^{2} [0 (0 - y z) - 0 (0 - z y) + x (y z + y z)]$

$= x^{2} y^{2} z^{2} [x (y z + y z)]$

$= x^{2} y^{2} z^{2} \times 2 x y z$

$= 2 x^{3} y^{3} z^{3}$

Hence, $| \begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{matrix} | = 2 x^{3} y^{3} z^{3} .$

4. $| \begin{matrix} 3 x & - x + y & - x + z \\ x - y & 3 y & z - y \\ x - z & y - z & 3 z \end{matrix} |$

Ans: Here, we have $| \begin{matrix} 3 x & - x + y & - x + z \\ x - y & 3 y & z - y \\ x - z & y - z & 3 z \end{matrix} |$

Applying $[C_{1} \to C_{1} + C_{2} + C_{3}]$ , we get

$= | \begin{matrix} x + y + z & - x + y & - x + z \\ x + y + z & 3 y & z - y \\ x + y + z & y - z & 3 z \end{matrix} |$

Now, take common $(x + y + z)$ from $C_{1}$

$= (x + y + z) | \begin{matrix} 1 & - x + y & - x + z \\ 1 & 3 y & z - y \\ 1 & y - z & 3 z \end{matrix} |$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

$= (x + y + z) | \begin{matrix} 1 & - x + y & - x + z \\ 0 & 2 y + x & x - y \\ 0 & x - z & 2 z + x \end{matrix} |$

Now, expanding along $R_{1}$

$= (x + y + z) [(2 y + x) (2 z + x) - (x - y) (x - z) - 0 + 0]$

$= (x + y + z) [4 y z + 2 x y + 2 x z + x^{2} - (x^{2} - x z - x y + y z)]$

$= (x + y + z) [4 y z + 2 x y + 2 x z + x^{2} - x^{2} + x z + x y - y z]$

$= (x + y + z) [3 y z + 3 x y + 3 x z]$

$= 3 (x + y + z) (y z + x y + x z)$

Hence, $| \begin{matrix} 3 x & - x + y & - x + z \\ x - y & 3 y & z - y \\ x - z & y - z & 3 z \end{matrix} | = 3 (x + y + z) (y z + x y + x z) .$

5. $| \begin{matrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{matrix} |$

Ans: Here, we have $| \begin{matrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{matrix} |$

Applying $[C_{1} \to C_{1} + C_{2} + C_{3}]$ , we get

$= | \begin{matrix} 3 x + 4 & x & x \\ 3 x + 4 & x + 4 & x \\ 3 x + 4 & x & x + 4 \end{matrix} |$

Now, taking common $(3 x + 4)$ from $C_{1}$

$= (3 x + 4) | \begin{matrix} 1 & x & x \\ 1 & x + 4 & x \\ 1 & x & x + 4 \end{matrix} |$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

$= (3 x + 4) | \begin{matrix} 1 & x & x \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} |$

Now, expanding along $R_{1}$

$= (3 x + 4) [(16 - 0) - 0 + 0]$

$= 16 (3 x + 4)$

Hence, $| \begin{matrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{matrix} | = 16 (3 x + 4)$ .

6. $| \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$

Ans: Here, we have $| \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$

Applying $[R_{1} \to R_{1} + R_{2} + R_{3}]$ , we get

$= | \begin{matrix} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$

Now, taking common $(a + b + c)$ from $R_{1}$

$= (a + b + c) | \begin{matrix} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$

Applying $[C_{2} \to C_{2} - C_{1} a n d C_{3} \to C_{3} - C_{1}]$ , we get

$= (a + b + c) | \begin{matrix} 1 & 0 & 0 \\ 2 b & - (a + b + c) & 0 \\ 2 c & 0 & - (a + b + c) \end{matrix} |$

Now, expanding along $R_{1}$

$= (a + b + c) [{(a + b + c)}^{2} - 0]$

$= {(a + b + c)}^{3}$

Hence, $| \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} | = {(a + b + c)}^{3} .$

Using the properties of determinants in Exercises 7 to 9.

7. $| \begin{matrix} y^{2} z^{2} & yz & y + z \\ z^{2} x^{2} & zx & z + x \\ x^{2} y^{2} & xy & x + y \end{matrix} | = 0$

Ans: Here, we have L.H.S = $| \begin{matrix} y^{2} z^{2} & y z & y + z \\ z^{2} x^{2} & z x & z + x \\ x^{2} y^{2} & x y & x + y \end{matrix} |$

Applying $[R_{1} \to x R_{1}, R_{2} \to y R_{2} a n d R_{3} \to z R_{3}]$ ,

And it can be written as,

⇒ L.H.S $= \frac{1}{x y z} | \begin{matrix} x y^{2} z^{2} & x y z & x y + x z \\ y z^{2} x^{2} & y z x & y z + x y \\ z x^{2} y^{2} & z x y & z x + z y \end{matrix} |$

Taking common (xyz) from $C_{1} a n d C_{2}$ , we get

⇒ L.H.S $= \frac{1}{x y z} \times {(x y z)}^{2} | \begin{matrix} y z & 1 & x y + x z \\ z x & 1 & y z + x y \\ x y & 1 & z x + z y \end{matrix} |$

⇒ L.H.S $= x y z | \begin{matrix} y z & 1 & x y + x z \\ z x & 1 & y z + x y \\ x y & 1 & z x + z y \end{matrix} |$

Now, applying $[C_{1} \to C_{1} + C_{3}]$ , we get

⇒ L.H.S $= x y z | \begin{matrix} x y + y z + x z & 1 & x y + x z \\ x y + y z + x z & 1 & y z + x y \\ x y + y z + x z & 1 & z x + z y \end{matrix} |$

Taking common ( $x y + y z + x z)$ from $C_{1}$

⇒ L.H.S $= x y z (x y + y z + x z) | \begin{matrix} 1 & 1 & x y + x z \\ 1 & 1 & y z + x y \\ 1 & 1 & z x + z y \end{matrix} |$

⇒ L.H.S $= x y z (x y + y z + x z) \times 0$

(when two columns are same, then value of determinants is 0)

⇒ L.H.S = 0

⇒ L.H.S $= R . H . S$

Hence Proved.

8. $| \begin{matrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{matrix} | = 4 xyz$

Ans: Here, we have L.H.S $= | \begin{matrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{matrix} |$

Applying $[C_{1} \to C_{1} + C_{2} + C_{3}]$ , we get

$= | \begin{matrix} 2 (y + z) & z & y \\ 2 (x + z) & z + x & x \\ 2 (x + y) & x & x + y \end{matrix} |$

Now, taking common $(2)$ from $C_{1}$

$= 2 | \begin{matrix} (y + z) & z & y \\ (x + z) & z + x & x \\ (x + y) & x & x + y \end{matrix} |$

Applying $[C_{1} \to C_{1} - C_{2}]$ , we get

$= 2 | \begin{matrix} y & z & y \\ 0 & z + x & x \\ y & x & x + y \end{matrix} |$

Applying $[R_{3} \to R_{3} - R_{1}]$ , we get

$= 2 | \begin{matrix} y & z & y \\ 0 & z + x & x \\ 0 & x - z & x \end{matrix} |$

Now, expanding along $R_{1}$

$= 2 [y {x z + x^{2} - (x^{2} - x z)} - 0 - 0]$

$= 2 [y {x z + x^{2} - x^{2} + x z}]$

$= 2 \times 2 x y z$

$= 4 x y z$

$= R . H . S$

Thus, L.H.S = R.H.S

Hence proved.

9. $| \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix} | = {(a - 1)}^{3}$

Ans: Here, we have L.H.S $= | \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix} |$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

$= | \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ - (a^{2} - 1) & - (a + 1) & 0 \\ - (a^{2} + 2 a - 3) & - (2 a - 2) & 0 \end{matrix} |$

$= | \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ - (a - 1) (a + 1) & - (a - 1) & 0 \\ - (a + 3) (a - 1) & - 2 (a - 1) & 0 \end{matrix} |$

Now, taking common $(a - 1)$ from $R_{2}$ and $R_{3}$

$= {(a - 1)}^{2} | \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ - (a + 1) & - 1 & 0 \\ - (a + 3) & - 2 & 0 \end{matrix} |$

Now, expanding along $R_{1}$

$= {(a - 1)}^{2} [0 - 0 + 1 {2 a + 2 - (a + 3)}]$

$= {(a - 1)}^{2} (2 a + 2 - a - 3)$

$= {(a - 1)}^{2} (a - 1)$

$= {(a - 1)}^{3}$

$= R . H . S$

Thus, L.H.S = R.H.S

Hence proved.

10. If $A + B + C = 0$ , then prove that $| \begin{matrix} 1 & c o s C & c o s B \\ c o s C & 1 & c o s A \\ c o s B & c o s A & 1 \end{matrix} | = 0$

Ans: Here, we have $A + B + C = 0$

Now, L.H.S = $| \begin{matrix} 1 & c o s C & c o s B \\ c o s C & 1 & c o s A \\ c o s B & c o s A & 1 \end{matrix} |$

Expanding along $R_{1}$ , we get

$= 1 (1 - C o s^{2} A) - \cos C (\cos C - \cos A . \cos B) + \cos B (\cos A . \cos C - \cos B)$

$= s i n^{2} A - c o s^{2} C + \cos A . \cos B . \cos C + \cos A . \cos B . \cos C - c o s^{2} B$

$= s i n^{2} A - c o s^{2} B + 2 \cos A . \cos B . \cos C - c o s^{2} C$

$= - \cos (A + B) . \cos (A - B) + 2 \cos A . \cos B . \cos C - c o s^{2} C$

[ $c o s^{2} B - s i n^{2} A = \cos (A + B) . \cos (A - B)]$

$= - \cos (- C) . \cos (A - B) + \cos C (2 \cos A . \cos B - \cos C)$

$= - \cos C (\cos A . \cos B + \sin A . \sin B) + \cos C (2 \cos A . \cos B - \cos C)$

$= - \cos C (\cos A . \cos B + \sin A . \sin B - 2 \cos A . \cos B + \cos C)$

$= - \cos C (\sin A . \sin B - \cos A . \cos B + \cos C)$

$= \cos C (\cos A . \cos B - \sin A . \sin B - \cos C)$

$= \cos C [\cos (A + B) - \cos C]$

$= \cos C [\cos (- C) - \cos C]$

$= \cos C [\cos C - \cos C]$

= 0

= R.H.S

Thus, L.H.S = R.H.S

Hence proved.

11. If the co-ordinates of the vertices of an equilateral triangle with sides of length $^{'} a^{'}$ are $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ , then ${| \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |}^{2} = \frac{3 a^{4}}{4} .$

Ans: Since, we know that area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}) a n d (x_{3}, y_{3})$ is given by $Δ = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |$

$\Rightarrow Δ^{2} = \frac{1}{4} {| \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |}^{2}$ ……………… eq (i)

And we know that area of an equilateral triangle with side a,

$\Rightarrow Δ = \frac{\sqrt{3}}{4} \times a^{2}$

$\Rightarrow Δ^{2} = {(\frac{\sqrt{3}}{4} \times a^{2})}^{2}$

$\Rightarrow Δ^{2} = \frac{3}{16} \times a^{4}$ ………………. Eq (ii)

From eq (i) and (ii), we get

$\Rightarrow \frac{1}{4} {| \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |}^{2} = \frac{3}{16} \times a^{4}$

$\Rightarrow {| \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |}^{2} = \frac{3 a^{4}}{4}$

Hence proved.

12. Find the value of $θ$ satisfying $| \begin{matrix} 1 & 1 & \sin 3 θ \\ - 4 & 3 & \cos 2 θ \\ 7 & - 7 & - 2 \end{matrix} | = 0$

Ans: Here, we have $| \begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix} | = 0$

Applying $[C_{1} \to C_{1} - C_{2}]$ , we get

⇒ $| \begin{matrix} 0 & 1 & s i n 3 θ \\ - 7 & 3 & c o s 2 θ \\ 14 & - 7 & - 2 \end{matrix} | = 0$

Taking common 7 from $C_{1}$

⇒ $7 | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0$

⇒ $| \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0$

Now, expanding along $R_{1}$

⇒ $0 (- 6 + 7 c o s 2 θ) - 1 (2 - 2 c o s 2 θ) + s i n 3 θ (7 - 6) = 0$

⇒ $- 2 + 2 c o s 2 θ + s i n 3 θ = 0$

⇒ $- 2 + 2 (1 - 2 s i n^{2} θ) + 3 s i n θ - 4 s i n^{3} = 0$

⇒ $- 2 + 2 - 4 s i n^{2} θ + 3 s i n θ - 4 s i n^{3} = 0$

⇒ $4 s i n^{3} + 4 s i n^{2} θ - 3 s i n θ = 0$

⇒ $s i n θ (4 s i n^{2} + 4 s i n θ - 3) = 0$

⇒ $s i n θ (4 s i n^{2} + 6 s i n θ - 2 s i n θ - 3) = 0$

⇒ $s i n θ {2 s i n θ (2 s i n θ + 3) - 1 (2 s i n θ + 3)} = 0$

⇒ $s i n θ (2 s i n θ + 3) (2 s i n θ - 1) = 0$

⇒ $s i n θ = 0, s i n θ = - \frac{3}{2}, s i n θ = \frac{1}{2}$

Now, Case 1: When $\sin θ = 0$

⇒ $\sin θ = {S i n 0}^{\circ}$

⇒ $θ = n π$ , where $n \in Z$

Case 2: When $\sin θ = - \frac{3}{2}$

Here, it is not possible because $\sin θ \in [- 1, 1]$

Case 3: when $\sin θ = \frac{1}{2}$

⇒ $\sin θ = S i n \frac{π}{6}$

⇒ $θ = n π + {(- 1)}^{n} \frac{π}{6}$ , where $n \in Z$

Hence, values of $θ$ are $n π$ or $n π + {(- 1)}^{n} \frac{π}{6}$ , where $n \in Z .$

13. If $| \begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0$ , then find the value of $x$ .

Ans: Here, we have $| \begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0$

Applying $[C_{1} \to C_{1} + C_{2} + C_{3}]$ , we get

⇒ $| \begin{matrix} 12 + x & 4 + x & 4 + x \\ 12 + x & 4 - x & 4 + x \\ 12 + x & 4 + x & 4 - x \end{matrix} | = 0$

Now, taking common $(12 + x)$ from $C_{1}$

⇒ $(12 + x) | \begin{matrix} 1 & 4 + x & 4 + x \\ 1 & 4 - x & 4 + x \\ 1 & 4 + x & 4 - x \end{matrix} | = 0$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

$\Rightarrow (12 + x) | \begin{matrix} 1 & 4 + x & 4 + x \\ 0 & - 2 x & 0 \\ 0 & 0 & - 2 x \end{matrix} | = 0$

Now, expanding along $C_{1}$

$\Rightarrow (12 + x) (4 x^{2}) = 0$

$\Rightarrow (12 + x) = 0 o r (4 x^{2}) = 0$

$\Rightarrow x = - 12 o r x = 0$

Hence, the values of $x$ are -12 and 0.

14. If $a_{1}, a_{2}, a_{3},$ …… $a_{r}$ are in G.P, then prove that the determinant $| \begin{matrix} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{matrix} |$ is independent of $r .$

Ans: Let the first term of given G.P be A and common ratio be R.

Then, nth term, $a_{n} = A R^{n - 1}$

Therefore,

$= | \begin{matrix} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{matrix} |$

$= | \begin{matrix} A R^{r} & A R^{r + 4} & A R^{r + 8} \\ A R^{r + 6} & A R^{r + 10} & A R^{r + 14} \\ A R^{r + 10} & A R^{r + 16} & A R^{r + 20} \end{matrix} |$

Taking common $A R^{r}, A R^{r + 6}$ and $A R^{r + 10}$ from $R_{1}, R_{2}$ and $R_{3}$ respectively.

$= A R^{r} . A R^{r + 6}$ . $A R^{r + 10} | \begin{matrix} 1 & R^{4} & R^{8} \\ 1 & R^{4} & R^{8} \\ 1 & R^{6} & R^{10} \end{matrix} |$

= 0 (two rows are identical, hence value of determinant is 0)

Hence, the determinant $| \begin{matrix} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{matrix} |$ is independent of $r .$

15. Show that the points $(a + 5, a - 4), (a - 2, a + 3)$ and $(a, a)$ do not lie on a straight line for any value of $a$ .

Ans: Here, given points are $(a + 5, a - 4), (a - 2, a + 3)$ and $(a, a)$ .

Now, $Δ = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |$

⇒ $Δ = \frac{1}{2} | \begin{matrix} a + 5 & a - 4 & 1 \\ a - 2 & a + 3 & 1 \\ a & a & 1 \end{matrix} |$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

⇒ $Δ = \frac{1}{2} | \begin{matrix} a + 5 & a - 4 & 1 \\ - 7 & 7 & 0 \\ - 5 & 4 & 0 \end{matrix} |$

Now, expanding along $C_{3}$

⇒ $Δ = \frac{1}{2} [- 28 + 35]$

⇒ $Δ = \frac{7}{2}$ sq. units

⇒ $Δ \neq 0$

Here, the area of this triangle is not equal to 0.

Hence, given points form a triangle i.e., points do not lie in a straight line.

16. Show that the $Δ ABC$ is an isosceles triangle if the determinant,

$Δ = | \begin{matrix} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ co s^{2} A + \cos A & co s^{2} B + \cos B & co s^{2} C + \cos C \end{matrix} | = 0$

Ans: Here, we have $| \begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix} | = 0$

Now, applying $[C_{2} \to C_{2} - C_{1}] and [C_{3} \to C_{3} - C_{1}]$ , we get

⇒ $| \begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix} | = 0$

⇒ $| \begin{matrix} 1 & 0 & 0 \\ 1 + c o s A & c o s B - c o s A & c o s C - c o s A \\ c o s^{2} A + c o s A & c o s^{2} B - c o s^{2} A + c o s B - c o s A & c o s^{2} C - c o s^{2} A + c o s C - c o s A \end{matrix} | = 0$

⇒ $| \begin{matrix} 1 & 0 & 0 \\ 1 + c o s A & c o s B - c o s A & c o s C - c o s A \\ c o s^{2} A + c o s A & (c o s B - c o s A) (c o s B + c o s A + 1) & (c o s C - c o s A) (c o s C + c o s A + 1) \end{matrix} | = 0$

Now, take common $(\cos B - \cos A)$ and $(\cos C - \cos A)$ from $C_{1}$ and $C_{2}$ respectively, we get

⇒ $(c o s B - c o s A) (c o s C - c o s A) | \begin{matrix} 1 & 0 & 0 \\ 1 + c o s A & 1 & 1 \\ c o s^{2} A + c o s A & (c o s B + c o s A + 1) & (c o s C + c o s A + 1) \end{matrix} | = 0$

Expanding along $R_{1}$ ,

⇒ $(c o s B - c o s A) (c o s C - c o s A) (c o s C + c o s A + 1 - c o s B - c o s A - 1) = 0$

⇒ $(\cos B - \cos A) (\cos C - \cos A) (\cos C - \cos B) = 0$

⇒ $(c o s B - c o s A) = 0 o r (c o s C - c o s A) = 0 o r (c o s C - c o s B) = 0$

⇒ $\cos B = \cos A o r \cos C = \cos A o r \cos C = \cos B$

⇒ $B = A o r C = A o r C = B$

Hence, $Δ A B C$ is an isosceles triangle.

17. Find $A^{- 1}$ if $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ and show that $A^{- 1} = \frac{A^{2} - 3 I}{2}$ .

Ans: Here, we have $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$

Here, minors of elements are

⇒ $M_{11} = | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1$ , $M_{12} = | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = - 1$ , $M_{13} = | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | = 1$

⇒ $M_{21} = | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = - 1$ , $M_{22} = | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1$ , $M_{23} = | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = - 1$

⇒ $M_{31} = | \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} | = 1$ , $M_{32} = | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = - 1$ , $M_{33} = | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1$

Now, cofactor of an element $a_{i j} i s A_{i j} = {(- 1)}^{i + j} M_{i j}$ .

⇒ $A_{11} = {(- 1)}^{1 + 1} M_{11} = - 1$ , $A_{12} = 1$ , $A_{13} = 1$

⇒ $A_{21} = 1$ , $A_{22} = - 1$ , $A_{23} = 1$

⇒ $A_{31} = 1$ , $A_{32} = 1$ , $A_{33} = - 1$

Thus, matrix formed by cofactors, say $P = [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]$

Now, Adjoint matrix, $a d j A =$ transpose of matrix P

⇒ $a d j A = {[P]}^{T}$

⇒ $a d j A = {[\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]}^{T}$

⇒ $a d j A = [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]$

And $| A | = | \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} |$

⇒ $| A | = 0 (0 - 1) - 1 (0 - 1) + 1 (1 - 0)$

⇒ $| A | = 1 + 1 = 2$

Now, L.H.S $= A^{- 1}$

$= \frac{a d j A}{| A |}$

$= \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]$

And R.H.S $= \frac{A^{2} - 3 I}{2}$

$= \frac{1}{2}$ $[\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ $[\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ $- \frac{1}{2} [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix}]$

$= \frac{1}{2} [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}]$ $- \frac{1}{2} [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix}]$

$= \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]$

Thus, L.H.S = R.H.S

Hence proved.

Long Answer (L.A)

18. If $A = [\begin{matrix} 1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1 \end{matrix}]$ , find $A^{- 1} .$ Using $A^{- 1}$ , solve the system of linear equations $x - 2 y = 10, 2 x - y - z = 8, - 2 y + z = 7.$

Ans: Here, we have $A = [\begin{matrix} 1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1 \end{matrix}]$ ……. Eq(i)

Now, $| A | = | \begin{matrix} 1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1 \end{matrix} |$

⇒ $| A | = 1 (- 1 - 2) - 2 (- 2 - 0) + 0 (2 - 0)$

⇒ $| A | = - 3 + 4 = 1$

Here, minors of elements are

⇒ $M_{11} = | \begin{matrix} - 1 & - 2 \\ - 1 & 1 \end{matrix} | = - 3$ , $M_{12} = | \begin{matrix} - 2 & - 2 \\ 0 & 1 \end{matrix} | = - 2$ , $M_{13} = | \begin{matrix} - 2 & - 1 \\ 0 & - 1 1 \end{matrix} | = 2$

⇒ $M_{21} = | \begin{matrix} 2 & 0 \\ - 1 & 1 \end{matrix} | = 2$ , $M_{22} = | \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} | = 1$ , $M_{23} = | \begin{matrix} 1 & 2 \\ 0 & - 1 \end{matrix} | = - 1$

⇒ $M_{31} = | \begin{matrix} 2 & 0 \\ - 1 & - 2 \end{matrix} | = - 4$ , $M_{32} = | \begin{matrix} 1 & 0 \\ - 2 & - 2 \end{matrix} | = - 2$ , $M_{33} = | \begin{matrix} 1 & 2 \\ - 2 & - 1 \end{matrix} | = 3$

Now, cofactor of an element $a_{i j} i s A_{i j} = {(- 1)}^{i + j} M_{i j}$ .

⇒ $A_{11} = {(- 1)}^{1 + 1} M_{11} = - 3$ , $A_{12} = 2$ , $A_{13} = 2$

⇒ $A_{21} = - 2$ , $A_{22} = 1$ , $A_{23} = 1$

⇒ $A_{31} = - 4$ , $A_{32} = 2$ , $A_{33} = 3$

Thus, matrix formed by cofactors, say $P = [\begin{matrix} - 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3 \end{matrix}]$

Now, Adjoint matrix, $a d j A =$ transpose of matrix P

⇒ $a d j A = {[P]}^{T}$

⇒ $a d j A = {[\begin{matrix} - 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3 \end{matrix}]}^{T}$

⇒ $a d j A = [\begin{matrix} - 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{matrix}]$

Now, $A^{- 1}$ $= \frac{a d j A}{| A |}$

⇒ $A^{- 1}$ $= [\begin{matrix} - 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{matrix}]$ ……….. eq(ii)

Hence, the required $A^{- 1}$ is $[\begin{matrix} - 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{matrix}]$

Also, we have the system of linear equations as

$x - 2 y = 10, 2 x - y - z = 8, - 2 y + z = 7$

These equations can be written as,

$[\begin{matrix} 1 & - 2 & 0 \\ 2 & - 1 & - 1 \\ 0 & - 2 & 1 \end{matrix}]$ $[\begin{matrix} x \\ y \\ z \end{matrix}]$ $= [\begin{matrix} 10 \\ 8 \\ 7 \end{matrix}]$

Let $C = [\begin{matrix} 1 & - 2 & 0 \\ 2 & - 1 & - 1 \\ 0 & - 2 & 1 \end{matrix}]$ $, X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ & $D = [\begin{matrix} 10 \\ 8 \\ 7 \end{matrix}]$

⇒ $C X = D$

We know that, ${(A^{T})}^{- 1} = {(A^{- 1})}^{T}$

⇒ $C^{T} = [\begin{matrix} 1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1 \end{matrix}] = A$ (from eq (i))

⇒ $X = C^{- 1} D = {(A^{- 1})}^{T} D$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}]$ = $[\begin{matrix} - 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3 \end{matrix}]$ $[\begin{matrix} 10 \\ 8 \\ 7 \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}]$ $= [\begin{matrix} - 30 + 16 + 14 \\ - 20 + 8 + 7 \\ - 40 + 16 + 21 \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}]$ $= [\begin{matrix} 0 \\ - 5 \\ - 3 \end{matrix}]$

On comparing, we get

⇒ $x = 0, y = - 5, z = - 3$

Hence, values of $x, y$ and $z$ are 0, $- 5$ and $- 3$ respectively.

19. Using the matrix method, solve the system of equations 3x+2y-2z=3, x+2y+3z=6, 2x-y+z=2.

Ans: Here, we have $3 x + 2 y - 2 z = 3, x + 2 y + 3 z = 6, 2 x - y + z = 2$

These, equations can be written as,

⇒ $[\begin{matrix} 3 & 2 & - 2 \\ 1 & 2 & 3 \\ 2 & - 1 & 1 \end{matrix}]$ $[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 3 \\ 6 \\ 2 \end{matrix}]$

Let $A = [\begin{matrix} 3 & 2 & - 2 \\ 1 & 2 & 3 \\ 2 & - 1 & 1 \end{matrix}]$ $X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ & $B = [\begin{matrix} 3 \\ 6 \\ 2 \end{matrix}]$

⇒ $A X = B$

⇒ $X = A^{- 1} B$ …………….. eq (i),

Now, $| A | = | \begin{matrix} 3 & 2 & - 2 \\ 1 & 2 & 3 \\ 2 & - 1 & 1 \end{matrix} |$

⇒ $| A | = 3 (2 + 3) - 2 (1 - 6) - 2 (- 1 - 4)$

⇒ $| A | = 15 + 10 + 10$

⇒ $| A | = 35$

Here, minors of elements are

⇒ $M_{11} = | \begin{matrix} 2 & 3 \\ - 1 & 1 \end{matrix} | = 5$ , $M_{12} = | \begin{matrix} 1 & 3 \\ 2 & 1 \end{matrix} | = - 5$ , $M_{13} = | \begin{matrix} 1 & 2 \\ 2 & - 1 \end{matrix} | = - 5$

⇒ $M_{21} = | \begin{matrix} 2 & - 2 \\ - 1 & 1 \end{matrix} | = 0$ , $M_{22} = | \begin{matrix} 3 & - 2 \\ 2 & 1 \end{matrix} | = 7$ , $M_{23} = | \begin{matrix} 3 & 2 \\ 2 & - 1 \end{matrix} | = - 7$

⇒ $M_{31} = | \begin{matrix} 2 & - 2 \\ 2 & 3 \end{matrix} | = 10$ , $M_{32} = | \begin{matrix} 3 & - 2 \\ 1 & 3 \end{matrix} | = 11$ , $M_{33} = | \begin{matrix} 3 & 2 \\ 1 & 2 \end{matrix} | = 4$

Now, cofactor of an element $a_{i j} i s A_{i j} = {(- 1)}^{i + j} M_{i j}$ .

⇒ $A_{11} = {(- 1)}^{1 + 1} M_{11} = 5$ , $A_{12} = 5$ , $A_{13} = - 5$

⇒ $A_{21} = 0$ , $A_{22} = 7$ , $A_{23} = 7$

⇒ $A_{31} = 10$ , $A_{32} = - 11$ , $A_{33} = 4$

Thus, matrix formed by cofactors, say $P = [\begin{matrix} 5 & 5 & - 5 \\ 0 & 7 & 7 \\ 10 & - 11 & 4 \end{matrix}]$

Now, Adjoint matrix, $a d j A =$ transpose of matrix P

⇒ $a d j A = {[P]}^{T}$

⇒ $a d j A = {[\begin{matrix} 5 & 5 & - 5 \\ 0 & 7 & 7 \\ 10 & - 11 & 4 \end{matrix}]}^{T}$

⇒ $a d j A = [\begin{matrix} 5 & 0 & 10 \\ 5 & 7 & - 11 \\ - 5 & 7 & 4 \end{matrix}]$

Now, $A^{- 1}$ $= \frac{a d j A}{| A |}$

⇒ $A^{- 1}$ $= \frac{1}{35} [\begin{matrix} 5 & 0 & 10 \\ 5 & 7 & - 11 \\ - 5 & 7 & 4 \end{matrix}]$

Now, $X = A^{- 1} B$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}]$ $= \frac{1}{35} [\begin{matrix} 5 & 0 & 10 \\ 5 & 7 & - 11 \\ - 5 & 7 & 4 \end{matrix}] [\begin{matrix} 3 \\ 6 \\ 2 \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{35} [\begin{matrix} 15 + 0 + 20 \\ 15 + 42 - 22 \\ - 15 + 42 + 8 \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}]$ $= \frac{1}{35} [\begin{matrix} 35 \\ 35 \\ 35 \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}]$ $= [\begin{matrix} \frac{35}{35} \\ \frac{35}{35} \\ \frac{35}{35} \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]$

On comparing, we get

⇒ $x = 1, y = 1, z = 1$

Hence, values of $x, y$ and $z$ are 1, $1$ and $1$ respectively.

20. Given $A = [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}]$ , $B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}]$ , find BA and use this to solve the system of equations $y + 2 z = 7, x - y = 3, 2 x + 3 y + 4 z = 17.$

Ans: Here, we have $A = [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}], B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}]$

⇒ $B A = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}]$ $[\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}]$

⇒ $B A = [\begin{matrix} 2 + 4 + 0 & 2 - 2 + 0 & - 4 + 4 + 0 \\ 4 - 12 + 8 & 4 + 6 - 4 & - 8 - 12 + 20 \\ 0 - 4 + 4 & 0 + 2 - 2 & 0 - 4 + 10 \end{matrix}]$

⇒ $B A = [\begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix}]$

⇒ $B A = 6 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

⇒ $B A = 6 I$

⇒ $B^{- 1} = \frac{A}{6}$

⇒ $B^{- 1} = \frac{1}{6} [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}]$ ……………. Eq (i)

Also, we have $x - y = 3, 2 x + 3 y + 4 z = 17, y + 2 z = 7$

These, equations can be written as,

⇒ $[\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}]$ $[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 3 \\ 17 \\ 7 \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}]$ $= {[\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}]}^{- 1}$ $[\begin{matrix} 3 \\ 17 \\ 7 \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}]$ $= \frac{1}{6} [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}]$ $[\begin{matrix} 3 \\ 17 \\ 7 \end{matrix}]$ (from eq (i))

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{6} [\begin{matrix} 6 + 34 - 28 \\ - 12 + 34 - 28 \\ 6 - 17 + 35 \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{6} [\begin{matrix} 12 \\ - 6 \\ 24 \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} \frac{12}{6} \\ - \frac{6}{6} \\ \frac{24}{6} \end{matrix}]$

⇒ $[\begin{matrix} x \\ y \\ z \end{matrix}]$ $= [\begin{matrix} 2 \\ - 1 \\ 4 \end{matrix}]$

On comparing, we get

⇒ $x = 2, y = - 1, z = 4$

Hence, values of $x, y$ and $z$ are 2, $- 1$ and $4$ respectively.

21. If $a + b + c \neq 0$ and $| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} | = 0$ , then prove that $a = b = c .$

Ans: Here, we have $a + b + c \neq 0$ and $| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} | = 0$

⇒ $| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} | = 0$

Applying $[C_{1} \to C_{1} + C_{2} + C_{3}]$ , we get

⇒ $| \begin{matrix} a + b + c & b & c \\ a + b + c & c & a \\ a + b + c & a & b \end{matrix} | = 0$

Now, taking common $(a + b + c)$ from $C_{1}$

⇒ $(a + b + c) | \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{matrix} | = 0$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

$\Rightarrow (a + b + c) | \begin{matrix} 1 & b & c \\ 0 & c - b & a - c \\ 0 & a - b & b - c \end{matrix} | = 0$

Now, expanding along $C_{1}$

$\Rightarrow (a + b + c) [(c - b) (b - c) - (a - b) (a - c)] = 0$

$\Rightarrow (a + b + c) [b c - c^{2} - b^{2} + b c - (a^{2} - a c - a b + b c)] = 0$

$\Rightarrow (a + b + c) (2 b c - c^{2} - b^{2} - a^{2} + a c + a b - b c) = 0$

$\Rightarrow (a + b + c) (b c - c^{2} - b^{2} - a^{2} + a c + a b) = 0$

$\Rightarrow - (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c) = 0$

$\Rightarrow - \frac{1}{2} (a + b + c) (2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 a c) = 0$

$\Rightarrow (a + b + c) [a^{2} + a^{2} + b^{2} + b^{2} + c^{2} + c^{2} - 2 a b - 2 b c - 2 a c] = 0$

$\Rightarrow (a + b + c) [(a^{2} + b^{2} - 2 a b) + (b^{2} + c^{2} - 2 b c) + c^{2} + a^{2} - 2 a c] = 0$

$\Rightarrow (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}] = 0$

Here, $(a + b + c) \neq 0$

Therefore,

$\Rightarrow {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} = 0$

It is possible when ${(a - b)}^{2} = 0, {(b - c)}^{2} = 0, {(c - a)}^{2} = 0$

$\Rightarrow a - b = 0, b - c = 0, c - a = 0$

$\Rightarrow a = b, b = c, c = a$

Thus, $a = b = c .$

Hence proved.

22. Prove that $| \begin{matrix} bc - a^{2} & ca - b^{2} & ab - c^{2} \\ ca - b^{2} & ab - c^{2} & bc - a^{2} \\ ab - c^{2} & bc - a^{2} & ca - b^{2} \end{matrix} |$ is divisible by $(a + b + c)$ and find the quotient.

Ans: Here, we have $| \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} \\ c a - b^{2} & a b - c^{2} & b c - a^{2} \\ a b - c^{2} & b c - a^{2} & c a - b^{2} \end{matrix} |$

Applying $[C_{1} \to C_{1} - C_{2}] and [C_{2} \to C_{2} - C_{3}]$ , we get

$= | \begin{matrix} b c - a^{2} - c a + b^{2} & c a - b^{2} - a b + c^{2} & a b - c^{2} \\ c a - b^{2} - a b + c^{2} & a b - c^{2} - b c + a^{2} & b c - a^{2} \\ a b - c^{2} - b c + a^{2} & b c - a^{2} - c a + b^{2} & c a - b^{2} \end{matrix} |$

$= | \begin{matrix} (b - a) (a + b + c) & (c - b) (a + b + c) & a b - c^{2} \\ (c - b) (a + b + c) & (a - c) (a + b + c) & b c - a^{2} \\ (a - c) (a + b + c) & (b - a) (a + b + c) & c a - b^{2} \end{matrix} |$

Taking $(a + b + c)$ common from $C_{1}$ and $C_{2}$

$= {(a + b + c)}^{2} | \begin{matrix} (b - a) & (c - b) & a b - c^{2} \\ (c - b) & (a - c) & b c - a^{2} \\ (a - c) & (b - a) & c a - b^{2} \end{matrix} |$

Applying $R_{1} \to R_{1} + R_{2} + R_{3}$

$= {(a + b + c)}^{2} | \begin{matrix} 0 & 0 & a b + b c + c a - a^{2} - b^{2} - c^{2} \\ (c - b) & (a - c) & b c - a^{2} \\ (a - c) & (b - a) & c a - b^{2} \end{matrix} |$

Taking common $(a b + b c + c a - a^{2} - b^{2} - c^{2})$ from $R_{1}$

$= {(a + b + c)}^{2} (a b + b c + c a - a^{2} - b^{2} - c^{2}) | \begin{matrix} 0 & 0 & 1 \\ (c - b) & (a - c) & b c - a^{2} \\ (a - c) & (b - a) & c a - b^{2} \end{matrix} |$

$= - {(a + b + c)}^{2} (a^{2} + b^{2} + c^{2} - a b - b c - a c) | \begin{matrix} 0 & 0 & 1 \\ (c - b) & (a - c) & b c - a^{2} \\ (a - c) & (b - a) & c a - b^{2} \end{matrix} |$

$= - (a + b + c) (a^{3} + b^{3} + c^{3} - 3 a b c) | \begin{matrix} 0 & 0 & 1 \\ (c - b) & (a - c) & b c - a^{2} \\ (a - c) & (b - a) & c a - b^{2} \end{matrix} |$

Expanding along $R_{1}$

$= - (a + b + c) (a^{3} + b^{3} + c^{3} - 3 a b c) [(c - b) (b - a) - (a - c) (a - c)]$

$= - (a + b + c) (a^{3} + b^{3} + c^{3} - 3 a b c) [b c - a c - b^{2} + a b - (a^{2} + c^{2} - 2 a c)]$

$= - (a + b + c) (a^{3} + b^{3} + c^{3} - 3 a b c) (b c - a c - b^{2} + a b - a^{2} - c^{2} + 2 a c)$

$= - (a + b + c) (a^{3} + b^{3} + c^{3} - 3 a b c) (- a^{2} - b^{2} - c^{2} + a b + b c + a c)$

$= \frac{1}{2} (a + b + c) (a^{3} + b^{3} + c^{3} - 3 a b c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]$

Hence, the given determinant is divisible by $(a + b + c)$ .

And quotient is $\frac{1}{2} (a^{3} + b^{3} + c^{3} - 3 a b c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]$ .

23. If $x + y + z = 0,$ prove that $| \begin{matrix} xa & yb & zc \\ yc & za & xb \\ zb & xc & ya \end{matrix} | = xyz | \begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix} |$ .

Ans: Here, we have $x + y + z = 0$

Now, L.H.S $= | \begin{matrix} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{matrix} |$

Expanding along $R_{1}$

$= x a (a^{2} y z - x^{2} b c) - y b (y^{2} a c - b^{2} x z) + z c (c^{2} x y - z^{2} a b)$

$= a^{3} x y z - x^{3} a b c - y^{3} a b c + b^{3} x y z + c^{3} x y z - z^{3} a b c$

$= x y z (a^{3} + b^{3} + c^{3}) - a b c (x^{3} + y^{3} + z^{3})$

$= x y z (a^{3} + b^{3} + c^{3}) - a b c \times 3 x y z$

[if $x + y + z = 0$ , $x^{3} + y^{3} + z^{3} = 3 x y z]$

$= x y z (a^{3} + b^{3} + c^{3} - 3 a b c)$

Now, R.H.S $= x y z | \begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix} |$

Applying $[C_{1} \to C_{1} + C_{2} + C_{3}]$ , we get

$= x y z | \begin{matrix} a + b + c & b & c \\ a + b + c & a & b \\ a + b + c & c & a \end{matrix} |$

Taking common $(a + b + c)$ from $C_{1}$

$= x y z (a + b + c) | \begin{matrix} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{matrix} |$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

$= x y z (a + b + c) | \begin{matrix} 1 & b & c \\ 0 & a - b & b - c \\ 0 & c - b & a - c \end{matrix} |$

Now, expanding along $C_{1}$ , we get

$= x y z (a + b + c) [(a - b) (a - c) - (c - b) (b - c)]$

$= x y z (a + b + c) [a^{2} - a c - a b + b c - (b c - c^{2} - b^{2} + b c)]$

$= x y z (a + b + c) (a^{2} - a c - a b + b c - b c + c^{2} + b^{2} - b c)$

$= x y z (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)$

$= x y z (a^{3} + b^{3} + c^{3} - 3 a b c)$

Thus, L.H.S = R.H.S

Hence proved.

Objective Type Questions (M.C.Q)

Choose the correct answer from given four options in each of the exercise from 24 to 37.

24. If $| \begin{matrix} 2 x & 5 \\ 8 & x \end{matrix} | =$ $| \begin{matrix} 6 & - 2 \\ 7 & 3 \end{matrix} |$ , then value of $x$ is

(A) 3

(B) $\pm 3$

(D) 6

Ans: The correct answer is option (C).

Here, we have $| \begin{matrix} 2 x & 5 \\ 8 & x \end{matrix} | = | \begin{matrix} 6 & - 2 \\ 7 & 3 \end{matrix} |$

⇒ $2 x^{2} - 40 = 18 + 14$

⇒ $2 x^{2} - 40 = 32$

⇒ $2 x^{2} = 72$

⇒ $x^{2} = 36$

⇒ $x = \sqrt{36}$

⇒ $x = \pm 6$

Hence, option (C) is the correct answer.

25. The value of determinant $| \begin{matrix} a - b & b + c & a \\ b - a & c + a & b \\ c - a & a + b & c \end{matrix} |$

(A) $a^{3} + b^{3} + c^{3}$

(B) $3 bc$

(D) none of these

Ans: Option (D) is the correct answer.

Here, we have $| \begin{matrix} a - b & b + c & a \\ b - a & c + a & b \\ c - a & a + b & c \end{matrix} |$

Applying $C_{3} \to C_{2} + C_{3}$

$= | \begin{matrix} a - b & b + c & a \\ b - a & c + a & b \\ c - a & a + b & c \end{matrix} |$

$= | \begin{matrix} a - b & b + c & a + b + c \\ b - a & c + a & a + b + c \\ c - a & a + b & a + b + c \end{matrix} |$

Taking common $(a + b + c)$ from $C_{3}$ , we get

$= (a + b + c) | \begin{matrix} a - b & b + c & 1 \\ b - a & c + a & 1 \\ c - a & a + b & 1 \end{matrix} |$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

$= (a + b + c) | \begin{matrix} a - b & b + c & 1 \\ - 2 (a - b) & a - b & 0 \\ c - 2 a + b & a - c & 0 \end{matrix} |$

Taking common $(a - b)$ from $R_{2}$

$= (a + b + c) (a - b) | \begin{matrix} a - b & b + c & 1 \\ - 2 & 1 & 0 \\ c - 2 a + b & a - c & 0 \end{matrix} |$

Expanding along $C_{3}$

$= (a + b + c) (a - b) (- 2 a + 2 c - c + 2 a - b)$

$= (a + b + c) (a - b) (c - b)$

Hence, option (D) is correct.

26. The area of a triangle with the vertices $(- 3, 0), (3, 0)$ and $(0, k)$ is 9 sq. units. The value of $k$ will be

(A) 9

(B) 3

(D) 6

Ans: The correct answer is option (B).

we know that, area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is given by

⇒ $Δ = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |$

⇒ $9 = \frac{1}{2} | \begin{matrix} - 3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} |$

⇒ $18 = | \begin{matrix} - 3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} |$

Expanding along $C_{2}$

⇒ $18 = - k (- 3 - 3)$

⇒ $18 = 6 k$

⇒ $\frac{18}{6} = k$

⇒ $3 = k$

Thus, the value of $k$ is 3.

Hence, option (B) is correct.

27. The determinant $| \begin{matrix} b^{2} - ab & b - c & bc - ac \\ ab - a^{2} & a - b & b^{2} - ab \\ bc - ac & c - a & ab - a^{2} \end{matrix} |$ equals

(A) $abc (b - c) (c - a) (a - b)$

(B) $(b - c) (c - a) (a - b)$

(D) None of these

Ans: The correct answer is option (D).

Here, we have $| \begin{matrix} b^{2} - a b & b - c & b c - a c \\ a b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{matrix} |$

$= | \begin{matrix} b (b - a) & b - c & c (b - a) \\ a (b - a) & a - b & b (b - a) \\ c (b - a) & c - a & a (b - a) \end{matrix} |$

Taking common $(b - a)$ from $C_{1}$ and $C_{3}$

$= {(b - a)}^{2} | \begin{matrix} b & b - c & c \\ a & a - b & b \\ c & c - a & a \end{matrix} |$

Applying $C_{2} \to C_{2} + C_{3}$

$= {(b - a)}^{2} | \begin{matrix} b & b & c \\ a & a & b \\ c & c & a \end{matrix} | = 0$

Here, two columns of determinant are identical. Thus, the value of this determinant is 0.

Hence, option (D) is the correct answer.

28. The number of distinct real roots of $| \begin{matrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{matrix} | = 0$ in the interval $- \frac{π}{4} ⩽ x ⩽ \frac{π}{4}$ is

(A) 0

(B) 2

(D) 3

Ans: The correct answer is option (C).

Here, we have $| \begin{matrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{matrix} | = 0$

Applying $[C_{1} \to C_{1} + C_{2} + C_{3}]$ , we get

⇒ $| \begin{matrix} 2 c o s x + s i n x & c o s x & c o s x \\ 2 c o s x + s i n x & s i n x & c o s x \\ 2 c o s x + s i n x & c o s x & s i n x \end{matrix} | = 0$

Taking common ( $2 \cos x + \sin x)$ from $C_{1}$

⇒ $(2 c o s x + s i n x) | \begin{matrix} 1 & c o s x & c o s x \\ 1 & s i n x & c o s x \\ 1 & c o s x & s i n x \end{matrix} | = 0$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

⇒ $(2 c o s x + s i n x) | \begin{matrix} 1 & c o s x & c o s x \\ 0 & s i n x - c o s x & 0 \\ 0 & 0 & s i n x - c o s x \end{matrix} | = 0$

Expanding along $C_{1}$

⇒ $(2 \cos x + \sin x) {(\sin x - \cos x)}^{2} = 0$

⇒ $(2 c o s x + s i n x) = 0 o r {(s i n x - c o s x)}^{2} = 0$

⇒ $s i n x = - 2 c o s x o r s i n x - c o s x = 0$

⇒ $t a n x = - 2 o r s i n x = c o s x$

⇒ $t a n x = - 2 o r t a n x = 1$

Also, we have $- \frac{π}{4} ⩽ x ⩽ \frac{π}{4}$

⇒ $- \tan \frac{π}{4} ⩽ \tan x ⩽ \tan \frac{π}{4}$

⇒ $- 1 ⩽ \tan x ⩽ 1$

Thus, $\tan x$ = $- 2$ is not possible.

Therefore, $\tan x$ = $1$

⇒ $x = \frac{π}{4}$

So, only one distinct real root exists.

Hence, option (C) is the correct answer.

29. If $A, B$ and $C$ are angles of a triangle, then the determinant $| \begin{matrix} - 1 & c o s C & c o s B \\ c o s C & - 1 & c o s A \\ c o s B & c o s A & - 1 \end{matrix} |$ is equal to

(A) 0

(B) -1

(D) None of these

Ans: The correct answer is option (A).

Here, we have $| \begin{matrix} - 1 & c o s C & c o s B \\ c o s C & - 1 & c o s A \\ c o s B & c o s A & - 1 \end{matrix} |$

Applying $[C_{1} \to a C_{1} + b C_{2} + c C_{3}]$ , we get

$= | \begin{matrix} - a + b . c o s C + c . c o s B & c o s C & c o s B \\ a c o s C - b + c . c o s A & - 1 & c o s A \\ a . c o s B + b . c o s A - c & c o s A & - 1 \end{matrix} |$

Also, by projection rule in a triangle, we know that

⇒ $a = b . \cos C + c . \cos B$

⇒ $b = a \cos C + c . \cos A$

⇒ $c = a . \cos B + b . \cos A$

Using above equations in column first, we get

$= | \begin{matrix} - a + a & c o s C & c o s B \\ b - b & - 1 & c o s A \\ c - c & c o s A & - 1 \end{matrix} |$

$= | \begin{matrix} 0 & c o s C & c o s B \\ 0 & - 1 & c o s A \\ 0 & c o s A & - 1 \end{matrix} | = 0$

Since, a determinant having all elements of any column or row gives the value of determinant is zero.

Hence, option (A) is the correct answer.

30. Let $f (t) = | \begin{matrix} \cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t \end{matrix} |$ , then $lim_{t \to 0} \frac{f (t)}{t^{2}}$ is equal to

(A) 0

(B) -1

(D) 3

Ans: The correct answer is option (A).

Here, we have $f (t) = | \begin{matrix} c o s t & t & 1 \\ 2 s i n t & t & 2 t \\ s i n t & t & t \end{matrix} |$

Expanding along $R_{1}$

$\Rightarrow f (t) = c o s t (t^{2} - 2 t^{2}) - t (2 t s i n t - 2 t s i n t) + 1 (2 t s i n t - t s i n t)$

$\Rightarrow f (t) = - t^{2} \cos t - 0 + t \sin t$

$\Rightarrow f (t) = - t^{2} \cos t + t \sin t$

Now, $lim_{t \to 0} \frac{f (t)}{t^{2}} = \underset{t \to 0}{l i m} (\frac{- t^{2} \cos t + t \sin t}{t^{2}})$

$\Rightarrow lim_{t \to 0} \frac{f (t)}{t^{2}} = \underset{t \to 0}{l i m} (- \cos t + \frac{\sin t}{t})$

$\Rightarrow lim_{t \to 0} \frac{f (t)}{t^{2}} = - \underset{t \to 0}{l i m} \cos t + \underset{t \to 0}{l i m} (\frac{\sin t}{t})$

$\Rightarrow lim_{t \to 0} \frac{f (t)}{t^{2}} = - \underset{t \to 0}{l i m} \cos 0 + 1$ $[\underset{t \to 0}{l i m} (\frac{\sin t}{t}) = 1]$

$\Rightarrow lim_{t \to 0} \frac{f (t)}{t^{2}} = - 1 + 1$

$\Rightarrow lim_{t \to 0} \frac{f (t)}{t^{2}} = 0$

Hence, option (A) is the correct answer.

31. The maximum value of $Δ = | \begin{matrix} 1 & 1 & 1 \\ 1 & 1 + s i n θ & 1 \\ 1 + c o s θ & 1 & 1 \end{matrix} |$ is ( $θ \in R) .$

(A) $\frac{1}{2}$

(B) $\frac{\sqrt{3}}{2}$

(D) $\frac{2 \sqrt{3}}{4}$

Ans: Here, we have $Δ = | \begin{matrix} 1 & 1 & 1 \\ 1 & 1 + s i n θ & 1 \\ 1 + c o s θ & 1 & 1 \end{matrix} |$

Applying $[C_{1} \to C_{1} - C_{3}] and [C_{2} \to C_{2} - C_{3}]$ , we get

⇒ $Δ = | \begin{matrix} 0 & 0 & 1 \\ 0 & s i n θ & 1 \\ c o s θ & 0 & 1 \end{matrix} |$

Now, expanding along $R_{1}$

⇒ $Δ = 1 (0 - s i n θ . c o s θ)$

⇒ $Δ = - \sin θ . \cos θ$

⇒ $Δ = - \frac{2}{2} s i n θ . c o s θ$

⇒ $Δ = - \frac{1}{2} s i n 2 θ$

Also, we know that, $- 1 ⩽ \sin 2 θ ⩽ 1$

⇒ $\frac{1}{2} ⩾ - \frac{1}{2} \sin 2 θ ⩾ - \frac{1}{2}$

⇒ $- \frac{1}{2} ⩽ Δ ⩽ \frac{1}{2}$

Thus, maximum value of given determinant is $\frac{1}{2}$

Hence, option (A) is the correct answer.

32. If $f (x) = | \begin{matrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{matrix} |$ , then

(A) $f (a) = 0$

(B) $f (b) = 0$

(D) $f (1) = 0$

Ans: The correct answer is option (C).

Here, we have $f (x) = | \begin{matrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{matrix} |$

⇒ $f (a)$ = $| \begin{matrix} 0 & 0 & a - b \\ 2 a & 0 & a - c \\ a + b & a + c & 0 \end{matrix} | = 2 a (a - b) (a + c) \neq 0$

⇒ $f (b) = | \begin{matrix} 0 & b - a & 0 \\ b + a & 0 & b - c \\ 2 b & b + c & 0 \end{matrix} | = - 2 b (b - a) (b - c) \neq 0$

⇒ $f (0) = | \begin{matrix} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{matrix} | = a b c - a b c = 0$

⇒ $f (1) = | \begin{matrix} 0 & 1 - a & 1 - b \\ 1 + a & 0 & 1 - c \\ 1 + b & 1 + c & 0 \end{matrix} | \neq 0$

Hence, option (C) is the correct answer.

33. If $A = [\begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix}]$ , then $A^{- 1}$ exists if

(A) $λ = 2$

(B) $λ \neq 2$

(D) None of these

Ans: The correct answer is option (D).

Here, we have $A = [\begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix}]$

Expanding along $R_{1}$

⇒ $| A | = 2 (6 - 5) - λ (0 - 5) - 3 (0 - 2)$

⇒ $| A | = 2 + 5 λ + 6$

We know that, $A^{- 1}$ exists, if A is non- singular matrix i.e., $| A | \neq 0$

Therefore,

⇒ $| A | = 2 + 5 λ + 6 \neq 0$

⇒ $5 λ + 8 \neq 0$

⇒ $5 λ \neq - 8$

⇒ $λ \neq - \frac{8}{5}$

Hence, option (D) is the correct answer.

34. If A and B are invertible matrices, then which of the following is not correct?

(A) $adj A = | A | . A^{- 1}$

(B) $\det {(A)}^{- 1} = {[\det (A)]}^{- 1}$

(D) ${(A + B)}^{- 1} = B^{- 1} + A^{- 1}$

Ans: The correct answer is option (D).

Since, A and B are invertible matrices. So, we can say that

${(A B)}^{- 1} = B^{- 1} A^{- 1}$ (it is correct)

Also, $A^{- 1} = \frac{1}{| A |} (a d j A)$

⇒ $a d j A = | A | . A^{- 1}$ (it is correct)

And $A A^{- 1} = I$

⇒ $| A A^{- 1} | = | I |$

⇒ $| A | . | A^{- 1} | = 1$

⇒ $| A^{- 1} | = \frac{1}{| A |}$

$\Rightarrow d e t {(A)}^{- 1} = {[d e t (A)]}^{- 1}$ (it is correct)

Now, ${(A + B)}^{- 1} = B^{- 1} + A^{- 1}$

L.H.S = ${(A + B)}^{- 1}$

= $\frac{1}{| A + B |} (a d j A + B)$

R.H.S = $B^{- 1} + A^{- 1}$

= $\frac{1}{| B |} (a d j B) + \frac{1}{| A |} (a d j A)$

Thus, L.H.S $\neq$ R.H.S (it is incorrect)

Hence, option (D) is the correct answer.

35. If $x, y, z$ are all different from zero and $| \begin{matrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{matrix} | = 0$ , then value $x^{- 1} + y^{- 1} + z^{- 1}$ is

(A) $xyz$

(B) $x^{- 1} . y^{- 1} . z^{- 1}$

(D) $- 1$

Ans: The correct answer is option (D).

Here, we have $| \begin{matrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{matrix} | = 0$

Applying $C_{1} \to C_{1} - C_{2}$ and $C_{2} \to C_{2} - C_{3}$

⇒ $| \begin{matrix} x & 0 & 1 \\ - y & y & 1 \\ 0 & - z & 1 + z \end{matrix} | = 0$

Expanding along $R_{1}$

⇒ $x (y + y z + z) + y z = 0$

⇒ $x y + x y z + x z + y z = 0$

⇒ $x y + x z + y z = - x y z$

⇒ $\frac{x y + x z + y z}{x y z} = - 1$

⇒ $\frac{1}{z} + \frac{1}{y} + \frac{1}{x} = - 1$

⇒ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = - 1$

⇒ $x^{- 1} + y^{- 1} + z^{- 1} = - 1$

Hence, option (D) is the correct answer.

36. The value of the determinant $| \begin{matrix} x & x + y & x + 2 y \\ x + 2 y & x & x + y \\ x + y & x + 2 y & x \end{matrix} |$ is

(A) $9 x^{2} (x + y)$

(B) $9 y^{2} (x + y)$

(D) $7 x^{2} (x + y)$

Ans: The correct answer is option (B).

Here, we have $| \begin{matrix} x & x + y & x + 2 y \\ x + 2 y & x & x + y \\ x + y & x + 2 y & x \end{matrix} |$

Applying $C_{1} \to C_{1} + C_{2} + C_{3}$ , we get

= $| \begin{matrix} 3 (x + y) & x + y & x + 2 y \\ 3 (x + y) & x & x + y \\ 3 (x + y) & x + 2 y & x \end{matrix} |$

Taking common $3 (x + y)$ from $C_{1}$

$= 3 (x + y) | \begin{matrix} 1 & x + y & x + 2 y \\ 1 & x & x + y \\ 1 & x + 2 y & x \end{matrix} |$

Applying $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$

$= 3 (x + y) | \begin{matrix} 1 & x + y & x + 2 y \\ 0 & - y & - y \\ 0 & y & - 2 y \end{matrix} |$

Expanding along $C_{1}$

$= 3 (x + y) (2 y^{2} + y^{2})$

$= 3 (x + y) \times 3 y^{2}$

$= 9 y^{2} (x + y)$

Hence, option (B) is the correct answer.

37. There are two values of $a$ which makes determinant, $Δ = | \begin{matrix} 1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2 a \end{matrix} | = 86,$ then sum of these number is

(A) 4

(B) 5

(D) 9

Ans: The correct answer is option (C).

Here, we have $Δ = | \begin{matrix} 1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2 a \end{matrix} | = 86$

⇒ $| \begin{matrix} 1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2 a \end{matrix} | = 86$

Expanding along $R_{1}$

⇒ $1 (2 a^{2} + 4) + 2 (4 a - 0) + 5 (8 - 0) = 86$

⇒ $2 a^{2} + 4 + 8 a + 40 = 86$

⇒ $2 a^{2} + 8 a + 44 = 86$

⇒ $2 a^{2} + 8 a + 44 - 86 = 0$

⇒ $2 a^{2} + 8 a - 42 = 0$

⇒ $a^{2} + 4 a - 21 = 0$

⇒ $a^{2} + 7 a - 3 a - 21 = 0$

⇒ $a (a + 7) - 3 (a + 7) = 0$

⇒ $(a + 7) (a - 3) = 0$

⇒ $(a + 7) = 0 o r (a - 3) = 0$

⇒ $a = - 7 o r a = 3$

Here, values of $a$ are $- 7$ and $3 .$

Therefore, sum of these values = $- 7 + 3 = - 4$

Hence, option (C) is the correct answer.

Fill in the blanks

38. If A is a matrix of order, $3 \times 3,$ then $| 3 A | =$ …………………

Ans: Here, A is a matrix of order, $3 \times 3$ .

And we know that, $| λ A | = λ^{n} | A |$ , here n is the order of the matrix and $λ$ is a constant.

Therefore,

⇒ $| 3 A | = 3^{3} | A | = 27 | A |$

Hence, If A is a matrix of order, $3 \times 3,$ then $| 3 A | = 27 | A |$ .

39. If A is invertible matrix of order $3 \times 3$ , then $| A^{- 1} | =$ …………….

Ans: Here, A is invertible matrix of order $3 \times 3$

Therefore,

$\Rightarrow A A^{- 1} = I$

⇒ $| A A^{- 1} | = | I |$

⇒ $| A | . | A^{- 1} | = 1$

⇒ $| A^{- 1} | = \frac{1}{| A |}$

Hence, If A is invertible matrix of order $3 \times 3$ , then $| A^{- 1} | = \frac{1}{| A |}$ .

40. If $x, y, z \in R,$ then the value of determinant $| \begin{matrix} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} |$ is equal to ………. .

Ans: Here, we have $| \begin{matrix} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} |$ , where $x, y, z \in R,$

Applying $C_{1} \to C_{1} - C_{2}$ ,we get

$= | \begin{matrix} {(2^{x} + 2^{- x})}^{2} - {(2^{x} - 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} - {(3^{x} - 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} - {(4^{x} - 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} |$

$= | \begin{matrix} 4 . 2^{x} . 2^{- x} & {(2^{x} - 2^{- x})}^{2} & 1 \\ 4 . 3^{x} . 3^{- x} & {(3^{x} - 3^{- x})}^{2} & 1 \\ 4 . 4^{x} . 4^{- x} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} |$ [ ${(a + b)}^{2} - {(a - b)}^{2} = 4 a b]$

$= | \begin{matrix} 4 & {(2^{x} - 2^{- x})}^{2} & 1 \\ 4 & {(3^{x} - 3^{- x})}^{2} & 1 \\ 4 & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} |$

Taking common 4 from $C_{1}$

$= | \begin{matrix} 1 & {(2^{x} - 2^{- x})}^{2} & 1 \\ 1 & {(3^{x} - 3^{- x})}^{2} & 1 \\ 1 & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | = 0$

Since, a determinant having all elements of any column or row gives the value of determinant is zero.

Hence, then the value of determinant $| \begin{matrix} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} |$ is equal to 0.

41. If $\cos 2 θ = 0$ , then ${| \begin{matrix} 0 & \cos θ & \sin θ \\ \cos θ & \sin θ & 0 \\ \sin θ & 0 & \cos θ \end{matrix} |}^{2} =$ …………….

Ans: Here, we have $c o s 2 θ = 0$

⇒ $c o s 2 θ = \cos 90^{\circ}$

⇒ $2 θ = 90^{\circ}$

⇒ $θ = 45^{\circ}$ …………….. (i)

Also, we have ${| \begin{matrix} 0 & c o s θ & s i n θ \\ c o s θ & s i n θ & 0 \\ s i n θ & 0 & c o s θ \end{matrix} |}^{2}$

Expanding along $R_{1}$

$=$ ${[0 - c o s θ (c o s^{2} θ - 0) + s i n θ (0 - s i n^{2} θ)]}^{2}$

$= {[- c o s^{3} θ - s i n^{3} θ]}^{2}$

$= {[c o s^{3} θ + s i n^{3} θ]}^{2}$

$= {[{(\frac{1}{\sqrt{2}})}^{3} + {(\frac{1}{\sqrt{2}})}^{3}]}^{2}$

$= {[\frac{1}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}}]}^{2}$

$= {[\frac{1}{\sqrt{2}}]}^{2}$

$= \frac{1}{2}$

Hence, If $c o s 2 θ = 0$ , then ${| \begin{matrix} 0 & c o s θ & s i n θ \\ c o s θ & s i n θ & 0 \\ s i n θ & 0 & c o s θ \end{matrix} |}^{2} = \frac{1}{2}$ .

42. If A is a matrix of order $3 \times 3$ , then ${(A^{2})}^{- 1} =$ …………..

Ans: If A is a matrix of order $3 \times 3$ , then ${(A^{2})}^{- 1} =$ ${(A^{- 1})}^{2}$ .

43. If A is a matrix of order $3 \times 3$ , then the number of minors in the determinant of A are …………….. .

Ans: If A is a matrix of order $3 \times 3$ , then the number of minors in determinant of A are 9. because, in a $3 \times 3$ matrix, there are 9 elements.

44. The sum of the products of elements of any row with the co-factors of corresponding elements is equal to ………….. .

Ans: The sum of the products of elements of any row with the co-factors of corresponding elements is equal to the value of determinant.

Let $Δ = | \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} |$

Expanding along $R_{1}$

⇒ $Δ = a_{11} (a_{22} . a_{33} - a_{23} . a_{32}) - a_{12} (a_{21} . a_{33} - a_{23} . a_{31}) + a_{13} (a_{21} . a_{32} - a_{22} . a_{31})$

⇒ $Δ = a_{11} (a_{22} . a_{33} - a_{23} . a_{32}) + a_{12} (a_{23} . a_{31} - a_{21} . a_{33}) + a_{13} (a_{21} . a_{32} - a_{22} . a_{31})$

⇒ $Δ = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}$

⇒ $Δ =$ sum of the products of elements of $R_{1}$ with their corresponding cofactors.

45. If $x = - 9$ is a root of $| \begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix} | = 0$ , then the other two roots are ……..

Ans: Here, we have $x = - 9$ is a root of $| \begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix} | = 0$

⇒ $| \begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix} | = 0$

Expanding along $R_{1}$ , we get

⇒ $x (x^{2} - 12) - 3 (2 x - 14) + 7 (12 - 7 x) = 0$

⇒ $x^{3} - 12 x - 6 x + 42 + 84 - 49 x = 0$

⇒ $x^{3} - 67 x + 126 = 0$

⇒ $x^{3} + 9 x^{2} - 9 x^{2} - 81 x + 14 x + 126 = 0$

⇒ $x^{2} (x + 9) - 9 x (x + 9) + 14 (x + 9) = 0$

⇒ $(x + 9) (x^{2} - 9 x + 14) = 0$

⇒ $(x + 9) (x - 7) (x - 2) = 0$

⇒ $x = - 9, 7, 2$

Hence, other two roots of $| \begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix} | = 0$ are 7 and 2.

46. $| \begin{matrix} 0 & x y z & x - z \\ y - x & 0 & y - z \\ z - x & z - y & 0 \end{matrix} | =$ ………….

Ans: Here, we have $| \begin{matrix} 0 & x y z & x - z \\ y - x & 0 & y - z \\ z - x & z - y & 0 \end{matrix} |$

Applying $C_{1} \to C_{1} - C_{3}$ , we get

$= | \begin{matrix} z - x & x y z & x - z \\ z - x & 0 & y - z \\ z - x & z - y & 0 \end{matrix} |$

Taking common $(z - x)$ from $C_{1}$

$= (z - x) | \begin{matrix} 1 & x y z & x - z \\ 1 & 0 & y - z \\ 1 & z - y & 0 \end{matrix} |$

Applying $[R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}]$ , we get

$= (z - x) | \begin{matrix} 1 & x y z & x - z \\ 0 & - x y z & y - x \\ 0 & z - y - x y z & z - x \end{matrix} |$

Expanding along $C_{1}$

$= (z - x) [- x y z (z - x) - (y - x) (z - y - x y z)]$

$= (z - x) [- x y z^{2} + x^{2} y z - (y z - y^{2} - x y^{2} z - x z + x y + x^{2} y z)]$

$= (z - x) [- x y z^{2} - y z + y^{2} + x y^{2} z + x z - x y]$

$= (z - x) [y^{2} - y z + x y^{2} z - x y z^{2} + x z - x y]$

$= (z - x) [y (y - z) + x y z (y - z) - x (y - z)]$

$= (z - x) (y - z) (y + x y z - x)$

Hence, $| \begin{matrix} 0 & x y z & x - z \\ y - x & 0 & y - z \\ z - x & z - y & 0 \end{matrix} | = (z - x) (y - z) (y + x y z - x) .$

47. If $f (x)$ = $| \begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix} |$ = $A + B x + C x^{2}$ …, then A = ………….

Ans: Here, we have $f (x) = | \begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix} |$

Taking common ${(1 + x)}^{17}, {(1 + x)}^{23}, {(1 + x)}^{41}$ from $R_{1}, R_{2} and R_{3}$ respectively, we get

$= {(1 + x)}^{17} . {(1 + x)}^{23} . {(1 + x)}^{41} | \begin{matrix} 1 & {(1 + x)}^{2} & {(1 + x)}^{6} \\ 1 & {(1 + x)}^{6} & {(1 + x)}^{11} \\ 1 & {(1 + x)}^{2} & {(1 + x)}^{6} \end{matrix} |$

= 0.

Here, $R_{1} & R_{3}$ are identical,

Thus, $f (x) = | \begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix} | = 0 .$

Hence, the value of A is 0.

State True or False for the statements of the following Exercises:

48. ${(A^{3})}^{- 1} = {(A^{- 1})}^{3}$ , where A is square matrix and $| A | \neq 0.$

Ans: Here, the given statement is true.

Since, ${(A^{n})}^{- 1} = {(A^{- 1})}^{n}$ , where $n \in N .$

49. ${(aA)}^{- 1} = \frac{1}{a} A^{- 1}$ , where $a$ is any real number and A is a square matrix.

Ans: Since, we know that, if A is a non- singular square matrix, then for any scalar a (non – zero), aA is invertible such that

⇒ $(a A) = (\frac{1}{a} A^{- 1}) = (a . \frac{1}{a}) (A A^{- 1})$ i.e., (aA) is inverse of $(\frac{1}{a} A^{- 1})$ or ${(a A)}^{- 1} = \frac{1}{a} A^{- 1}$ , where a is any non -zero scalar false.

In the above statement, a is any real number. So, we can conclude that above statement is false.

50. $| A^{- 1} | \neq {| A |}^{- 1},$ where A is non – singular matrix.

Ans: Since, $| A^{- 1} | = {| A |}^{- 1}$ , where A is non – singular matrix.

Hence, the given statement is false.

51. If A and B are the matrices of order 3 and |A|=5, |B|=3, then |3AB| $= 27 \times 5 \times 3 = 405.$

Ans: Here, A and B are the matrices of order 3 and $| A | = 5, | B | = 3$

⇒ $| 3 A B | = 3^{3} \times | A | \times | B |$

⇒ $| 3 A B | = 27 \times 5 \times 3$

⇒ $| 3 A B | = 405$

Hence, the given statement is true.

52. If the value of a third order determinant is 12, then the value of the determinants formed by replacing each element by its co – factor will be 144.

Ans: let A be the determinant, $| A | = 12$

Also, we know that, if A is the square matrix of order n,

then $| A d j A | = {| A |}^{n - 1}$

⇒ $| A d j A | = {(12)}^{3 - 1}$

⇒ $| A d j A | = {(12)}^{2}$

⇒ $| A d j A | = 144$

Hence, the given statement is true.

53. $| \begin{matrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{matrix} | = 0$ , where $a, b, c$ are in $A . P .$

Ans: Here, we have $| \begin{matrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{matrix} |$

Applying $C_{1} \to C_{1} - C_{2}$ , we get

$= | \begin{matrix} - 1 & x + 2 & x + a \\ - 1 & x + 3 & x + b \\ - 1 & x + 4 & x + c \end{matrix} |$

Applying $R_{2} \to R_{2} R_{1} a n d R_{3} \to R_{3} R_{1}$ , we get

$= | \begin{matrix} - 1 & x + 2 & x + a \\ 0 & 1 & b - a \\ 0 & 2 & c - a \end{matrix} |$

Expanding along $C_{1}$

$= - 1 [(c - a) - 2 (b - a)]$

$= - 1 (c - a - 2 b + 2 a]$

$= - 1 (c - 2 b + a]$

Since, $a, b$ and $c$ are in A.P.

Therefore, $2 b = a + c$

$= - 1 (2 b - 2 b)$

= 0.

Hence, the given statement is true.

54. $| adj . A | = {| A |}^{2}$ , where A is a square matrix of order two.

Ans: Here, A is square matrix of order two

We know that,

$\Rightarrow A . A d j A = | A | I$

$\Rightarrow | A . A d j A | = | | A | I |$

$\Rightarrow | A | | A d j A | = {| A |}^{n}$

$\Rightarrow | A d j A | = \frac{{| A |}^{n}}{| A |}$

$\Rightarrow | A d j A | = {| A |}^{n - 1}$

Here, $n = 2$

$\Rightarrow | A d j A | = {| A |}^{2 - 1}$

$\Rightarrow | A d j A | = | A |$

Hence, the given statement is false.

55. The determinant $| \begin{matrix} \sin A & \cos A & \sin A + \cos B \\ \sin B & \cos A & \sin B + \cos B \\ \sin C & \cos A & \sin C + \cos B \end{matrix} |$ is equal to zero.

Ans: Here, we have $| \begin{matrix} \sin A & \cos A & \sin A + \cos B \\ \sin B & \cos A & \sin B + \cos B \\ \sin C & \cos A & \sin C + \cos B \end{matrix} |$

$= | \begin{matrix} \sin A & \cos A & \sin A \\ \sin B & \cos A & \sin B \\ \sin C & \cos A & \sin C \end{matrix} |$ + $| \begin{matrix} \sin A & \cos A & \cos B \\ \sin B & \cos A & \cos B \\ \sin C & \cos A & \cos B \end{matrix} |$

Since, the value of the determinant having two identical rows and columns is zero. Therefore,

$= 0 + | \begin{matrix} \sin A & \cos A & \cos B \\ \sin B & \cos A & \cos B \\ \sin C & \cos A & \cos B \end{matrix} |$

$= | \begin{matrix} \sin A & \cos A & \cos B \\ \sin B & \cos A & \cos B \\ \sin C & \cos A & \cos B \end{matrix} |$

Taking common $\cos A$ and $\cos B$ from $C_{2}$ and $C_{3}$

$= \cos A . \cos B | \begin{matrix} s i n A & 1 & 1 \\ s i n B & 1 & 1 \\ s i n C & 1 & 1 \end{matrix} |$

$= 0$

Since, the value of the determinant having two identical rows and columns is zero.

Hence, the given statement is true.

56. If the determinant $| \begin{matrix} x + a & p + u & l + f \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{matrix} |$ splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.

Ans: Here, we have $| \begin{matrix} x + a & p + u & l + f \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{matrix} |$

Now, splitting first row in the sum of two determinant

$=$ $| \begin{matrix} x & p & l \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{matrix} |$ + $| \begin{matrix} a & u & f \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{matrix} |$

Now, splitting second row in the sum of two determinants $= | \begin{matrix} x & p & l \\ y & q & m \\ z + c & r + w & n + h \end{matrix} |$ + $| \begin{matrix} x & p & l \\ b & v & g \\ z + c & r + w & n + h \end{matrix} |$ + $| \begin{matrix} a & u & f \\ y & q & m \\ z + c & r + w & n + h \end{matrix} |$ + $| \begin{matrix} a & u & f \\ b & v & g \\ z + c & r + w & n + h \end{matrix} |$

Similarly, we can split these 4 determinants in 8 determinants by splitting each one in two determinants.

Hence, the given statement is true.

57. Let $Δ = | \begin{matrix} a & p & x \\ b & q & y \\ c & r & z \end{matrix} | = 16,$ then $Δ_{1} = | \begin{matrix} p + x & a + x & a + p \\ q + y & b + y & b + q \\ r + z & c + z & c + r \end{matrix} | = 32.$

Ans: Here, we have $Δ = | \begin{matrix} a & p & x \\ b & q & y \\ c & r & z \end{matrix} | = 16$ …….. (i)

Now, $Δ_{1} = | \begin{matrix} p + x & a + x & a + p \\ q + y & b + y & b + q \\ r + z & c + z & c + r \end{matrix} |$

Applying $C_{1} \to C_{1} + C_{2} + C_{3}$ , we get

⇒ $Δ_{1} = | \begin{matrix} 2 (p + x + a) & a + x & a + p \\ 2 (q + y + b) & b + y & b + q \\ 2 (r + z + c) & c + z & c + r \end{matrix} |$

Taking common 2 from $C_{1}$

⇒ $Δ_{1} = 2 | \begin{matrix} (p + x + a) & a + x & a + p \\ (q + y + b) & b + y & b + q \\ (r + z + c) & c + z & c + r \end{matrix} |$

Applying $C_{1} \to C_{1} - C_{2}$ ,and $C_{2} \to C_{2} - C_{3}$ we get

⇒ $Δ_{1} = 2 | \begin{matrix} p & x - p & a + p \\ q & y - q & b + q \\ r & z - r & c + r \end{matrix} |$

Applying $C_{2} \to C_{1} + C_{2}$

⇒ $Δ_{1} = 2 | \begin{matrix} p & x & a + p \\ q & y & b + q \\ r & z & c + r \end{matrix} |$

⇒ $\frac{Δ_{1}}{2} = | \begin{matrix} p & x & a \\ q & y & b \\ r & z & c \end{matrix} | + | \begin{matrix} p & x & p \\ q & y & q \\ r & z & r \end{matrix} |$

⇒ $\frac{Δ_{1}}{2} = | \begin{matrix} p & x & a \\ q & y & b \\ r & z & c \end{matrix} | + 0$

Since, the value of the determinant having two identical rows and columns is zero.

⇒ $Δ_{1} = 2 | \begin{matrix} p & x & a \\ q & y & b \\ r & z & c \end{matrix} |$

⇒ $Δ_{1} = 2 . {(- 1)}^{2} | \begin{matrix} a & p & x \\ b & q & y \\ c & r & z \end{matrix} |$

⇒ $Δ_{1} = 2 \times 16 = 32$ (from eq (i))

Hence, the given statement is true.

58. The maximum value of $| \begin{matrix} 1 & 1 & 1 \\ 1 & 1 + \sin θ & 1 \\ 1 & 1 & 1 + \cos θ \end{matrix} |$ is $\frac{1}{2} .$

Ans: Here, we have $Δ = | \begin{matrix} 1 & 1 & 1 \\ 1 & 1 + s i n θ & 1 \\ 1 & 1 & 1 + c o s θ \end{matrix} |$

Applying $[C_{1} \to C_{1} - C_{3}] and [C_{2} \to C_{2} - C_{3}]$ , we get

⇒ $Δ = | \begin{matrix} 0 & 0 & 1 \\ 0 & s i n θ & 1 \\ - c o s θ & - c o s θ & 1 \end{matrix} |$

Now, expanding along $R_{1}$

⇒ $Δ = 1 (0 + s i n θ . c o s θ)$

⇒ $Δ = \sin θ . \cos θ$

⇒ $Δ = \frac{2}{2} s i n θ . c o s θ$

⇒ $Δ = \frac{1}{2} s i n 2 θ$

Also, we know that, $- 1 ⩽ \sin 2 θ ⩽ 1$

⇒ $- \frac{1}{2} ⩽ \frac{1}{2} \sin 2 θ ⩽ \frac{1}{2}$

⇒ $- \frac{1}{2} ⩽ Δ ⩽ \frac{1}{2}$

Thus, maximum value of given determinant is $\frac{1}{2}$

Hence, the given statement is true.

Determinants

Determinants are the fourth Chapter of NCERT Exemplar Solutions for Class 12 Mathematics. Introduction to Determinants, then a Determinant Matrix of order one, two, or three, properties of the Determinant, minors and cofactor, adjoint and inverse of a Matrix, applications of Matrices, and solution of a System of Linear Equations using the inverse of a Matrix are all covered in this Chapter.

Students can also utilize Vedantu NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions) PDF from the links on the Vedantu site to learn and comprehend all of the fundamentals of Determinants in a simple way.

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FAQs on NCERT Exemplar for Class 12 Maths Chapter-4 (Book Solutions)

1. How will NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions) help you?

NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions) will assist you in understanding how to calculate the Determinant of different Square Matrices, Co-factors, and the Inverse of Matrices, among other things. Students can use NCERT Exemplar Class 12 Maths Chapter 4 solutions to gain a better understanding of Matrices and how to answer them uniquely. Chapter 4 of NCERT Class 12 Maths solutions can be highly important in terms of scoring for students in the 12th grade. Learners can access the topic and solutions offline using the NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions) download.

2. What are the applications of the NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants?

Determinants are Matrices that are answered in Chapter 3 of the NCERT Maths book for 12th grade. Studying Determinants is about preparing for further education in any discipline of Math, science, economics, and so on, not just passing tests. Students will study Determinants, their elements, and how to compute Determinants of various square Matrices in NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions).

This Chapter will not only help you boost the score in your final Exams but also has very important applications in various other fields including Physics and other principles to determine various values in the calculation.

3. What are the different sub topics covered in the NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions)?

The different sub topics covered in the NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions) includes:

4.1 The Beginning
4.2 Factor to consider
4.2.1 Determinant of a one-dimensional Matrix
4.2.2 Determinant of a two-dimensional Matrix
4.2.3 The Determinant of a 3x3 Matrix
4.3 Determinant properties
4.4 Triangle surface area
Minors and co-factors (4.5)
4.6 A Matrix's adjoint and inverse
4.6.1 Matrix adjacency
4.7 Matrix and Determinant Applications
4.7.1 Using the inverse of Matrices to solve a system of linear equations

4. What are the various advantages of NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions)?

Vedantu's solution will assist you in locating a detailed NCERT Exemplar for Class 12 Maths Chapter 4 Determinants (Book Solutions) answers to the questions presented in the NCERT exemplar book. We don't miss stages or try to solve questions in a quick or easy method instead, our professional guides will assist you in answering each question in the most minute detail possible.

The wording used to answer the questions is straightforward to comprehend. Not only are the problems thoroughly answered, but we also keep a marking system in mind to help our students improve their grades.

NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions) will assist you in building confidence in your ability to solve more difficult problems. Students will be helped to understand the issues and their complexities by detailed procedures written in simple language.

5. What are some of the most important concepts covered in the NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions)?

Some of the most important concepts students learn from the NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions) includes a better understanding of how Determinants interact with Matrices. The students are taught and practise Matrices, as well as how to use them to represent linear equations. Students can get the Determinants of various square Matrices in the following ways:

The order one Determinant - The order two Determinant - The order 3x3 Determinants. The Determinants and their properties up to order 3 will be covered, as well as how to get the area of a triangle by transforming the vertices into Determinants. Students will learn how to extend Determinants in the form of cofactors and minors in certain themes. NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions) also covers the principles of Determinants of different Matrices.

NCERT Exemplar for Class 12 Maths Chapter-4 (Book Solutions)

NCERT Exemplar for Class 12 Maths - Determinants - Free PDF Download

Access NCERT Exemplar Solutions for Class 12 Mathematics Chapter 4 - Determinants

Solved Examples

Short Answer (S. A)

Long Answer (L.A.)

Objective Type Questions

Choose the correct answer from the given four options in each of Examples 10 and 11.

Fill in the blank in each of the examples 12 to 14.

State whether the statements in Examples 15 to 18 is True or False.

Exercise 4.3

Using the properties of determinants in Exercise 1 to 6, evaluate

Using the properties of determinants in Exercises 7 to 9.

Long Answer (L.A)

Objective Type Questions (M.C.Q)

Choose the correct answer from given four options in each of the exercise from 24 to 37.

Fill in the blanks

Determinants

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FAQs on NCERT Exemplar for Class 12 Maths Chapter-4 (Book Solutions)