If $a\ne 6,b,c$ satisfy $\left| \begin{matrix}
a & 2b & 2c \\
3 & b & c \\
4 & a & b \\
\end{matrix} \right|=0$, then $abc=$
A. $a+b+c$
B. 0
C. ${{b}^{3}}$
D. $ab+b-c$
Answer
Verified
435.9k+ views
Hint: To solve this question we will use the properties of matrix. We will first simplify the given matrix by using the operations of matrix and determinants then we put the values obtained equal to zero, then we will find the value of $abc$ by simplifying the obtained equation.
Complete step by step solution:
We have been given that $a\ne 6,b,c$ satisfy $\left| \begin{matrix}
a & 2b & 2c \\
3 & b & c \\
4 & a & b \\
\end{matrix} \right|=0$.
We have to find the value of $abc$.
Now, let us first solve the given matrix. We know that we can solve the given matrix by multiplying the element by $2\times 2$ determinant. The determinant of a $3\times 3$ matrix is calculated for a matrix having 3 rows and 3 columns. Then we will get
$\Rightarrow a\left( b\times b-a\times c \right)-2b\left( 3\times b-4\times c \right)+2c\left( 3\times a-4\times b \right)$
Now, simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow a\left( {{b}^{2}}-ac \right)-2b\left( 3b-4c \right)+2c\left( 3a-4b \right) \\
& \Rightarrow a{{b}^{2}}-{{a}^{2}}c-6{{b}^{2}}+8bc+6ac-8bc \\
& \Rightarrow a{{b}^{2}}-{{a}^{2}}c-6{{b}^{2}}+6ac \\
\end{align}$
We have given that $a\ne 6,b,c$ satisfy matrix value equal to zero.
Then we will get
$\begin{align}
& \Rightarrow a{{b}^{2}}-{{a}^{2}}c-6{{b}^{2}}+6ac=0 \\
& \Rightarrow a{{b}^{2}}-6{{b}^{2}}-{{a}^{2}}c+6ac=0 \\
& \Rightarrow {{b}^{2}}\left( a-6 \right)-ac\left( a-6 \right)=0 \\
& \Rightarrow \left( a-6 \right)\left( {{b}^{2}}-ac \right)=0 \\
\end{align}$
If $a\ne 6$ then $\left( {{b}^{2}}-ac \right)=0$ then simplifying the obtained equation we will get
$\begin{align}
& \Rightarrow {{b}^{2}}-ac=0 \\
& \Rightarrow {{b}^{2}}=ac \\
& \Rightarrow abc={{b}^{3}} \\
\end{align}$
Hence we get the value of $abc$ as ${{b}^{3}}$.
Option C is the correct answer.
Note:
Students must remember the condition given in the question that $a\ne 6$. If we consider the factor $a-6=0$ then we will get the value $a=6$. So we need to ignore the factor. In matrices, determinants are the special numbers calculated from the square matrix. Square matrix should have an equal number of rows and columns.
Complete step by step solution:
We have been given that $a\ne 6,b,c$ satisfy $\left| \begin{matrix}
a & 2b & 2c \\
3 & b & c \\
4 & a & b \\
\end{matrix} \right|=0$.
We have to find the value of $abc$.
Now, let us first solve the given matrix. We know that we can solve the given matrix by multiplying the element by $2\times 2$ determinant. The determinant of a $3\times 3$ matrix is calculated for a matrix having 3 rows and 3 columns. Then we will get
$\Rightarrow a\left( b\times b-a\times c \right)-2b\left( 3\times b-4\times c \right)+2c\left( 3\times a-4\times b \right)$
Now, simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow a\left( {{b}^{2}}-ac \right)-2b\left( 3b-4c \right)+2c\left( 3a-4b \right) \\
& \Rightarrow a{{b}^{2}}-{{a}^{2}}c-6{{b}^{2}}+8bc+6ac-8bc \\
& \Rightarrow a{{b}^{2}}-{{a}^{2}}c-6{{b}^{2}}+6ac \\
\end{align}$
We have given that $a\ne 6,b,c$ satisfy matrix value equal to zero.
Then we will get
$\begin{align}
& \Rightarrow a{{b}^{2}}-{{a}^{2}}c-6{{b}^{2}}+6ac=0 \\
& \Rightarrow a{{b}^{2}}-6{{b}^{2}}-{{a}^{2}}c+6ac=0 \\
& \Rightarrow {{b}^{2}}\left( a-6 \right)-ac\left( a-6 \right)=0 \\
& \Rightarrow \left( a-6 \right)\left( {{b}^{2}}-ac \right)=0 \\
\end{align}$
If $a\ne 6$ then $\left( {{b}^{2}}-ac \right)=0$ then simplifying the obtained equation we will get
$\begin{align}
& \Rightarrow {{b}^{2}}-ac=0 \\
& \Rightarrow {{b}^{2}}=ac \\
& \Rightarrow abc={{b}^{3}} \\
\end{align}$
Hence we get the value of $abc$ as ${{b}^{3}}$.
Option C is the correct answer.
Note:
Students must remember the condition given in the question that $a\ne 6$. If we consider the factor $a-6=0$ then we will get the value $a=6$. So we need to ignore the factor. In matrices, determinants are the special numbers calculated from the square matrix. Square matrix should have an equal number of rows and columns.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success
Master Class 11 Computer Science: Engaging Questions & Answers for Success
Master Class 11 Maths: Engaging Questions & Answers for Success
Master Class 11 Social Science: Engaging Questions & Answers for Success
Master Class 11 Economics: Engaging Questions & Answers for Success
Master Class 11 Business Studies: Engaging Questions & Answers for Success
Trending doubts
10 examples of friction in our daily life
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
State and prove Bernoullis theorem class 11 physics CBSE
Pigmented layer in the eye is called as a Cornea b class 11 biology CBSE
What problem did Carter face when he reached the mummy class 11 english CBSE