
A is a \[3 \times 3\] matrix and B is its adjoint matrix. If the determinant of B is 64, then the det A is
A) \[4\]
B) \[ \pm 4\]
C) \[ \pm 8\]
D) \[8\]
Answer
556.2k+ views
Hint:
Here, we will use the basic concept of determinant to get the value of det A. We will use the basic formula for the determinant of the adjoint matrix in terms of the determinant of the main matrix and then by solving that we will get the value of det A.
Formula used:
\[\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}\], where \[n\] is the order of the matrix A.
Complete step by step solution:
Given matrix A is of order \[3 \times 3\] and B is the adjoint matrix.
Therefore determinant of matrix B is \[\left| B \right| = \left| {adj\,A} \right|\].
We know that determinant of adjoint matrix is equal to \[\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}\]
Also, \[n = 3\] as the order of matrix A is 3. Therefore the equation becomes
\[ \Rightarrow \left| {adj\,A} \right| = {\left| A \right|^{3 - 1}} = {\left| A \right|^2}\]
It is given that the determinant of the adjoint matrix B is equal to 64. Therefore, we get
\[ \Rightarrow 64 = {\left| A \right|^2}\]
Now by solving this we will get the value of the det A, we get
\[ \Rightarrow {\left| A \right|^2} = 64\]
\[ \Rightarrow \left| A \right| = \pm 8\]
Hence, the value of the det A is \[ \pm 8\].
So, option C is the correct option.
Note:
We should note know the formula of the adjoint matrix in terms of the main matrix. We should note that while doing the square root of a perfect square number then the sign \[ \pm \] is applied to the answer.
The determinant of a matrix is a scalar value of the matrix which is computed from the elements of that matrix. The determinant is only applied to the square matrices and the square matrix is the matrix that has a number of rows of the matrix equal to the number of the column of the matrix. The determinant of a matrix is denoted as \[\left| A \right|\].
Here, we will use the basic concept of determinant to get the value of det A. We will use the basic formula for the determinant of the adjoint matrix in terms of the determinant of the main matrix and then by solving that we will get the value of det A.
Formula used:
\[\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}\], where \[n\] is the order of the matrix A.
Complete step by step solution:
Given matrix A is of order \[3 \times 3\] and B is the adjoint matrix.
Therefore determinant of matrix B is \[\left| B \right| = \left| {adj\,A} \right|\].
We know that determinant of adjoint matrix is equal to \[\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}\]
Also, \[n = 3\] as the order of matrix A is 3. Therefore the equation becomes
\[ \Rightarrow \left| {adj\,A} \right| = {\left| A \right|^{3 - 1}} = {\left| A \right|^2}\]
It is given that the determinant of the adjoint matrix B is equal to 64. Therefore, we get
\[ \Rightarrow 64 = {\left| A \right|^2}\]
Now by solving this we will get the value of the det A, we get
\[ \Rightarrow {\left| A \right|^2} = 64\]
\[ \Rightarrow \left| A \right| = \pm 8\]
Hence, the value of the det A is \[ \pm 8\].
So, option C is the correct option.
Note:
We should note know the formula of the adjoint matrix in terms of the main matrix. We should note that while doing the square root of a perfect square number then the sign \[ \pm \] is applied to the answer.
The determinant of a matrix is a scalar value of the matrix which is computed from the elements of that matrix. The determinant is only applied to the square matrices and the square matrix is the matrix that has a number of rows of the matrix equal to the number of the column of the matrix. The determinant of a matrix is denoted as \[\left| A \right|\].
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