Question

$A$ is a $3 \times 3$ and $B$ is its adjoint matrix. If the determinant of $B$ is $64$ , then det $A$ is:
A. $4$
B. $\pm 4$
C. $\pm 8$
D. $8$

Answer 1

Here we are given the matrix $A{\text{ and }}B$ and $A$ is $3 \times 3$ matrix and adjoint of $A{\text{ is }}B$
So we can say that $B = adj(A)$
After taking the determinant to both sides and applying the properties of the adjoint of the matrix, we can easily approach the solution and get the required value of the determinant of the matrix $A$ .

Complete step by step solution:
Here we are given that $A$ is a $3 \times 3$ and $B$ is its adjoint matrix. The determinant of $B$ is $64$ and now we need to find the determinant of $A$
So we are given that adjoint of the matrix $A$ is $B$
So we can write:
$B = adj(A)$ $- - - - - (1)$
Taking the determinant to both the sides of the equation (1) we get that:
$\left| B \right| = \left| {adj(A)} \right|$
We are given that $\left| B \right| = 64$
Hence we can substitute the value of it in the above equation:
$64 = \left| {adj(A)} \right|$
$\left| {adj(A)} \right| = 64$ $- - - - (2)$
Now we need to use the property of the adjoint which states that:
The determinant of the adjoint of the matrix is always equal to the determinant of the matrix which is raised to the power $(n - 1)$ where $n$ is the order of the matrix. Here the order of the matrix is three as it is given in the question.
So we can write:
$\left| {adj(A)} \right| = {\left| A \right|^{n - 1}}$
Put $n = 3$
$\left| {adj(A)} \right| = {\left| A \right|^2}$ $- - - - (3)$
From the equation (2) we have got the value of the determinant of the adjoint of the matrix $A$ and hence we can substitute its value in the equation (3) and get the required value of the determinant of $A$
$\left| {adj(A)} \right| = {\left| A \right|^2}$
${\left| A \right|^2} = 64$
$\left| A \right| = \sqrt {64} = \pm 8$

Hence we get C is the correct option.

Note:
Here the student can make the mistake by marking the answer as $8$ instead of $\pm 8$ because he may consider $\left| A \right|$ to be the modulus and take the positive term out of $+ 8{\text{ and - 8}}$ but we must remember that it is not the modulus but the determinant which can be both negative as well as positive.