Question

If A is a square matrix such that $A\left( AdjA \right)=\left( \begin{matrix} 4 & 0 & 0 \\\ 0 & 4 & 0 \\\ 0 & 0 & 4 \\\ \end{matrix} \right)$ then $\det \left( AdjA \right)=$
A. 4
B. 16
C. 64
D. 256

Answer 1

Hint: We will be using the concepts of matrices and determinants to solve the problem. We will be using the properties of matrices that $\left| A \right|{{I}_{n}}=A\left( adj\left( A \right) \right)$ to relate the data given to us with what we have to find then we will further take its discriminant and substitute the value to find the final answer.
Complete step-by-step answer:
Now, we have been given that $A\left( AdjA \right)=\left( \begin{matrix} 4 & 0 & 0 \\\ 0 & 4 & 0 \\\ 0 & 0 & 4 \\\ \end{matrix} \right)$
Now, we know that the inverse of a square matrix A is given by;
${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}.................\left( 1 \right)$
Where $\left| A \right|$ is determinant of A now multiplying by A in (1) we have;
$A\ {{A}^{-1}}=\dfrac{A\left( adjA \right)}{\left| A \right|}$
Now, we know that $A\ {{A}^{-1}}={{I}_{n}}$ , where n is the order of the matrix.
Therefore,
$\left| A \right|{{I}_{n}}=A\left( adj\left( A \right) \right)$
Now, we will take determinant of both sides,
$\begin{aligned} & \left| \left| A \right|{{I}_{n}} \right|=\left| A\left( adjA \right) \right| \\\ & {{\left| A \right|}^{n}}=\left| A\ adjA \right| \\\ \end{aligned}$
Now, we know that;
$A\left( adjA \right)=\left( \begin{matrix} 4 & 0 & 0 \\\ 0 & 4 & 0 \\\ 0 & 0 & 4 \\\ \end{matrix} \right)$
So, we have;
$\begin{aligned} & \left| A\ adjA \right|=4\left( 4\times 4-0 \right) \\\ & =4\left( 4\times 4 \right) \\\ & =4\times 4\times 4 \\\ & =64 \\\ \end{aligned}$
So, we have;
${{\left| A \right|}^{n}}=64$
Where n is the order of matrix which is equal to 3 therefore;
$\begin{aligned} & {{\left| A \right|}^{3}}=64 \\\ & \left| A \right|=4..............\left( 2 \right) \\\ \end{aligned}$
Now, we know that $\det \left( adjA \right)$ is ${{\left| A \right|}^{n-1}}$.
$\det \left( Adj\left( A \right) \right)={{\left| A \right|}^{n-1}}$
Where n is the order of the square matrix and equal to 3.
$\begin{aligned} & \det \left( AdjA \right)={{\left| A \right|}^{3-1}} \\\ & ={{\left| A \right|}^{2}} \\\ \end{aligned}$
Now, we know from (2) that $\left| A \right|=4$ therefore,
$\begin{aligned} & \det \left( Adj\ A \right)={{4}^{2}} \\\ & =16 \\\ \end{aligned}$
Therefore the correct option is (B).

Note: To solve these type of question one must remember few identities for square matrix like;
$\begin{aligned} & \left| adj\ A \right|={{\left| A \right|}^{n-1}} \\\ & \left| \left| A \right|{{I}_{n}} \right|={{\left| A \right|}^{n}} \\\ \end{aligned}$