Question

If for any 2 × 2 square matrix A, A (adjA) = $\left[ \begin{matrix} 8 & 0 \\\ 0 & 8 \\\ \end{matrix} \right]$ then write the value of det [A].

Answer 1

Take a general 2 × 2 square matrix A = $\left[ \begin{matrix} \text{q} & \text{b} \\\ \text{c} & \text{d} \\\ \end{matrix} \right]$ then find its adjoint and multiply both of them to get the solution.

Complete step by step solution: Given: A (adjA) = $\left[ \begin{matrix} 8 & 0 \\\ 0 & 8 \\\ \end{matrix} \right]$
First of all we should know how to find adjoint of a given matrix.
Let $\text{A}=[{{\text{Q}}_{\text{ij}}}]$ be a square matrix of order n. The adjoint of a matrix A is the transpose of the cofactor matrix of A. It is denoted by adj A.
so now, let A = $\left[ \begin{matrix} \text{q} & \text{b} \\\ \text{c} & \text{d} \\\ \end{matrix} \right]$ be a 2 × 2 square matrix.
We know,
Cofactor of ${{\text{a}}_{\text{ij}}}={{(-1)}^{\text{i}+\text{j}}}$ × the number we get by removing column and row of designated element in a matrix
Now, cofactor of Q = ${{(-1)}^{\text{1}+\text{1}}}$d = d
cofactor of b = ${{(-1)}^{\text{1}+\text{2}}}$c = −c
cofactor of c = ${{(-1)}^{\text{2}+\text{1}}}$b = −b
and cofactor of d = ${{(-1)}^{\text{2}+\text{2}}}$Q = a
Now, cofactor matrix of $\text{A}=\left[ \begin{matrix} \text{d} & -\text{c} \\\ -\text{b} & \text{a} \\\ \end{matrix} \right]$
Now adj$\text{A}=\left[ \begin{matrix} \text{d} & -\text{c} \\\ -\text{b} & \text{a} \\\ \end{matrix} \right]$ {since adjA is transpose of co-factor matrix}
so, adj$\text{A}=\left[ \begin{matrix} \text{d} & -\text{b} \\\ -\text{c} & \text{a} \\\ \end{matrix} \right]$
Now we will have to find A(adj A)
$\text{A}(\text{adjA})=\left[ \begin{matrix} \text{a} & \text{b} \\\ \text{c} & \text{d} \\\ \end{matrix} \right]*\left[ \begin{matrix} \text{d} & -\text{b} \\\ -\text{c} & \text{a} \\\ \end{matrix} \right]$
⇒ $\left[ \begin{matrix} 8 & 0 \\\ 0 & 8 \\\ \end{matrix} \right]=\left[ \begin{matrix} \text{a} & \text{b} \\\ \text{c} & \text{d} \\\ \end{matrix} \right]*\left[ \begin{matrix} \text{d} & -\text{b} \\\ -\text{c} & \text{a} \\\ \end{matrix} \right]$
⇒ $\left[ \begin{matrix} 8 & 0 \\\ 0 & 8 \\\ \end{matrix} \right]=\left[ \begin{matrix} \text{ad}-\text{bc} & -\text{ab+ab} \\\ \text{cd}-\text{cd} & -\text{bc+ad} \\\ \end{matrix} \right]$
⇒ $\left[ \begin{matrix} 8 & 0 \\\ 0 & 8 \\\ \end{matrix} \right]=\left[ \begin{matrix} \text{ad}-\text{bc} & \text{O} \\\ \text{O} & \text{ad}-\text{bc} \\\ \end{matrix} \right]$
From here on comparing the elements we get
ad − bd = 8 1
Now our objective is to find det A.
So, $\det \text{A}=\left| \begin{matrix} \text{a} & \text{b} \\\ \text{c} & \text{d} \\\ \end{matrix} \right|$
= ad − bc
Now, from equation 1
ad − bc = 8 which is equal to the value of det A.

so, |A| = 8 answer.

Note: Students often make mistakes in the part where two matrix are multiplied, so be careful with it.
There is a direct formula to calculate A ⋅ adjA which is
A ⋅ |adjA| = |A| ⋅ I
so putting value in these, we can also solve the problem.