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If for any $2 \times 2$ square matrix $A$, $A(\operatorname{adj} A)=\left[\begin{array}{ll}8 & 0 \\ 0 & 8\end{array}\right]$, then the value of $\operatorname{det}(A) .$
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Verified Answer
The correct answer is:
8
Let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \Rightarrow \operatorname{adj}(A)=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$
$$
\begin{aligned}
&\text { Now, } A(A d j A)=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \times\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] \\
&\Rightarrow \quad\left[\begin{array}{ll}
8 & 0 \\
0 & 8
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \times\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]=\left[\begin{array}{cc}
a d-b c & 0 \\
0 & a d-b c
\end{array}\right] \\
&\Rightarrow \quad a d-b c=8,-a b+a b=0, d c-d c=0 \\
&\Rightarrow \quad-b c+a d=8 \\
&\operatorname{det}(A)=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=a d-b c=8
\end{aligned}
$$
$$
\begin{aligned}
&\text { Now, } A(A d j A)=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \times\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] \\
&\Rightarrow \quad\left[\begin{array}{ll}
8 & 0 \\
0 & 8
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \times\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]=\left[\begin{array}{cc}
a d-b c & 0 \\
0 & a d-b c
\end{array}\right] \\
&\Rightarrow \quad a d-b c=8,-a b+a b=0, d c-d c=0 \\
&\Rightarrow \quad-b c+a d=8 \\
&\operatorname{det}(A)=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=a d-b c=8
\end{aligned}
$$
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