If $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$, then…
If $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$, then $A(\operatorname{adj} A)=$
- $\left[\begin{array}{ccc}-1 / 3 & 0 & 0 \\ 0 & -1 / 3 & 0 \\ 0 & & -1 / 3\end{array}\right]$
- $\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$
- $\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$
- $\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & 2 \\ 3 & 2 & 4\end{array}\right]$
Solution
$\begin{aligned}
& A=\left[\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right] \\
& \therefore|A|=1(4,-4)-2(-4-2)+3(-2-1)=12-9=3
\end{aligned}$
We know that $\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathrm{I}$
$\therefore A(\operatorname{adj} A)=3\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right]$
Asked in: MHT CET 2021 (23 Sep Shift 2)
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