$$ \text { }\left[\begin{array}{ccc} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 3 & 0 & 2 \end{array}\right] $$ $$…
$$
\text { }\left[\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 1 & 2 \\
3 & 0 & 2
\end{array}\right]
$$
$$
\left|\begin{array}{ll}
2022 & 2024 \\
2021 & 2023
\end{array}\right|
$$
is equal to
- $\left[\begin{array}{ccc}8 & 4 & 11 \\ 4 & -1 & 3 \\ 9 & 6 & 13\end{array}\right]$
- $\left[\begin{array}{ccc}8 & 4 & 13 \\ 4 & -1 & 3 \\ 9 & 6 & 12\end{array}\right]$
- $\left[\begin{array}{ccc}8 & 4 & 13 \\ 4 & -1 & 3 \\ 9 & 6 & 13\end{array}\right]$
- $\left[\begin{array}{lll}8 & 4 & 11 \\ 4 & 1 & 13 \\ 9 & 6 & 13\end{array}\right]$
Solution
Let
$
\begin{aligned}
& A^{|B|}=\left[\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 1 & 2 \\
3 & 0 & 2
\end{array}\right]_{2021}^{2022} \quad \text {, where } \\
& A=\left[\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 1 & 2 \\
3 & 0 & 2
\end{array}\right], B=\left[\begin{array}{cc}
2022 & 2024 \\
2021 & 2023
\end{array}\right] \\
&
\end{aligned}
$
Then, $|B|=\left|\begin{array}{ll}2022 & 2024 \\ 2021 & 2023\end{array}\right|=$
$=2$
$
\therefore \quad A^{|B|}=A^2
$
$
A^2=\left[\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 1 & 2 \\
3 & 0 & 2
\end{array}\right]\left[\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 1 & 2 \\
3 & 0 & 2
\end{array}\right]=\left[\begin{array}{ccc}
8 & 4 & 13 \\
4 & -1 & 3 \\
9 & 6 & 13
\end{array}\right]
$
Asked in: AP EAMCET 2021 (23 Aug Shift 1)
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