JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 15)
Let $$A = \left[ {\matrix{
{{1 \over {\sqrt {10} }}} & {{3 \over {\sqrt {10} }}} \cr
{{{ - 3} \over {\sqrt {10} }}} & {{1 \over {\sqrt {10} }}} \cr
} } \right]$$ and $$B = \left[ {\matrix{
1 & { - i} \cr
0 & 1 \cr
} } \right]$$, where $$i = \sqrt { - 1} $$. If $$\mathrm{M=A^T B A}$$, then the inverse of the matrix $$\mathrm{AM^{2023}A^T}$$ is
$$\left[ {\matrix{
1 & { - 2023i} \cr
0 & 1 \cr
} } \right]$$
$$\left[ {\matrix{
1 & 0 \cr
{2023i} & 1 \cr
} } \right]$$
$$\left[ {\matrix{
1 & {2023i} \cr
0 & 1 \cr
} } \right]$$
$$\left[ {\matrix{
1 & 0 \cr
{ - 2023i} & 1 \cr
} } \right]$$
Explanation
$$
\begin{aligned}
& \mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{cc}
\frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\
\frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}}
\end{array}\right]\left[\begin{array}{cc}
\frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\
\frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}}
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] = I \\\\\
& \mathrm{B}^2=\left[\begin{array}{cc}
1 & -\mathrm{i} \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & -\mathrm{i} \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & -2 \mathrm{i} \\
0 & 1
\end{array}\right] \\\\
& \mathrm{B}^3=\left[\begin{array}{cc}
1 & -3 \mathrm{i} \\
0 & 1
\end{array}\right]
\end{aligned}
$$
Similarly,
$$ \begin{aligned} & \mathrm{B}^{2023}=\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \\\\ & \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \\\\ & \mathrm{M}^2=\mathrm{M} \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \mathrm{A}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{~A} \\\\ & \mathrm{M}^3=\mathrm{M}^2 \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{AA}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^3 \mathrm{~A} \end{aligned} $$
Similarly,
$$ \begin{aligned} & \mathrm{M}^{2023}= \mathrm{A}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{~A} \\\\ & \mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}=\mathrm{AA}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{AA}^{\mathrm{T}} = \mathrm{I}.\mathrm{B}^{2023}.\mathrm{I}=\mathrm{B}^{2023} \\\\ & =\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \end{aligned} $$
$$ \therefore $$ $$ \text { Inverse of }\left(\mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}\right) \text { is }\left[\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}\right] $$
Similarly,
$$ \begin{aligned} & \mathrm{B}^{2023}=\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \\\\ & \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \\\\ & \mathrm{M}^2=\mathrm{M} \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \mathrm{A}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{~A} \\\\ & \mathrm{M}^3=\mathrm{M}^2 \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{AA}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^3 \mathrm{~A} \end{aligned} $$
Similarly,
$$ \begin{aligned} & \mathrm{M}^{2023}= \mathrm{A}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{~A} \\\\ & \mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}=\mathrm{AA}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{AA}^{\mathrm{T}} = \mathrm{I}.\mathrm{B}^{2023}.\mathrm{I}=\mathrm{B}^{2023} \\\\ & =\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \end{aligned} $$
$$ \therefore $$ $$ \text { Inverse of }\left(\mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}\right) \text { is }\left[\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}\right] $$
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