The matrix $A^2+4 A-5 I$, where $I$ is identity matrix and $A=\left[\begin{array}{cc}1 & 2 \\ 4 &…
The matrix $A^2+4 A-5 I$, where $I$ is identity matrix and $A=\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]$, equals :
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$4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]$
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$4\left[\begin{array}{cc}0 & -1 \\ 2 & 2\end{array}\right]$
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$32\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]$
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$32\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]$
Solution
$\begin{aligned} & \mathrm{A}^2+4 \mathrm{~A}-5 \mathrm{I}=\mathrm{A} \times \mathrm{A}+4 \mathrm{~A}-5 \mathrm{I} \\ = & {\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right] \times\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]+4\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]-5\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] } \\ & =\left[\begin{array}{cc}9 & -4 \\ -8 & 17\end{array}\right]+\left[\begin{array}{cc}4 & 8 \\ 16 & -12\end{array}\right]-\left[\begin{array}{cc}5 & 0 \\ 0 & 5\end{array}\right] \\ & =\left[\begin{array}{cc}9+4-5 & -4+8-0 \\ -8+16-0 & 17-12-5\end{array}\right]=\left[\begin{array}{ll}8 & 4 \\ 8 & 0\end{array}\right] \\ & =4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]\end{aligned}$
Asked in: JEE Main 2013 (09 Apr Online)
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