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Consider the following in respect of the matrix $\mathrm{A}=\left(\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right):$
$1.$ $\mathrm{A}^{2}=-\mathrm{A}$
$2.$ $A^{3}=4 A$
Which of the above is/are correct?
Options:
$1.$ $\mathrm{A}^{2}=-\mathrm{A}$
$2.$ $A^{3}=4 A$
Which of the above is/are correct?
Solution:
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Verified Answer
The correct answer is:
2 only
$A=\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]$
$\begin{aligned} A \cdot A &=\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right] \\ &=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=-2\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right] \\ A^{2} &=-2 A \end{aligned}$
$A^{2}, A=-2\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]$
$\quad=-2\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=4\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]$
$A^{3}=4 A$
Hence $A^{2} \neq-A, A^{3}=4 A$
$\begin{aligned} A \cdot A &=\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right] \\ &=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=-2\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right] \\ A^{2} &=-2 A \end{aligned}$
$A^{2}, A=-2\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]$
$\quad=-2\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=4\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]$
$A^{3}=4 A$
Hence $A^{2} \neq-A, A^{3}=4 A$
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