Question
If $A =\left[\begin{array}{cc}\sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan 45^{\circ} & \sin 90^{\circ}\end{array}\right]$ and $B =\left[\begin{array}{cc}0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ}\end{array}\right]$ Find $A^2$
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Answer
$\begin{aligned} & \text { Given } \\ & A=\left[\begin{array}{cc}\sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan 45^{\circ} & \sin 90^{\circ}\end{array}\right] \\ & \text { and } \\ & B=\left[\begin{array}{cc}0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ}\end{array}\right] \\ & A=\left[\begin{array}{cc}\sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan 45^{\circ} & \sin 90^{\circ}\end{array}\right]=\left[\begin{array}{ll}2 & 0 \\ 3 & 1\end{array}\right] \ldots\left(\because \sec 60^{\circ}=2, \cos 90^{\circ}=0, \tan 45^{\circ}=1, \sin 90^{\circ}=1\right) \\ & B=\left[\begin{array}{cc}0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ}\end{array}\right]=\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right] \ldots\left(\because \cot 45^{\circ}=1\right) \\ & A^2=A \times A=\left[\begin{array}{cc}2 & 0 \\ -3 & 1\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ -3 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}4+0 & 0+0 \\ -6-3 & 0+1\end{array}\right] \\ & =\left[\begin{array}{cc}4 & 0 \\ -9 & 1\end{array}\right] . \\ & \end{aligned}$
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