If $A=\begin{bmatrix}3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4\end{bmatrix}$ and $B=A^{3}$, then $B^{-1}=$

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Aicharya · Accepted Answer

The given matrix $A=\left[\begin{array}{ccc}3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4\end{array}ight]$ and $\begin{aligned} B & =A^{3}=\left[\begin{array}{ccc} 3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4 \end{array}ight]\left[\begin{array}{ccc} 3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4 \end{array}ight]\left[\begin{array}{ccc} 3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4 \end{array}ight] \ & =\left[\begin{array}{ccc} 27 & 0 & 0 \ 0 & 125 & 0 \ 0 & 0 & 64 \end{array}ight] \ 	herefore & \ B^{-1} & =\frac{1}{27 	imes 125 	imes 64}\left[\begin{array}{ccc} 125 	imes 64 & 0 & 0 \ 0 & 27 	imes 64 & 0 \ 0 & 0 & 27 	imes 125 \end{array}ight] \ & =\left[\begin{array}{ccc} \frac{1}{27} & 0 & 0 \ 0 & \frac{1}{125} & 0 \ 0 & 0 & \frac{1}{64} \end{array}ight] \end{aligned}$ Hence, option (c) is correct.

If $A=\begin{bmatrix}3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 4\end{bmatrix}$ and $B=A^{3}$, then $B^{-1}=$

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