If $A=\left[\begin{array}{lll}3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4\end{array}\right]$ and $B=A^3$, then $B^{-1}=$

Question

If $A=\left[\begin{array}{lll}3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4\end{array}\right]$ and $B=A^3$, then $B^{-1}=$, $\left[\begin{array}{ccc}-3 & 0 & 0 \ 0 & -5 & 0 \ 0 & 0 & -4\end{array}\right]$, $\left[\begin{array}{ccc}-27 & 0 & 0 \ 0 & -125 & 0 \ 0 & 0 & -64\end{array}\right]$, $\left[\begin{array}{ccc}\frac{1}{27} & 0 & 0 \ 0 & \frac{1}{125} & 0 \ 0 & 0 & \frac{1}{64}\end{array}\right]$, $\left[\begin{array}{ccc}\frac{-1}{27} & 0 & 0 \ 0 & \frac{-1}{125} & 0 \ 0 & 0 & \frac{-1}{64}\end{array}\right]$

MARKS App · Accepted Answer

The given matrix $A=\left[\begin{array}{lll}3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4\end{array}\right]$ and $\begin{aligned} B & =A^3=\left[\begin{array}{lll} 3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4 \end{array}\right]\left[\begin{array}{lll} 3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4 \end{array}\right]\left[\begin{array}{ccc} 3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4 \end{array}\right] \ & =\left[\begin{array}{ccc} 9 & 0 & 0 \ 0 & 25 & 0 \ 0 & 0 & 16 \end{array}\right]\left[\begin{array}{lll} 3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4 \end{array}\right]=\left[\begin{array}{ccc} 27 & 0 & 0 \ 0 & 125 & 0 \ 0 & 0 & 64 \end{array}\right] \ \therefore & \ B^{-1} & =\frac{1}{27 \times 125 \times 64}\left[\begin{array}{ccc} 125 \times 64 & 0 & 0 \ 0 & 27 \times 64 & 0 \ 0 & 0 & 527 \times 125 \end{array}\right] \ & =\left[\begin{array}{ccc} \frac{1}{27} & 0 & 0 \ 0 & \frac{1}{125} & 0 \ 0 & 0 & \frac{1}{64} \end{array}\right] \end{aligned}$ Hence, option (c) is correct.

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