If $A=\left[\begin{array}{ccc}k / 2 & 0 & 0 \ 0 & l / 3 & 0 \ 0 & 0 & m / 4\end{array} ight]$ and $A^{-1}=\left[\begin{array}{ccc}1 / 2 & 0 & 0 \ 0 & 1 / 3 & 0 \ 0 & 0 & 1 / 4\end{array} ight]$, then $k+l+m=$

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If $A=\left[\begin{array}{ccc}k / 2 & 0 & 0 \ 0 & l / 3 & 0 \ 0 & 0 & m / 4\end{array}ight]$ and $A^{-1}=\left[\begin{array}{ccc}1 / 2 & 0 & 0 \ 0 & 1 / 3 & 0 \ 0 & 0 & 1 / 4\end{array}ight]$, then $k+l+m=$

Aicharya · Accepted Answer

Given, $ A=\left[\begin{array}{ccc} k / 2 & 0 & 0 \ 0 & l / 3 & 0 \ 0 & 0 & m / 4 \end{array}ight] 	ext { and } A^{-1}=\left[\begin{array}{ccc} 1 / 2 & 0 & 0 \ 0 & 1 / 3 & 0 \ 0 & 0 & 1 / 4 \end{array}ight] $ Since, $\quad A A^{-1}=I$ $ \begin{aligned} & \Rightarrow \quad\left[\begin{array}{ccc} \frac{k}{2} & 0 & 0 \ 0 & \frac{l}{3} & 0 \ 0 & 0 & \frac{m}{4} \end{array}ight]\left[\begin{array}{lll} \frac{1}{2} & 0 & 0 \ 0 & \frac{1}{3} & 0 \ 0 & 0 & \frac{1}{4} \end{array}ight]=\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}ight] \ & \Rightarrow \quad\left[\begin{array}{ccc} \frac{k}{4} & 0 & 0 \ 0 & \frac{l}{9} & 0 \ 0 & 0 & \frac{m}{16} \end{array}ight]=\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}ight] \end{aligned} $ On comparing, we get $ \begin{aligned} & & \frac{k}{4} & =1, \frac{l}{9}=1 	ext { and } \frac{m}{16}=1 \ & & k & =4, l=9, m=16 \ 	herefore & & k+l+m & =4+9+16=29 \end{aligned} $

If $A=\left[\begin{array}{ccc}k / 2 & 0 & 0 \\ 0 & l / 3 & 0 \\ 0 & 0 & m / 4\end{array}\right]$ and…

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