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Question: Answered & Verified by Expert
If $A=\left[\begin{array}{ccc}k / 2 & 0 & 0 \\ 0 & l / 3 & 0 \\ 0 & 0 & m / 4\end{array}\right]$ and $A^{-1}=\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]$, then $k+l+m=$
MathematicsMatricesJEE Main
Options:
  • A 1
  • B 9
  • C 14
  • D 29
Solution:
1591 Upvotes Verified Answer
The correct answer is: 29
Given,
$$
A=\left[\begin{array}{ccc}
k / 2 & 0 & 0 \\
0 & l / 3 & 0 \\
0 & 0 & m / 4
\end{array}\right] \text { and } A^{-1}=\left[\begin{array}{ccc}
1 / 2 & 0 & 0 \\
0 & 1 / 3 & 0 \\
0 & 0 & 1 / 4
\end{array}\right]
$$
Since, $\quad A A^{-1}=I$
$$
\begin{aligned}
& \Rightarrow \quad\left[\begin{array}{ccc}
\frac{k}{2} & 0 & 0 \\
0 & \frac{l}{3} & 0 \\
0 & 0 & \frac{m}{4}
\end{array}\right]\left[\begin{array}{lll}
\frac{1}{2} & 0 & 0 \\
0 & \frac{1}{3} & 0 \\
0 & 0 & \frac{1}{4}
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
& \Rightarrow \quad\left[\begin{array}{ccc}
\frac{k}{4} & 0 & 0 \\
0 & \frac{l}{9} & 0 \\
0 & 0 & \frac{m}{16}
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$$
On comparing, we get
$$
\begin{aligned}
& & \frac{k}{4} & =1, \frac{l}{9}=1 \text { and } \frac{m}{16}=1 \\
& & k & =4, l=9, m=16 \\
\therefore & & k+l+m & =4+9+16=29
\end{aligned}
$$

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