Q93 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\text { If } x, y, z \text { are non zero real numbers, then inverse of matrix } A=\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right] \text { is }$
- A$x y z\left[\begin{array}{ccc} 1 / x & 0 & 0 \\ 0 & 1 / y & 0 \\ 0 & 0 & 1 / z \end{array}\right]$
- B$\dfrac{1}{x y z}\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
- C$\dfrac{1}{x y z}\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right]$
- Correct answerD$\left[\begin{array}{ccc} 1 / x & 0 & 0 \\ 0 & 1 / y & 0 \\ 0 & 0 & 1 / z \end{array}\right]$
Solution
The given matrix is a diagonal matrix $A = \text{diag}(x, y, z) = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$.
For a diagonal matrix $A = \text{diag}(a_1, a_2, a_3)$, the inverse $A^{-1}$ is given by $\text{diag}(1/a_1, 1/a_2, 1/a_3)$, provided all $a_i \neq 0$.
Since $x, y, z$ are non-zero real numbers, the inverse exists and is calculated as follows:
$A^{-1} = \begin{bmatrix} 1/x & 0 & 0 \\ 0 & 1/y & 0 \\ 0 & 0 & 1/z \end{bmatrix}$.
Answer: $\left[\begin{array}{ccc} 1 / x & 0 & 0 \\ 0 & 1 / y & 0 \\ 0 & 0 & 1 / z \end{array}\right]$
For a diagonal matrix $A = \text{diag}(a_1, a_2, a_3)$, the inverse $A^{-1}$ is given by $\text{diag}(1/a_1, 1/a_2, 1/a_3)$, provided all $a_i \neq 0$.
Since $x, y, z$ are non-zero real numbers, the inverse exists and is calculated as follows:
$A^{-1} = \begin{bmatrix} 1/x & 0 & 0 \\ 0 & 1/y & 0 \\ 0 & 0 & 1/z \end{bmatrix}$.
Answer: $\left[\begin{array}{ccc} 1 / x & 0 & 0 \\ 0 & 1 / y & 0 \\ 0 & 0 & 1 / z \end{array}\right]$
Hence, the correct answer is option: 'D'.