Question

If \( A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \) is a non-singular matrix, then find \( A^{-1} \) by elementary row transformations. Hence write the inverse of \( \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \).

Hint:

For diagonal matrices, the inverse is obtained by taking reciprocals of diagonal elements; use row operations to confirm.

Answer 1

For a diagonal matrix \( A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \), non-singular means \( x, y, z \neq 0 \). Augment with identity matrix: \( [A | I] = \begin{bmatrix} x & 0 & 0 & 1 & 0 & 0 \\ 0 & y & 0 & 0 & 1 & 0 \\ 0 & 0 & z & 0 & 0 & 1 \end{bmatrix} \).
Step 1: Divide row 1 by \( x \), row 2 by \( y \), row 3 by \( z \) (since \( x, y, z \neq 0 \)): \[ \begin{bmatrix} 1 & 0 & 0 & \frac{1}{x} & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{y} & 0 \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{z} \end{bmatrix}. \] Thus, \( A^{-1} = \begin{bmatrix} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{bmatrix} \).
For \( \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \), set \( x = 2 \), \( y = 1 \), \( z = -1 \): \[ A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & \frac{1}{-1} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}. \]