CUET Preparation Today
If x, y, z are non-zero numbers, then the inverse of matrix $A = \begin{bmatrix} x & 0 & 0\\ 0 & y & 0\\0&0&z \end{bmatrix}$ is |
$\begin{bmatrix}\frac{1}{x}&0&0\\0&\frac{1}{y}&0\\0&0&\frac{1}{z}\end{bmatrix}$ $\begin{bmatrix}\frac{1}{yz}&0&0\\0&\frac{1}{xz}&0\\0&0&\frac{1}{xy}\end{bmatrix}$ $\begin{bmatrix}\frac{1}{x^2yz}&0&0\\0&\frac{1}{xy^2z}&0\\0&0&\frac{1}{xyz^2}\end{bmatrix}$ $\begin{bmatrix}\frac{1}{xyz}&0&0\\0&\frac{1}{xyz}&0\\0&0&\frac{1}{xyz}\end{bmatrix}$ |
$\begin{bmatrix}\frac{1}{x}&0&0\\0&\frac{1}{y}&0\\0&0&\frac{1}{z}\end{bmatrix}$ |
The correct answer is Option (1) → $\begin{bmatrix}\frac{1}{x}&0&0\\0&\frac{1}{y}&0\\0&0&\frac{1}{z}\end{bmatrix}$ Given the matrix: $ A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} $ Since $A$ is a diagonal matrix and $x$, $y$, $z$ are non-zero, its inverse exists. The inverse of a diagonal matrix is obtained by taking the reciprocal of each non-zero diagonal entry: $ A^{-1} = \begin{bmatrix} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{bmatrix} $ |
