If $\begin{bmatrix} x-y & 0 \\ x+y & 1 \end{bmatrix}$ is an identity matrix and $\begin{bmatrix} x & y \\ z & x \end{bmatrix}$ is a singular matrix then:

An identity matrix has the form $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Since $\begin{bmatrix} x-y & 0 \\ x+y & 1 \end{bmatrix}$ is an identity matrix:

$x - y = 1$ ... (1)

$x + y = 0$ ... (2)

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Adding equations (1) and (2):

$x - y + x + y = 1 + 0$

$2x = 1$

$x = \frac{1}{2}$

From equation (2):

$\frac{1}{2} + y = 0$

$y = -\frac{1}{2}$

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A matrix is singular when its determinant equals 0.

For the matrix $\begin{bmatrix} x & y \\ z & x \end{bmatrix}$, the determinant is:

$\text{Det} = x^2 - yz$

Since the matrix is singular:

$x^2 - yz = 0$

$x^2 = yz$

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Substituting $x = \frac{1}{2}$ and $y = -\frac{1}{2}$:

$\left(\frac{1}{2}\right)^2 = \left(-\frac{1}{2}\right) \cdot z$

$\frac{1}{4} = -\frac{z}{2}$

$z = -\frac{1}{2}$

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The values are:

$x = \frac{1}{2}$

$y = -\frac{1}{2}$

$z = -\frac{1}{2}$

Therefore: $x > y = z$