CUET UG 2025 — Mathematics Algebra

An identity matrix has the form $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$[10​01​].

Since $$\begin{bmatrix} x-y & 0 \\ x+y & 1 \end{bmatrix}$$[x−yx+y​01​] is an identity matrix:

$x - y = 1$x−y=1 ... (1)

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

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Adding equations (1) and (2):

$$x - y + x + y = 1 + 0$$x−y+x+y=1+0

$$2x = 1$$2x=1

$$x = \frac{1}{2}$$x=21​

From equation (2):

$$\frac{1}{2} + y = 0$$21​+y=0

$$y = -\frac{1}{2}$$y=−21​

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A matrix is singular when its determinant equals 0.

For the matrix $$\begin{bmatrix} x & y \\ z & x \end{bmatrix}$$[xz​yx​], the determinant is:

$$\text{Det} = x^2 - yz$$Det=x2−yz

Since the matrix is singular:

$$x^2 - yz = 0$$x2−yz=0

$$x^2 = yz$$x2=yz

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Substituting $x = \frac{1}{2}$x=21​ and $y = -\frac{1}{2}$y=−21​:

$$\left(\frac{1}{2}\right)^2 = \left(-\frac{1}{2}\right) \cdot z$$(21​)2=(−21​)⋅z

$$\frac{1}{4} = -\frac{z}{2}$$41​=−2z​

$$z = -\frac{1}{2}$$z=−21​

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The values are:

$$x = \frac{1}{2}$$x=21​

$$y = -\frac{1}{2}$$y=−21​

$$z = -\frac{1}{2}$$z=−21​

Therefore: $x > y = z$x>y=z