If bmatrix x-y & 0 \\ x+y & 1 bmatrix is an identity matrix and bmatrix x & y \\ z & x bmatrix is a singular matrix then: Option 1: x > y = z Option 2: x = y > z Option 3: x = y < z Option 4: x ≠ y ≠ z

The answer is 'x > y = z'. View full solution:

Algebra > Matrices & Determinants

Medium

core

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:

$x > y = z$

$x = y > z$

$x = y < z$

$x \neq y \neq z$

✅ Correct Option: 1

Related questions:

2026: 18th May Shift 1

Match List-I with List-II

List-I	List-II
Matrix A	$\left\vert \text{adj}A \right\vert$
(A) $\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$	(I) $2$
(B) $\begin{bmatrix} 3 & 0 \\ 1 & 5 \end{bmatrix}$	(II) $5$
(C) $\begin{bmatrix} 6 & -1 \\ 5 & 3 \end{bmatrix}$	(III) $15$
(D) $\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$	(IV) $23$

Choose the correct answer from the options given below:

2022: 16 July Shift 2

If $\begin{vmatrix} (a-x)^2 & (a-y)^2 & (a-z)^2 \\ (b-x)^2 & (b-y)^2 & (b-z)^2 \\ (c-x)^2 & (c-y)^2 & (c-z)^2 \end{vmatrix} = \lambda(a-b)(b-c)(c-a) \cdot (x-y)(y-z)(z-x)$

then the value of $\lambda$ is:

2026: 31st May Shift 2

The number of all possible matrices of order $2 \times 2$ with each entry 0 or 1 having their determinants as positive value is :

2025: 13 May Shift 2

If $A$ is a square matrix and $I$ is an identity matrix of same order such that $A^2 = A$ , then $(2I + A)^2 - 5A$ is

2026: 30th May Shift 2

A is a square matrix of order 3 x 3, then

Match List-I with List-II

List-I	List-II
(A) $\left\vert \text{adj}A \right\vert$	(I) $\dfrac{1}{\left\vert A \right\vert}$
(B) $\left\vert A^{-1} \right\vert$	(II) $\left\vert A \right\vert^{4}$
(C) $\left\vert A.\text{adj}A \right\vert$	(III) $\left\vert A \right\vert^{3}$
(D) $\left\vert \text{adj}(\text{adj}A) \right\vert$	(IV) $\left\vert A \right\vert^{2}$

Choose the correct answer from the options given below:

2026: 30th May Shift 2

If $A = \begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}$ and $f(x) = x^2-2x-3$ , then $f(A)$ is

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:

Keyboard Shortcuts