If A^(-1) exists for the matrix A = bmatrix 1 & & -1 \\ -1 & 1 & 0 \\ & 1 & 1 bmatrix then Option 1: = -1 Option 2: ≠ -1 Option 3: ≠ 2 Option 4: ≠ 1

Answer: ≠ -1. Solution: For A^(-1) to exist, the determinant of A must be non-zero. (A) ≠ 0 Given: A = bmatrix 1 & & -1 \\ -1 & 1 & 0 \\ & 1 & 1 bmatrix Expanding along Row 1: (A) = 1 * vmatrix 1 & 0 \\ 1 & 1 vmatrix - * vmatrix -1 & 0 \\ & 1 vmatrix + (-1) * vmatrix -1 & 1 \\ & 1 vmatrix Computing each 2×2 determinant: vmatrix 1 & 0 \\ 1 & 1 vmatrix = (1)(1) - (0)(1) = 1 vmatrix -1 & 0 \\ & 1 vmatrix = (-1)(1) - (0)() = -1 vmatrix -1 & 1 \\ & 1 vmatrix = (-1)(1) - (1)() = -1 - Substituting back: (A) = 1(1) - (-1) + (-1)(-1 - ) = 1 + + 1 + = 2 + 2 = 2(1 + ) For A^(-1) to exist: (A) ≠ 0 2(1 + ) ≠ 0 1 + ≠ 0 ≠ -1 Therefore, A^(-1) exists when ≠ -1.

Algebra

Matrices & Determinants

Medium

core

If $A^{-1}$ exists for the matrix $A = \begin{bmatrix} 1 & \lambda & -1 \\ -1 & 1 & 0 \\ \lambda & 1 & 1 \end{bmatrix}$ then

$\lambda = -1$

$\lambda \neq -1$

$\lambda \neq 2$

$\lambda \neq 1$

✅ Correct Option: 2

For $A^{-1}$ to exist, the determinant of $A$ must be non-zero.

$\det(A) \neq 0$

Given: $A = \begin{bmatrix} 1 & \lambda & -1 \\ -1 & 1 & 0 \\ \lambda & 1 & 1 \end{bmatrix}$

Expanding along Row 1:

$\det(A) = 1 \cdot \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - \lambda \cdot \begin{vmatrix} -1 & 0 \\ \lambda & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} -1 & 1 \\ \lambda & 1 \end{vmatrix}$

Computing each 2×2 determinant:

$\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = (1)(1) - (0)(1) = 1$

$\begin{vmatrix} -1 & 0 \\ \lambda & 1 \end{vmatrix} = (-1)(1) - (0)(\lambda) = -1$

$\begin{vmatrix} -1 & 1 \\ \lambda & 1 \end{vmatrix} = (-1)(1) - (1)(\lambda) = -1 - \lambda$

Substituting back:

$\det(A) = 1(1) - \lambda(-1) + (-1)(-1 - \lambda)$

$= 1 + \lambda + 1 + \lambda$

$= 2 + 2\lambda$

$= 2(1 + \lambda)$

For $A^{-1}$ to exist:

$\det(A) \neq 0$

$2(1 + \lambda) \neq 0$

$1 + \lambda \neq 0$

$\lambda \neq -1$

Therefore, $A^{-1}$ exists when $\lambda \neq -1$ .

2026: 29 May Shift 1

If A is a non-singular matrix of order 3, then

Match List-I with List-II

List-I	List-II
Determinant of A	$\left\vert \text{adj}(\text{adj } A) \right\vert =$
(A) $\left\vert A \right\vert = 3$	(I) $1$
(B) $\left\vert A \right\vert = 4$	(II) $16$
(C) $\left\vert A \right\vert = 2$	(III) $81$
(D) $\left\vert A \right\vert = 1$	(IV) $256$

Choose the correct answer from the options given below:

2026: 19 May Shift 2

If A and B are invertible matrices of the same order, then,

Match List-I with List-II

LIST-I	LIST-II
A. $\left\vert A^T \right\vert$	I. $B^{-1}A^{-1}$
B. $A(\text{adj }A)$	II. $\text{adj }A$
C. $A^{-1}\left\vert A \right\vert$	III. $(\text{adj }A)A$
D. $(AB)^{-1}$	IV. $\left\vert A \right\vert$

Choose the correct answer from the options given below:

2026: 22 May Shift 1

For every square matrix $A$ of order $n\times n$ , which of the following statements are correct ?

(A) $(adjA)A = A(adjA) = |A|I$ , where $I$ is identity matrix of order n.

(B) $|kA|=k^n|A|$ , k is any scalar.

(D) $A^2=A$

Choose the correct answer from the options given below:

2025: 22 May Shift 1

The number of all possible matrices of order $2 \times2$ with each entry $0, 1$ or $2$ are.

2024: 16 May Shift 1

If $P=\left[\begin{array}{r}-1 \\ 2 \\ 1\end{array}\right]$ and $Q=\left[\begin{array}{lll}2 & -4 & 1\end{array}\right]$ are two matrices, then $(P Q)^{\prime}$ will be :

2026: 22 May Shift 2

If $A=\begin{bmatrix}k & 2\\2 & k\end{bmatrix}$ such that $\lvert A^3 \rvert = 125$ then which of the following statements are TRUE?

(A) $\lvert k.adj(3A) \rvert = 405$

(B) $\lvert k.adj(3A) \rvert = 2025$

(D) $\lvert adj(3A) \rvert = 45$

Choose the correct answer from the options given below:

If A−1A^{-1}A−1 exists for the matrix A=[1λ−1−110λ11]A = \begin{bmatrix} 1 & \lambda & -1 \\ -1 & 1 & 0 \\ \lambda & 1 & 1 \end{bmatrix}A=​1−1λ​λ11​−101​​ then

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $A^{-1}$ exists for the matrix $A = \begin{bmatrix} 1 & \lambda & -1 \\ -1 & 1 & 0 \\ \lambda & 1 & 1 \end{bmatrix}$ then