Algebra

Matrices & Determinants

Medium

common

The value of $\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$ is:

$(a-b)(c-b)(a-c)$

$(a-b)(b-c)(c-a)(a+b+c)$

$abc(a-b)(b-c)(c-a)$

$(b-a)(c-b)(a-c)$

✅ Correct Option: 1

This Vandermonde determinant equals $(a-b)(b-c)(c-a)$ , which is the same as $(a-b)(c-b)(a-c)$ . Checking with $a=0$ , $b=1$ , $c=2$ gives determinant 2, matching option 1.

2026: 29 May Shift 1

Match List-I with List-II

List-I (Singular matrix)	List-II (A value of $x$ )
(A) $A=\begin{bmatrix}x & 4 \\ 2 & 2x\end{bmatrix}$	(I) 3
(B) $A=\begin{bmatrix}3 & x \\ 3x & 9\end{bmatrix}$	(II) 4
(C) $A=\begin{bmatrix}x & 1 \\ 1 & x\end{bmatrix}$	(III) 2
(D) $A=\begin{bmatrix}x & 1 \\ 16 & x\end{bmatrix}$	(IV) 1

Choose the correct answer from the options given below:

2023: 22 May Shift 3

If $A = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{pmatrix}$ , then $A^2 =$

2026: 18 May Shift 1

For what values of $x$ and $y$ , are the following matrices equal?

$A=\begin{bmatrix}2x+1 & 3y\\ 0 & y^2-5y\end{bmatrix}$ , $B=\begin{bmatrix}x+3 & y^2+2\\ 0 & -6\end{bmatrix}$

2023: 15 June Shift 2

If $A = \begin{bmatrix} -2 & 6 \\ -5 & -1 \end{bmatrix}$ then $A^{-1}$ is :

2025: 15 May Shift 1

If A is a square matrix, then $(A^T - A)$ is-

2025: 22 May Shift 1

Let A be a non-singular square matrix of order 3 and $|adj A| = 5$ then $|A|$ is equal to

The value of ∣1aa21bb21cc2∣\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}​111​abc​a2b2c2​​ is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The value of $\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$ is: