CUET Mathematics - Match the LIST-I with LIST-II | | LIST-I<br>(Matrix) | | LIST-II<br>(Type of matrix) | |---|---|---|---| | A. | A = [a_ij]_m x m where a_ij = 0\ i ≠ j | I. | Scalar matrix | | B. | A = [a_ij]_m x 1 | II. | Diagonal matrix | | C. | A = [a_ij]_1 x n | III. | Column matrix | | D. | A = [a_ij]_m x m where a_ij = 0\ i ≠ j and a_ij = k\ i = j, k ≠ 0 | IV. | Row matrix | Choose the correct answer from the options given below: | PYQs + Solutions

CUET Mathematics 2026 30th May Shift 1

Algebra

Easy

common

A-II, B-III, C-IV, D-I

A-III, B-IV, C-I, D-II

A-IV, B-I, C-II, D-III

A-I, B-II, C-III, D-IV

✅ Correct Option: 1

Related questions:

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If $A = \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix}$ , $B = [-1 \quad 2 \quad 3]$ , then the value of $A'B'$ is

3 June Shift 1

If $A = \begin{bmatrix} 1 & 2 & 3 \\ -4 & -5 & -2 \end{bmatrix}$ , $B = \begin{bmatrix} 2 & -3 \\ 4 & -5 \\ 2 & -1 \end{bmatrix}$ and $BA = [b_{ij}]$ , then $(b_{23} - b_{31})$ is equal to

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Match List-I with List-II

List-I	List-II
(A) $\begin{bmatrix} 5 & 3 \\ -7 & 1 \end{bmatrix} + \begin{bmatrix} 7 & 3 \\ 7 & 2 \end{bmatrix}$	(I) Identity matrix
(B) $\begin{bmatrix} 5 & 3 \\ -7 & 1 \end{bmatrix} + \begin{bmatrix} -4 & -3 \\ -6 & 0 \end{bmatrix}$	(II) Upper Triangular Matrix
(C) $\begin{bmatrix} 5 & -3 \\ -2 & -1 \end{bmatrix} + \begin{bmatrix} -5 & 3 \\ 2 & 1 \end{bmatrix}$	(III) Zero Matrix
(D) $\begin{bmatrix} 5 & 13 \\ -2 & 6 \end{bmatrix} + \begin{bmatrix} -4 & -13 \\ 2 & -5 \end{bmatrix}$	(IV) Lower Triangular Matrix

Choose the correct answer from the options given below:

CUET Mathematics 2026 30th May Shift 1 PYQs

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