Match List-I with List-II, List-I, List-II, Matrix Product, Order of resultant matrix, (A) [a_ij]_(2 x 3) x [b_ij]_(3 x 4), (I) 2 x 4, (B) [a_ij]_(2 x 1) x [b_ij]_(1 x 3), (II) Not possible, (C) [a_ij]_(3 x 2) x [b_ij]_(3 x 2), (III) 3 x 3, (D) [a_ij]_(3 x 3) x [b_ij]_(3 x 3), (IV) 2 x 3, Choose the correct answer from the options given below: Option 1: (A) - (I), (B) - (II), (C) - (IV), (D) - (III) Option 2: (A) - (I), (B) - (II), (C) - (III), (D) - (IV) Option 3: (A) - (I), (B) - (IV), (C) - (II), (D) - (III) Option 4: (A) - (III), (B) - (II), (C) - (I), (D) - (IV)

Answer: (A) - (I), (B) - (IV), (C) - (II), (D) - (III). Solution: For matrix multiplication, the number of columns of the first matrix must equal the number of rows of the second matrix. If the first matrix is of order m x n and the second is of order n x p, the resultant matrix has order m x p. (A) [a_ij]_(2 x 3) x [b_ij]_(3 x 4) Columns of first matrix = 3, Rows of second matrix = 3 → Multiplication is possible. Resultant order = 2 x 4 (A) arrow (I) (B) [a_ij]_(2 x 1) x [b_ij]_(1 x 3) Columns of first matrix = 1, Rows of second matrix = 1 → Multiplication is possible. Resultant order = 2 x 3 (B) arrow (IV) (C) [a_ij]_(3 x 2) x [b_ij]_(3 x 2) Columns of first matrix = 2, Rows of second matrix = 3 → 2 ≠ 3, so multiplication is not possible. (C) arrow (II) (D) [a_ij]_(3 x 3) x [b_ij]_(3 x 3) Columns of first matrix = 3, Rows of second matrix = 3 → Multiplication is possible. Resultant order = 3 x 3 (D) arrow (III) (A) arrow (I), (B) arrow (IV), (C) arrow (II), (D) arrow (III)

Algebra

Matrices & Determinants

Easy

applied

Match List-I with List-II

List-I List-II
Matrix Product Order of resultant matrix
--- ---
(A) $[a_{ij}]_{2 \times 3} \times [b_{ij}]_{3 \times 4}$ (I) $2 \times 4$
(B) $[a_{ij}]_{2 \times 1} \times [b_{ij}]_{1 \times 3}$ (II) Not possible
(C) $[a_{ij}]_{3 \times 2} \times [b_{ij}]_{3 \times 2}$ (III) $3 \times 3$
(D) $[a_{ij}]_{3 \times 3} \times [b_{ij}]_{3 \times 3}$ (IV) $2 \times 3$

Choose the correct answer from the options given below:

List-I	List-II
Matrix Product	Order of resultant matrix
---	---
(A) $[a_{ij}]_{2 \times 3} \times [b_{ij}]_{3 \times 4}$	(I) $2 \times 4$
(B) $[a_{ij}]_{2 \times 1} \times [b_{ij}]_{1 \times 3}$	(II) Not possible
(C) $[a_{ij}]_{3 \times 2} \times [b_{ij}]_{3 \times 2}$	(III) $3 \times 3$
(D) $[a_{ij}]_{3 \times 3} \times [b_{ij}]_{3 \times 3}$	(IV) $2 \times 3$

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

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

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

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

✅ Correct Option: 3

For matrix multiplication, the number of columns of the first matrix must equal the number of rows of the second matrix. If the first matrix is of order $m \times n$ and the second is of order $n \times p$ , the resultant matrix has order $m \times p$ .

(A) $[a_{ij}]_{2 \times 3} \times [b_{ij}]_{3 \times 4}$

Columns of first matrix $= 3$ , Rows of second matrix $= 3$ → Multiplication is possible.

Resultant order $= 2 \times 4$

$(A) \rightarrow (I)$

(B) $[a_{ij}]_{2 \times 1} \times [b_{ij}]_{1 \times 3}$

Columns of first matrix $= 1$ , Rows of second matrix $= 1$ → Multiplication is possible.

Resultant order $= 2 \times 3$

$(B) \rightarrow (IV)$

Columns of first matrix $= 2$ , Rows of second matrix $= 3$ → $2 \neq 3$ , so multiplication is not possible.

$(C) \rightarrow (II)$

(D) $[a_{ij}]_{3 \times 3} \times [b_{ij}]_{3 \times 3}$

Columns of first matrix $= 3$ , Rows of second matrix $= 3$ → Multiplication is possible.

Resultant order $= 3 \times 3$

$(D) \rightarrow (III)$

$(A) \rightarrow (I)$ , $(B) \rightarrow (IV)$ , $(C) \rightarrow (II)$ , $(D) \rightarrow (III)$

2025: 3 June Shift 1

The integral value of k for which the system of linear equations $kx + y + 2z = 0$ , $ky = x - 3z$ and $2x + y + kz = 0$ has a non-zero solution is

2025: 2 June Shift 1

If $x, y$ and $z$ are real number such that $x + y + z = 0$ , then value of $\begin{vmatrix}3x & -x+y & -x+z\\x-y & 3y & z-y\\x-z & y-z & 3z\end{vmatrix}$ is

2025: 14 May Shift 2

The values of $\lambda$ for which the system of equation $x + 2y + z = 14, - x + y + z = 10, x + \lambda y + z = 2$ has unique solution is

2025: 14 May Shift 2

Let A be any square matrix of order 3 and $B = \begin{bmatrix} 0 & -4 & 2 \\ 4 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}$ . Then the matrix $ABA^T$ is a

2026: 29 May Shift 1

Let $A=\begin{bmatrix}1 & 0 \\ -1 & 7\end{bmatrix}$ and $B=\begin{bmatrix}0 & 4 \\ -1 & 7\end{bmatrix}$ then

Match List-I with List-II

List-I	List-II
Determinants	Value
(A) $\left\vert A^{2} \right\vert$	(I) $28$
(B) $\left\vert AB^{T} \right\vert$	(II) $7$
(C) $\left\vert \text{adj } A \right\vert$	(III) $16$
(D) $\left\vert 2B \right\vert$	(IV) $49$

Choose the correct answer from the options given below:

2026: 22 May Shift 1

Which of the following are correct?

(A) $A=\begin{bmatrix}-9 & 4 & -3\\-1 & 0 & 4\\2 & 2 & 0\end{bmatrix}$ ; $a_{22}+a_{32}=-9$ where $a_{ij}$ denotes the $(i,j)^{th}$ element of matrix A.

(B) $A=\begin{bmatrix}2 & -2\\-2 & 2\end{bmatrix}$ ; $A^2=pA$ , then $p=4$

(D) $\begin{bmatrix}0 & -5 & 3\\5 & 0 & -7\\-3 & 7 & 0\end{bmatrix}$ is a symmetric matrix

Choose the correct answer from the options given below:

Solution

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