Algebra

Matrices & Determinants

Medium

core

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

✅ Correct Option: 3

Matrix B is 3×2 and matrix A is 2×3.

The number of columns in B equals the number of rows in A, so BA can be computed.

The result BA will be a 3×3 matrix.

$BA = \begin{bmatrix} 2 & -3 \\ 4 & -5 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ -4 & -5 & -2 \end{bmatrix}$

Row 1 of B with each column of A:

$(1,1): 2(1) + (-3)(-4) = 2 + 12 = 14$

$(1,2): 2(2) + (-3)(-5) = 4 + 15 = 19$

$(1,3): 2(3) + (-3)(-2) = 6 + 6 = 12$

Row 2 of B with each column of A:

$(2,1): 4(1) + (-5)(-4) = 4 + 20 = 24$

$(2,2): 4(2) + (-5)(-5) = 8 + 25 = 33$

$(2,3): 4(3) + (-5)(-2) = 12 + 10 = 22$

Row 3 of B with each column of A:

$(3,1): 2(1) + (-1)(-4) = 2 + 4 = 6$

$(3,2): 2(2) + (-1)(-5) = 4 + 5 = 9$

$(3,3): 2(3) + (-1)(-2) = 6 + 2 = 8$

$BA = \begin{bmatrix} 14 & 19 & 12 \\ 24 & 33 & 22 \\ 6 & 9 & 8 \end{bmatrix}$

From the resulting matrix:

$b_{23} = 22$ (row 2, column 3)

$b_{31} = 6$ (row 3, column 1)

$b_{23} - b_{31} = 22 - 6 = 16$

Therefore, $(b_{23} - b_{31}) = 16$

2026: 30 May Shift 2

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

Match List-I with List-II

List-I (Determinant)	List-II (Equal to)
(A) $\left\vert A^{-1}B \right\vert$	(I) $\dfrac{1}{\left\vert A \right\vert \left\vert B \right\vert}$
(B) $\left\vert AB^{-1} \right\vert$	(II) $\dfrac{\left\vert B \right\vert}{\left\vert A \right\vert}$
(C) $\left\vert A^{-1}B^{-1} \right\vert$	(III) $\left\vert A \right\vert \left\vert B \right\vert$
(D) $\left\vert AB \right\vert$	(IV) $\dfrac{\left\vert A \right\vert}{\left\vert B \right\vert}$

Choose the correct answer from the options given below:

2025: 30 May Shift 1

If $A = \begin{bmatrix} 0 & 0 & \sqrt{7} \\ 0 & \sqrt{7} & 0 \\ \sqrt{7} & 0 & 0 \end{bmatrix}$ , then $|\text{adj } A|$ is equal to

2025: 30 May Shift 2

The diagonal elements of a skew symmetric matrix are all

2025: 15 May Shift 2

Let $A = [a_{ij}]$ be a square matrix, where $a_{ij} = \begin{cases} 0, & \text{when } i = j \\ 1, & \text{otherwise} \end{cases}$ . If |adj A| = |A|², then which of the following statements are correct?

(A) A is a skew symmetric matrix.

(B) A is a non-singular matrix.

(D) A is a symmetric matrix.

Choose the correct answer from the options given below:

2026: 18 May Shift 2

If $\Delta$ represents the value of the determinant $\begin{vmatrix} 1 & \sin k\theta & 1 \\ -\sin k\theta & 1 & \sin k\theta \\ -1 & -\sin k\theta & 1 \end{vmatrix}$ , where $k$ is any real number then the value of $\Delta$ belongs to

2022: 16 July Shift 2

The values of $x$ and $y$ in the equation $2\begin{bmatrix} x & 1 \\ 4 & -3 \end{bmatrix} + 3\begin{bmatrix} -2 & 1 \\ 2 & y-2 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 14 & 3 \end{bmatrix}$ are respectively:

Solution

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