The value of vmatrix1 & _ba\_ab & 1vmatrix is: Option 1: 1 Option 2: _ba Option 3: _ab Option 4: 0

Answer: 0. Solution: The determinant is 1-_ba_ab. Since _ba_ab=1 by the change-of-base rule, the value is 1-1=0.

Algebra

Matrices & Determinants

Easy

core

The value of $\begin{vmatrix}1 & \log_ba\\\log_ab & 1\end{vmatrix}$ is:

$\log_ba$

$\log_ab$

✅ Correct Option: 4

The determinant is $1-\log_ba\cdot\log_ab$ . Since $\log_ba\cdot\log_ab=1$ by the change-of-base rule, the value is $1-1=0$ .

2026: 29 May Shift 2

If $A+B = \begin{bmatrix}7 & 0\\2 & 5\end{bmatrix}$ and $A-B = \begin{bmatrix}3 & 0\\0 & 3\end{bmatrix}$ , then

2025: 14 May Shift 2

If $A = \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}$ and $I$ is an identity matrix of order 2, then $A - 3I$ equals

2026: 15 May Shift 1

If $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ x & y & -1 \end{bmatrix}$ , then $A^2$ is

2026: 18 May Shift 1

Match List-I with List-II

List-I	List-II
A is given matrix	$\left\vert \text{adj}(\text{adj}A) \right\vert$
(A) Order of A $= 2 \times 2$ and $\left\vert A \right\vert = 3$	(I) $3$
(B) Order of A $= 3 \times 3$ and $\left\vert A \right\vert = 2$	(II) $1$
(C) Order of A $= 4 \times 4$ and $\left\vert A \right\vert = 1$	(III) $16$
(D) Order of A $= 2 \times 2$ and $\left\vert A \right\vert = 4$	(IV) $4$

Choose the correct answer from the options given below:

2025: 29 May Shift 2

The inverse of the matrix $\begin{bmatrix} 4 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 6 \end{bmatrix}$ is

2026: 18 May Shift 2

If $A=\begin{bmatrix} 1 & 0 \\ -1 & 5 \end{bmatrix}$ and I is indentity matrix of order 2, then which of following statements are TRUE?

A. $A^2-6A-mI=0$ if $m=-5$ .

B. $A^2-6A+mI=0$ if $m=-5$ .

C. $A$ is a non-singular matrix.

D. $A^2=A$

Choose the correct answer from the options given below: