Algebra

Matrices & Determinants

Medium

core

If $A$ is a square matrix and $I$ is an identity matrix of same order such that $A^2 = A$ , then $(2I + A)^2 - 5A$ is

$0$

$4I$

$5A$

$2I$

✅ Correct Option: 2

Given: $A$ is a square matrix, $I$ is an identity matrix of the same order, and $A^2 = A$ .

Expand $(2I + A)^2$ :

$(2I + A)^2 = (2I + A)(2I + A)$

$= 2I \cdot 2I + 2I \cdot A + A \cdot 2I + A \cdot A$

$= 4I + 2A + 2A + A^2$

$= 4I + 4A + A^2$

Using $A^2 = A$ :

$= 4I + 4A + A$

$= 4I + 5A$

Calculate the final expression:

$(2I + A)^2 - 5A = 4I + 5A - 5A$

$= 4I$

Final Answer: $4I$

2026: 22 May Shift 1

If $A=\begin{bmatrix}1 & 1 & 1\\1 & 0 & 3\\1 & -2 & 1\end{bmatrix}$ , then

Match List-I with List-II

List-I	List-II
(A) $\lvert A\rvert$	(I) 48
(B) $\lvert 2A\rvert$	(II) 36
(C) $\left\lvert\operatorname{adj}(A)\right\rvert$	(III) 12
(D) $2\lvert A\rvert$	(IV) 6

Choose the correct answer from the options given below:

2022: 25 May Shift 1

If the matrix $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$ , then $A^2$ is equal to:

2025: 14 May Shift 2

If the matrix $\begin{bmatrix} 0 & 1 & 4x\ \\ -1 & 0 & -5 \\ 2 & 5 & y \end{bmatrix}$ is skew-symmetric, then

2025: 14 May Shift 2

The number of all possible matrices of order $2 \times 3$ with entries -1 or 1 is

2026: 18 May Shift 2

The value of $\lambda$ for which the following system of equations has no solution is

$x+y+2z=4$

$2x+y+3z=5$

$x+3y+\lambda z=7$

2023: 22 May Shift 3

If $P = \begin{bmatrix} 1 & x & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$ is the adjoint of 3x3 matrix A and $|A|$ is 4, then $x$ is equal to :

If AAA is a square matrix and III is an identity matrix of same order such that A2=AA^2 = AA2=A, then (2I+A)2−5A(2I + A)^2 - 5A(2I+A)2−5A is