Matrices- 1 - JEE Maths Free MCQ Test with solutions

2^3x3 = 2⁹ = 512.

The number of elements in a 3 X 3 matrix is the product 3 X 3=9.

Each element can either be a 0 or a 1.

Given this, the total possible matrices that can be selected is 2⁹=512

In linear algebra, the identity matrix, or sometimes ambiguously called a unit matrix, of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. It is denoted by I_n, or simply by I if the size is immaterial or can be trivially determined by the context

The product of any matrix with itself can be found only when it is a square matrix.i.e. m = n.

If A and B are square matrices of same order , then , product of the matrices is not commutative.Therefore , the given result is true only when AB = BA.

Coefficient matrix

For no solution, |A| = 0 and the augmented matrix rank > rank(A).

Set |A| = 0 → λ – 2 = 0 → λ = 2

Check inconsistency:
If λ = 2, equations 2 and 3 become:

x + 2y + 2z = 3
x + 2y + 2z = 4 → contradiction

Hence no solution when λ = 2.

, therefore matrix A has 4 elements in each row

Let matrix A is of order m x n , and matrix B is of order p x q . then , the product AB is defined only when n = p. that’s why, If A and B are any two matrices, then AB may or may not be defined.

Adj.(KA) = Kⁿ⁻¹ Adj.A , where K is a scalar and A is a n x n matrix.

Option C is correct: x = ±6.

Determinant on the left: (2x)(x) - 5 × 8 = 2x² - 40.

Determinant on the right: 6 × 3 - (-2) × 7 = 18 + 14 = 32.

Equate the determinants: 2x² - 40 = 32.

Add 40 to both sides: 2x² = 72.

Divide by 2: x² = 36.

Taking square roots: x = ±6. Therefore the correct option is C.

Here, matrix P is of order 2 × 3 and matrix Q is of order 2 × 2 ,
then , the product PQ is defined only when : no. of columns in P = no. of rows in Q.
And the order of resulting matrix is given by : rows in P x columns in Q. = 2 x 2

A lower triangular matrix is given by : 
 ,

here , a_ij = 0 if i is less than j and

aij ≠ 0 if i is greater than j.

Write the system as AX = B, where

Check consistency using rank:

Since

the system is inconsistent.

Therefore, no solution.

Option C is correct.

Let A = [cosθ - sinθ; sinθ cosθ].

Then A^T = [cosθ sinθ; -sinθ cosθ].

So A^T + A = [2 cosθ 0; 0 2 cosθ] = 2 cosθ I₂.

Given A^T + A = I₂, we get 2 cosθ I₂ = I₂, hence 2 cosθ = 1 and cosθ = 1/2.

All solutions of cosθ = 1/2 are θ = 2nπ ± π/3, n ∈ Z. Therefore option C is the correct choice.

To determine the number of possible 2×2 matrices where each entry can be 0, 1, or 2, we calculate the total number of combinations for each of the four independent entries.

• Each entry has 3 choices.
• There are 4 independent entries in a 2×2 matrix.
• The total number of combinations is calculated as 3⁴.
• Therefore, the total number of possible matrices is 81.

Since A^k = 0, the matrix A is nilpotent.
For a nilpotent matrix, the matrix geometric series holds:

Given in the question:

So, 

Comparing powers, we get:

p = -1

Upper Triangular matrix is given by  :
.
 Here, a_ij=0 , if i is greater than j.and a_ij ≠ 0, if I is less than j.

Let A and B be two matrices such that AB = A and BA = B Now, consider AB = A Take Transpose on both side (AB)^T = A^T 
⇒ A^T = B^T ⋅ A^T ...(1)
Now, BA = B 
Take, Transpose on both side (BA)^T = B^T 
⇒ B^T = A^T⋅B^T…(2)
Now, from equation (1) and (2). we have A^T = (A^T . B^T)A^T 
A^T=A^T(B^TA^T)
= A^T(AB)^T(∵(AB)^T = B^T = B^TA^T)
= A^T ⋅ A^T 
Thus, A^T = (A^T)²

For any square matrix A, both (A + A^T) is symmetric and (A - A^T) is skew-symmetric. Therefore, options A and B are true.

• (A + A^T) is symmetric.
• (A - A^T) is skew-symmetric.

To check the nature of the solutions, compute the determinant of the coefficient matrix:

Since |A| = 0, the system is either inconsistent or has infinitely many solutions.

Now check consistency:

Which is false → contradiction.

So the system is inconsistent.

(In general) and in a diagonal matrix non-diagonal elements

Given System:

Coefficient matrix:

For a non-zero solution,

∣A∣ = 0

Now, we will find the determinant:

So, option (a) is the correct answer.

We have,

and