Test: Matrices - JAMB MCQ

Given that, A is a square matrix and A² = A.
Consider (I + A)³, where I is the identity matrix.
Using the identity of (a + b)³ = a³ + b³ + 3ab (a + b), we get;
(I + A)³ = I3 + A³ + 3A²I + 3AI²
= I + A²(A) + 3AI + 3A
= I + A² + 3A + 3A
= 7A + I {since it is given that A² = A}
So, (I + A)³ = 7A + I….(1)
Now,
(I + A)³ – 7A = 7A + I – 7A [From (1)]
= I

We know that a matrix 3 × 3 contains 9 elements.
Given that each entry of this 3 × 3 matrix is either 0 or 2.
Thus, by simple counting principle, we can calculate the total number of possible matrices as:
Total number of possible matrices = Total number of ways in which 9 elements can take possible values
= 2⁹
= 512

Given,

Now, by equating the corresponding elements of these two matrices, we get;
2p + q = 4….(1)
p – 2q = -3….(2)
5r – s = 11….(3)
4r + 3s = 24….(4)
By equ (1) × 2 + equ (2), we get;
4p + 2q + p – 2q = 8 – 3
5p = 5
p = 1
Substituting p = 1 in (1),
2 + q = 4
q = 4 – 2 = 2
By equ (3) × 3 + equ (4), we get;
15r – 3s + 4r + 3s = 33 + 24
19r = 57
r = 3
Substituting r = 3 in (3),
15 – s = 11
s = 15 – 11 = 4
Now,
p + q – r + 2s = 1 + 2 – 3 + 2(4) = 8

For any two matrices A and B,
AB = BA and AB ≠ BA are not valid unless they follow the condition of matrix multiplication.
Also, AB = O is not true in all cases.

Given that A is a skew-symmetric matrix, so A′ = -A.
Consider the transpose of A².
(A²)′ = (AA)′
= A′A′
=(-A)(-A)
= A²
⇒ (A²)′ = A²
Therefore, A² is a symmetric matrix.

Given,
A = [a_ij] is a square matrix of order 2 such that a_ij = 1, when i ≠ j and a_ij = 0, when i = j.
So, a₁₁ = 0, a₁₂ = 1, a₂₁ = 1 and a₂₂ = 0.
Thus,

Given that, the order of matrix A is 3 × m, and the order of B is 3 × n.
Also, m = n.
So, the order of matrix A and B is the same, i.e. 3 × m.
Thus, subtraction of matrices is possible and (5A – 3B) also has the same order, i.e. 3 × n.

Let the given matrix be:

Let us find the transpose of A.

Therefore, A is a symmetric matrix.

Given that A and B are symmetric matrices of the same order.
Let’s find the transpose of (AB′ –BA′).
(AB′ –BA′)′ = (AB′)′ – (BA′)′
= (BA′ – AB′)
= – (AB′ –BA′)
As (AB′ –BA′)′ = – (AB′ –BA′), the matrix (AB′ –BA′) is skew symmetric.

Given,

By equating the corresponding elements,
-4k = 24
k = -6
Also, 2k = 3a
2(-6) = 3a
3a = -12
a = -4
And
3k = 2b
3(-6) = 2b
2b = -18
b = -9
Therefore, k = -6, a = -4, and b = -9.