Test: Matrix Multiplication (May 4) - JEE MCQ

Multiply and find the pattern.
 
Observing the pattern,

Hence option 2 is correct.

Since all the entries of the matric are 1 so when it will be multiplied by itself then we will get the pattern in the entries of the resultant matrices.
Given:




Referring to the pattern formed,
∴ ��⁵⁰ = 3⁴⁹×��.
So, the correct answer is option 1.

Given:

This is only possible when a = b
So, B should be of the form 
As a ∈ N so there are infinitely many B's.
So, the correct answer is option 4.

Concept:
If  are two matrices. then

Given:

2x² - 9x + 32x = 0
2x² + 23x = 0
x(2x + 23) = 0
⇒ x = 0 or x = - (23/2)

The number of solutions can be determined by finding out the rank of the Augmented matrix and the rank of the Coefficient matrix.

• If rank(Augmented matrix) = rank(Coefficient matrix) = no. of variables then no of solutions = 1.
• If rank(Augmented matrix)  ≠ rank(Coefficient matrix) then no of solutions = 0.
• If rank(Augmented matrix) = rank(Coefficient matrix) < no. of variables, no of solutions = infinite.

The augmented matrix for the system of equations is

If λ = 6 and μ ≠ 20 then
Rank (A | B) = 3 and Rank (A) = 2
∵ Rank (A | B) ≠ Rank (A)
∴ Given the system of equations has no solution for λ = 6 and μ ≠ 20

Two matrices A_{m × n} and B_{p × q} 
if AB and BA are defined then p = n and q = m
Given:
AB and BA are defined.
so the order of the matrix B is B_{n × m}

The number of solutions can be determined by finding out the rank of the Augmented matrix and the rank of the Coefficient matrix.

• If rank(Augmented matrix) = rank(Coefficient matrix) = no. of variables then no of solutions = 1.
• If rank(Augmented matrix)  ≠ rank(Coefficient matrix) then no of solutions = 0.
• If rank(Augmented matrix) = rank(Coefficient matrix) < no. of variables, no of solutions = infinite.

The augmented matrix for the system of equations is

Performing: R₃ → R₃ – R₂

If λ = 6 and μ ≠ 20 then
Rank (A | B) = 3 and Rank (A) = 2
∵ Rank (A | B) ≠ Rank (A)
∴ Given the system of equations has no solution for λ = 6 and μ ≠ 20

The product of two matrices A and B is defined if the number of columns of matrix A is equal to the number of rows of matrix B.

Given: 
The order of the matrix A is 2 × 3 whereas the order of the matrix B is 3 × 3,
Then Product of matrix A and matrix B while be of the order 2 × 3

If  is a square matrix of order 2, then determinant of A is given by:  |A| = (a­₁₁ × a₂₂) – (a₁₂ – a₂₁).

Since P and Q are upper triangular matrix, therefore determinant of these matrices are the product of their diagonal elements.
det(��) = 6×1×2=12 
det(��) = 2×5×5=50 
The product of determinant P and Q,
det(P)×det(Q) = 12×50 = 600 
∴ The product of determinant P and Q is 600