250+ TOP MCQs on Finding Inverse and Rank of Matrix and Answers

Linear Algebra Interview Questions and Answers focuses on “Finding Inverse and Rank of Matrix”.

1. If A (α, β) = (begin{bmatrix}
cosalpha & sinalpha & 0\
-sinalpha & cosalpha & 0\
0 & 0 & e^{beta}\
end{bmatrix}) then A(α, β)^-1 is equal to?
a) A (-α, β)
b) A (-α, -β)
c) A (α, -β)
d) A (α, β)
Answer: b
Explanation: Here (|A| = cosα(e^β cosα) – sinα(e^β.-sinα) = e^β )
(And, adjA = begin{bmatrix}
e^{beta} cosalpha & e^{beta} sinalpha & 0\
-e^{beta} sinalpha & e^{beta} cosalpha & 0\
0 & 0 & 1\
end{bmatrix})
( ∴ A(α,β)^{-1} = frac{1}{e^β} begin{bmatrix}
e^{beta} cosalpha & e^{beta} sinalpha & 0\
-e^{beta} sinalpha & e^{beta} cosalpha & 0\
0 & 0 & 1\
end{bmatrix})
( = A (-α,-β).)

2. If (A = begin{bmatrix}
a+ib & c+id\
-c+id & a-ib\
end{bmatrix} and , a^2+b^2+c^2+d^2=1) then find A^-1.
a) (begin{bmatrix}
a+ib & -c+id\
-a+id & a-ib\
end{bmatrix} )
b) (begin{bmatrix}
a-ib & c-id\
-c-id & a+ib\
end{bmatrix} )
c) (begin{bmatrix}
a-ib & -c-id\
c-id & a+ib\
end{bmatrix} )
d) (begin{bmatrix}
a+ib & c+id\
-c+id & a-ib\
end{bmatrix} )
Answer: c
Explanation: We have
(|A| = (a+ib) (a-ib) – (-c+id) (c+id) )
( = a^2+b^2+c^2+d^2 )
(s=1 )
Also (adj (A) = begin{bmatrix}
a-ib & -c-id\
c-id & a+ib\
end{bmatrix} )
∴ A^-1 = (frac{1}{|A|} .adj (A) = frac{1}{1} begin{bmatrix}
a-ib & -c-id\
c-id & a+ib\
end{bmatrix} )
(begin{bmatrix}
a-ib & -c-id\
c-id & a+ib\
end{bmatrix} ).

3. The inverse of a symmetric matrix (if it exists) is?
a) A symmetric matrix
b) A skew symmetric matrix
c) A diagonal matrix
d) A triangular matrix
Answer: a
Explanation: Let A be an invertible matrix.
We have AA^-1 = A^-1A = I_N
(AA^-1)’ = (A^-1A)’ = (I_N)’
(A^-1)’ A’ = A’ (A^-1)’ = I_N
(A^-1)’ A = A (A^-1)’ = I_N
(A^-1)’ = A^-1 (inverse of a matrix is unique).

4. Find the rank of the matrix (begin{bmatrix}
4 & 2 & -1 & 2\
1 & -1 & 2 & 1\
2 & 2 & -2 & 0\
end{bmatrix} )?
a) 0
b) 1
c) 2
d) 3
Answer: d
Explanation: Say,
A = (begin{bmatrix}
4 & 2 & -1 & 2\
1 & -1 & 2 & 1\
2 & 2 & -2 & 0\
end{bmatrix} )
( ~ begin{bmatrix}
1 & -1 & 2 & 1\
4 & 2 & -1 & 2\
2 & 2 & -2 & 0\
end{bmatrix} )R₂↔R₁
( ~ begin{bmatrix}
1 & -1 & 2 & 1\
0 & 6 & -9 & -2\
0 & 4 & -6 & -2\
end{bmatrix}) R₂-4R₁, R₃-2R₁
( ~ begin{bmatrix}
1 & -1 & 2 & 1\
0 & 6 & -9 & -2\
0 & 0 & 0 & -4\
end{bmatrix}) 6R₃-4R₂
(∴ ρ(A)= 3.)

5. Rank of the following matrix is A = (begin{bmatrix}
3 & 2 & -1\
4 & 2 & 6\
7 & 4 & 5\
end{bmatrix} ) is?
a) 1
b) 2
c) 3
d) 0
Answer: b
Explanation: (|A| = 3(10-24) + 2(42-20) – (16-14) = 10)
(∵ |A| = 0)
( ~ begin{bmatrix}
3 & 2\
4 & 2\
end{bmatrix} ≠0 = -2 )
(∴ ρ (A) = 2.)

6. If every minor of order ‘r’ of a matrix is zero then ρ (A) =?
a) >r
b) =r
c) ≤r
d) Answer: d
Explanation: By the definition of ‘Rank of a matrix’
A matrix is said to have rank ‘r’ if
(i) At least one minor of order r is non-zero
(ii) All minors of order r+1 is zero
∴ The given matrix (ii) condition seems to be applied
Hence, rank of matrix ρ (A) = < r.

7. Find the inverse of the matrix by using Cayley Hamilton Theorem.
A = ( ~ begin{bmatrix}
1 & 0 & 0\
0 & 1 & 1\
0 & -2 & 4\
end{bmatrix} )
a) (frac{1}{6} begin{bmatrix}
1 & 0 & 0\
0 & -1 & -1\
0 & 2 & -4\
end{bmatrix} )
b) (frac{1}{6} begin{bmatrix}
1 & 0 & 0\
0 & 1 & 1\
0 & 2 & 4\
end{bmatrix} )
c) (frac{1}{6} begin{bmatrix}
1 & 0 & 0\
0 & -1 & -1\
0 & -2 & -4\
end{bmatrix} )
d) (frac{1}{6} begin{bmatrix}
1 & 0 & 0\
0 & 1 & 1\
0 & -2 & 4\
end{bmatrix} )
Answer: a
Explanation: By Cayley Hamliton Theorem
We have λ³ – S₁ λ² + S₂ λ – |A| = 0 characteristics equation of matrix A
Also A satisfies above equation according to theorem.
∴ A³– S₁A²+ S₂A – 6 = 0……… (i) Where S₁=6, S₂=6 & |A|=6
∴ A³ – 6A² + 6A – 6 = 0
Multiplying equation (i) by A^-1
A^-1 (A³) – 6 A^-1 (A²) + A^-1 (A) – 6 A^-1 = 0
=> A² – 6A + 6I – 6 A^-1 = (frac{1}{6}) A^-1
=> A^-1 = (frac{1}{6} Bigg{begin{bmatrix}
1 & 0 & 0\
0 & -1 & 5\
0 & -10 & 14\
end{bmatrix} – begin{bmatrix}
6 & 0 & 0\
0 & 6 & 6\
0 & -12 & -24\
end{bmatrix} + begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 1\
end{bmatrix} Bigg} )
(⇨ A^{-1} = frac{1}{6} begin{bmatrix}
1 & 0 & 0\
0 & -1 & -1\
0 & 2 & -4\
end{bmatrix} ).

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