# Linear Algebra Questions and Answers – Finding Inverse and Rank of Matrix

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

1. If A (α, β) = $$\begin{bmatrix} cos\alpha & sin\alpha & 0\\ -sin\alpha & cos\alpha & 0\\ 0 & 0 & e^{\beta}\\ \end{bmatrix}$$ then A(α, β)-1 is equal to?
a) A (-α, β)
b) A (-α, -β)
c) A (α, -β)
d) A (α, β)

Explanation: Here $$|A| = cosα(e^β cosα) – sinα(e^β.-sinα) = e^β$$
$$And, adjA = \begin{bmatrix} e^{\beta} cos\alpha & e^{\beta} sin\alpha & 0\\ -e^{\beta} sin\alpha & e^{\beta} cos\alpha & 0\\ 0 & 0 & 1\\ \end{bmatrix}$$
$$∴ A(α,β)^{-1} = \frac{1}{e^β} \begin{bmatrix} e^{\beta} cos\alpha & e^{\beta} sin\alpha & 0\\ -e^{\beta} sin\alpha & e^{\beta} cos\alpha & 0\\ 0 & 0 & 1\\ \end{bmatrix}$$
$$= A (-α,-β).$$

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) $$\begin{bmatrix} a+ib & c+id\\ -c+id & a-ib\\ \end{bmatrix}$$

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

Explanation: Let A be an invertible matrix.
We have AA-1 = A-1A = IN
(AA-1)’ = (A-1A)’ = (IN)’
(A-1)’ A’ = A’ (A-1)’ = IN
(A-1)’ A = A (A-1)’ = IN
(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

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

Explanation: $$|A| = 3(10-24) + 2(42-20) – (16-14) = 10$$
$$∵ |A| = 0$$
$$~ \begin{bmatrix} 3 & 2\\ 4 & 2\\ \end{bmatrix} ≠0 = -2$$
$$∴ ρ (A) = 2.$$
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6. If every minor of order ‘r’ of a matrix is zero then ρ (A) =?
a) >r
b) =r
c) ≤r
d) <r

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) $$\frac{1}{6} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & -2 & 4\\ \end{bmatrix}$$

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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