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)
View Answer
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) \(\begin{bmatrix}
a+ib & c+id\\
-c+id & a-ib\\
\end{bmatrix} \)
View Answer
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
View Answer
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
View Answer
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
View Answer
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) <r
View Answer
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) \(\frac{1}{6} \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 1\\
0 & -2 & 4\\
\end{bmatrix} \)
View Answer
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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