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

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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

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

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
View Answer

Answer: a
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

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
\(∴ ρ(A)= 3.\)
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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

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) <r
View Answer

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) \(\frac{1}{6} \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 1\\
0 & -2 & 4\\
\end{bmatrix} \)
View Answer

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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn