Mathematics Questions and Answers – Invertible Matrices

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Invertible Matrices”.

1. Which among the following is inverse of the matrix A=\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\) ?
a) \(\begin{bmatrix}\frac{1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)
b) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)
c) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\1&\frac{-2}{13}\end{bmatrix}\)
d) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&-2\end{bmatrix}\)
View Answer

Answer: b
Explanation: Consider the matrix A=\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\)
Using elementary row operation, we write A=IA.
\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\)=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)A
\(\begin{bmatrix}-13&0\\5&1\end{bmatrix}\)=\(\begin{bmatrix}1&-3\\0&1\end{bmatrix}\)A   (Applying R1→R1-3R2)
\(\begin{bmatrix}1&0\\5&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\0&1\end{bmatrix}\)A   (Applying \(R_1 \rightarrow -\frac{R_1}{13}\))
\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)A   (Applying R2→R2-5R1)
\(A^{-1}=\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\).
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2. Which of the following matrices will not have an inverse?
a) \(\begin{bmatrix}2&4\\-1&1\end{bmatrix}\)
b) \(\begin{bmatrix}1&5&2\\6&4&2\\1&3&2\end{bmatrix}\)
c) \(\begin{bmatrix}1&2\\1&1\end{bmatrix}\)
d) \(\begin{bmatrix}1&2&5\\3&6&4\end{bmatrix}\)
View Answer

Answer: d
Explanation: The matrix A=\(\begin{bmatrix}1&2&5\\3&6&4\end{bmatrix}\) will not have an inverse as it is a rectangular matrix. Rectangular matrix does not possess an inverse matrix.

3. If A and B are invertible matrices of the same order, then (AB)-1=B-1 A-1.
a) True
b) False
View Answer

Answer: a
Explanation: The given statement is true.
(AB) (AB)-1=I (Using the formula AA-1=I)
Multiplying both sides by A-1, we get
A-1 (AB) (AB)-1=A-1 I
(A-1 A)B(AB)-1=A-1
IB(AB-1)=A-1
B(AB-1)=A-1
⇒B-1 B(AB-1)=B-1 A-1
(AB-1)=B-1 A-1
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4. A matrix A is invertible if it has all zeroes in one or more rows on L.H.S.
a) True
b) False
View Answer

Answer: b
Explanation: The given statement is false. A matrix is non-invertible if it has all zeroes in one or more rows on L.H.S. This is because after applying all the elementary operations on the matrix, we should get an identity matrix on the L.H.S. to obtain an inverse of the given matrix, which is not possible if we obtain all zeroes in one or more rows.

5. The inverse of the matrix A=\(\begin{bmatrix}1&2&4\\5&2&4\\3&6&2\end{bmatrix}\) is
a) \(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&1&\frac{-1}{10}\end{bmatrix}\)
b) \(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&1\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}\)
c) \(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}\)
d) \(\begin{bmatrix}\frac{-1}{4}&-\frac{1}{4}&0\\ \frac{1}{40}&\frac{1}{8}&\frac{-1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}\)
View Answer

Answer: c
Explanation: Consider the matrix A=\(\begin{bmatrix}1&2&4\\5&2&4\\3&6&2\end{bmatrix}\)
Using the elementary row operation, we write A=IA
\(\begin{bmatrix}1&2&4\\5&2&4\\3&6&2\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)A
Applying R1→R1-R2
\(R_1 \rightarrow \frac{R_1}{-4}\)
\(\begin{bmatrix}1&0&0\\5&2&4\\3&6&2\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\0&1&0\\0&0&1\end{bmatrix}\)A
Applying R2→R2-5R1 and R3→R3-3R1
\(\begin{bmatrix}1&0&0\\0&2&4\\0&6&2\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\\frac{5}{4}&\frac{-1}{4}&0\\\frac{3}{4}&\frac{-3}{4}&1\end{bmatrix}\)A
Applying R2→R2-2R3 and \(R_2 \rightarrow \frac{R_2}{-10}\)
\(\begin{bmatrix}1&0&0\\0&1&0\\0&6&2\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-5}{40}&\frac{1}{5}\\\frac{3}{4}&\frac{-3}{4}&1\end{bmatrix}\)A
Applying R3→R3-6R2 and \(R_2 \rightarrow \frac{R_2}{2}\)
\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}\)A
A-1=\(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}\).
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6. Which of the following is the inverse of the matrix A=\(\begin{bmatrix}8&1\\1&2\end{bmatrix}\)?
a) \(\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\\frac{1}{15}&\frac{8}{15}\end{bmatrix}\)
b) \(\begin{bmatrix}\frac{1}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{1}{15}\end{bmatrix}\)
c) \(\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}\)
d) \(\begin{bmatrix}\frac{2}{15}&\frac{1}{15}\\\frac{1}{15}&\frac{4}{15}\end{bmatrix}\)
View Answer

Answer: c
Explanation: Consider the matrix A=\(\begin{bmatrix}8&1\\1&2\end{bmatrix}\)
Using the elementary row operation, we write A=IA
Applying R2→8R2-R1 and R2→R2/15, we get
\(\begin{bmatrix}8&1\\0&1\end{bmatrix}\)=\(\begin{bmatrix}1&0\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}\)A
Applying R1→R1-R2 and R1→R1/8, we get
\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}\)A
A-1=\(\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}\).

7. Which among the below matrices has the inverse A-1=\(\begin{bmatrix}1&-\frac{5}{8}\\0&\frac{1}{8}\end{bmatrix}\)
a) \(\begin{bmatrix}1&5\\0&8\end{bmatrix}\)
b) \(\begin{bmatrix}1&5\\-1&8\end{bmatrix}\)
c) \(\begin{bmatrix}1&5\\0&16\end{bmatrix}\)
d) \(\begin{bmatrix}1&8\\0&8\end{bmatrix}\)
View Answer

Answer: a
Explanation: Consider the matrix A=\(\begin{bmatrix}1&5\\0&8\end{bmatrix}\)
Using the elementary column operations, we write A=AI
\(\begin{bmatrix}1&5\\0&8\end{bmatrix}\)=A\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
Applying C2→C2-5C1
\(\begin{bmatrix}1&0\\0&8\end{bmatrix}\)=A\(\begin{bmatrix}1&-5\\0&1\end{bmatrix}\)
Applying C2→C2/8
\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=A\(\begin{bmatrix}1&-\frac{5}{8}\\0&\frac{1}{8}\end{bmatrix}\)
A-1=\(\begin{bmatrix}1&-\frac{5}{8}\\0&\frac{1}{8}\end{bmatrix}\).
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8. Find the inverse of matrix A=\(\begin{bmatrix}5&1&3\\4&2&6\\5&4&2\end{bmatrix}\)
a) \(\begin{bmatrix}0&-\frac{1}{6}&0\\-\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}\)
b) \(\begin{bmatrix}\frac{1}{3}&-\frac{1}{6}&0\\-\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}\)
c) \(\begin{bmatrix}\frac{1}{3}&-\frac{1}{6}&0\\-\frac{11}{30}&1&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}\)
d) \(\begin{bmatrix}\frac{1}{3}&\frac{1}{6}&0\\\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&\frac{1}{10}\end{bmatrix}\)
View Answer

Answer: b
Explanation: Consider the matrix A=\(\begin{bmatrix}5&1&3\\4&2&6\\5&4&2\end{bmatrix}\)
Using the elementary row operations, we write A=IA
\(\begin{bmatrix}5&1&3\\4&2&6\\5&4&2\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)A
Applying R2→5R2-4R1 and R3→R3-R1
\(\begin{bmatrix}5&1&3\\0&6&18\\0&3&-1\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\-4&5&0\\-1&0&1\end{bmatrix}\)A
Applying R1→R2-6R1 and R3→R2-2R3
\(\begin{bmatrix}-30&0&0\\0&6&18\\0&0&20\end{bmatrix}\)=\(\begin{bmatrix}-10&5&0\\-4&5&0\\-2&5&-2\end{bmatrix}\)A
Applying R2→20R2-18R3
\(\begin{bmatrix}-30&0&0\\0&120&0\\0&0&20\end{bmatrix}\)=\(\begin{bmatrix}-10&5&0\\-44&10&36\\-2&5&-2\end{bmatrix}\)A
Applying R1→R1/(-30), R2→R2/120, R3→R3/20
\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{1}{3}&-\frac{1}{6}&0\\-\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}\)A
A-1=\(\begin{bmatrix}\frac{1}{3}&-\frac{1}{6}&0\\-\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}\).

9. Which of the following is not a property of invertible matrices if A and B are matrices of the same order?
a) (AB)-1=A-1 B-1
b) (AA-1)=(A-1 A)=I
c) (AB)-1=B-1 A-1
d) AB=BA=I
View Answer

Answer: a
Explanation: (AB)-1=A-1 B-1 is incorrect. The correct formula is (AB)-1=B-1 A-1. B-1 A-1 ≠ A-1 B-1 as matrix multiplication is not commutative.
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10. Find the inverse of A=\(\begin{bmatrix}5&3\\4&1\end{bmatrix}\).
a) \(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\)
b) \(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&\frac{5}{7}\end{bmatrix}\)
c) \(\begin{bmatrix}-\frac{1}{7}&-\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\)
d) \(\begin{bmatrix}0&\frac{3}{7}\\\frac{4}{7}&\frac{5}{7}\end{bmatrix}\)
View Answer

Answer: a
Explanation: Consider the matrix A=\(\begin{bmatrix}5&3\\4&1\end{bmatrix}\)
By using the elementary row operations, we write A=IA
\(\begin{bmatrix}5&3\\4&1\end{bmatrix}\)=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)A
Applying R1→R1-3R2 and R1→R1/(-7), we get
\(\begin{bmatrix}1&0\\4&1\end{bmatrix}\)=\(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\0&1\end{bmatrix}\)
Applying R2→R2-4R1, we get
\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=\(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\)A
⇒A-1=\(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\).

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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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 | Youtube | Instagram | Facebook | Twitter