Mathematics Questions and Answers – Transpose of a Matrix

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Transpose of a Matrix”.

1. Which of the following is not the property of transpose of a matrix?
a) (A’)’=A
b) (A+B)’=A’+B’
c) (AB)’=(BA)’
d) (kA)’=KA’
View Answer

Answer: c
Explanation: (AB)’=(BA)’is incorrect. We know that matrix multiplication is not commutative i.e. AB≠BA. Hence, its transpose will also not be commutative.
(AB)’=B’A’
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2. Find the transpose of A=\(\begin{bmatrix}1&-2\\-1&5\end{bmatrix}\).
a) A=\(\begin{bmatrix}-1&-2\\-1&-5\end{bmatrix}\)
b) A=\(\begin{bmatrix}1&2\\1&5\end{bmatrix}\)
c) A=\(\begin{bmatrix}-1&2\\-1&5\end{bmatrix}\)
d) A=\(\begin{bmatrix}1&-1\\-2&5\end{bmatrix}\)
View Answer

Answer: d
Explanation: A=\(\begin{bmatrix}1&-2\\-1&5\end{bmatrix}\). To find the transpose of the matrix, interchange the rows with columns and columns with rows.
Hence, A’=\(\begin{bmatrix}1&-1\\-2&5\end{bmatrix}\).

3. If A=\(\begin{bmatrix}2\\7\\8\end{bmatrix}\), B=\(\begin{bmatrix}-3&4&1\end{bmatrix}\), find (AB)’.
a) (AB)’=\(\begin{bmatrix}-6&-21&-24\\8&28&32\\2&7&8\end{bmatrix}\)
b) (AB)’=\(\begin{bmatrix}-6&8&2\\-21&-28&7\\-24&32&8\end{bmatrix}\)
c) (AB)’=\(\begin{bmatrix}6&21&24\\-8&28&7\\-2&7&-8\end{bmatrix}\)
d) (AB)’=\(\begin{bmatrix}-6&8&-21\\8&2&7\\-24&8&2\end{bmatrix}\)
View Answer

Answer: a
Explanation: Given that, A=\(\begin{bmatrix}2\\7\\8\end{bmatrix}\), B=\(\begin{bmatrix}-3&4&1\end{bmatrix}\)
AB=\(\begin{bmatrix}2\\7\\8\end{bmatrix}\)\(\begin{bmatrix}-3&4&1\end{bmatrix}\)=\(\begin{bmatrix}-6&8&2\\-21&28&7\\-24&32&8\end{bmatrix}\)
∴To find (AB)’, interchange the rows with columns and columns with rows of the matrix AB
(AB)’=\(\begin{bmatrix}-6&-21&-24\\8&28&32\\2&7&8\end{bmatrix}\)
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4. If A’=\(\begin{bmatrix}8&2\\6&4\end{bmatrix}\) and B’=\(\begin{bmatrix}9&5\\7&3\end{bmatrix}\). Find (A+2B)’.
a) \(\begin{bmatrix}26&20\\10&12\end{bmatrix}\)
b) \(\begin{bmatrix}26&12\\20&10\end{bmatrix}\)
c) \(\begin{bmatrix}26&10\\20&12\end{bmatrix}\)
d) \(\begin{bmatrix}26&20\\12&10\end{bmatrix}\)
View Answer

Answer: b
Explanation: Given that A’=\(\begin{bmatrix}8&2\\6&4\end{bmatrix}\) and B’=\(\begin{bmatrix}9&5\\7&3\end{bmatrix}\)
Calculating the transpose of A’ and B’, we get
A=\(\begin{bmatrix}8&6\\2&4\end{bmatrix}\) and B=\(\begin{bmatrix}9&7\\5&3\end{bmatrix}\)
⇒(A+2B)=\(\begin{bmatrix}8&6\\2&4\end{bmatrix}\)+2\(\begin{bmatrix}9&7\\5&3\end{bmatrix}\)
=\(\begin{bmatrix}8+18&6+14\\2+10&4+6\end{bmatrix}\)=\(\begin{bmatrix}26&20\\12&10\end{bmatrix}\)
Hence, (A+2B)’=\(\begin{bmatrix}26&12\\20&10\end{bmatrix}\).

5. If A=\(\begin{bmatrix}cos⁡x&-sin⁡x&-cos⁡x\\sin⁡x&-cos⁡x&sin⁡x \end{bmatrix}\). Find A’A.
a) \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
b) \(\begin{bmatrix}1&0&1\\1&0&1\\1&0&1\end{bmatrix}\)
c) \(\begin{bmatrix}1&0&1\\0&1&0\\1&0&1\end{bmatrix}\)
d) \(\begin{bmatrix}1&0&0\\1&1&0\\1&1&1\end{bmatrix}\)
View Answer

Answer: d
Explanation: Given that, A=\(\begin{bmatrix}cos⁡x&-sin⁡x&cos⁡x\\sin⁡x&cos⁡x&sin⁡x \end{bmatrix}\)
∴ A’=\(\begin{bmatrix}cos⁡x&sinx\\-sin⁡x&cos⁡x\\cos⁡x&sin⁡x \end{bmatrix}\)
⇒A’ A=\(\begin{bmatrix}cos⁡x&sin⁡x\\-sin⁡x&cos⁡x\\cos⁡x&sin⁡x \end{bmatrix}\)\(\begin{bmatrix}cos⁡ x &-sin⁡x&cos⁡x\\sin⁡x&cos⁡x &sin⁡x \end{bmatrix}\)
=\(\begin{bmatrix}cos^2⁡x+sin^2⁡x&-sin⁡x cos⁡x+sin⁡x cos⁡x&cos^2⁡x+sin^2⁡x\\-sin⁡x cos⁡x+sin⁡x cos⁡x&sin^2⁡x+cos^2⁡x&-sin⁡x cos⁡x+cos⁡x sin⁡x\\cos^2⁡x+sin^2 x&-cos⁡x sin⁡x+sin⁡x cos⁡x &cos^2⁡x+sin^2⁡x \end{bmatrix}\)
=\(\begin{bmatrix}1&0&1\\0&1&0\\1&0&1\end{bmatrix}\)
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6. Find the transpose of the matrix A=\(\begin{bmatrix}-1&2&\sqrt{3}\\-4&5&\sqrt{6}\\-7&8&-9\end{bmatrix}\)
a) \(\begin{bmatrix}1&-2&-\sqrt{3}\\4&-5&-\sqrt{6}\\7&-8&9\end{bmatrix}\)
b) \(\begin{bmatrix}-1&-4&-7\\2&5&8\\\sqrt{3}&\sqrt{6}&-9\end{bmatrix}\)
c) \(\begin{bmatrix}1&4&7\\-2&-5&-8\\-\sqrt{3}&-\sqrt{6}&9\end{bmatrix}\)
d) \(\begin{bmatrix}1&4&7\\-2&5&2\\1&8&9\end{bmatrix}\)
View Answer

Answer: b
Explanation: To find the transpose of the matrix of the given matrix, interchange the rows with columns and columns with rows.
Hence, we get A’=\(\begin{bmatrix}-1&-4&-7\\2&5&8\\\sqrt{3}&\sqrt{6}&-9\end{bmatrix}\)

7. If matrix A=\(\begin{bmatrix}4&1\\6&2\end{bmatrix}\) and B=\(\begin{bmatrix}-1&3\\2&1\\6&6\end{bmatrix}\), then find A’ B’.
a) \(\begin{bmatrix}14&14\\5&4\\6&18\end{bmatrix}\)
b) \(\begin{bmatrix}14&5\\14&4\\6&18\end{bmatrix}\)
c) \(\begin{bmatrix}14&14&60\\5&4&18\end{bmatrix}\)
d) \(\begin{bmatrix}14&14&18\\5&4&60\end{bmatrix}\)
View Answer

Answer: c
Explanation: Given that, A=\(\begin{bmatrix}4&1\\6&2\end{bmatrix}\) and B=\(\begin{bmatrix}-1&3\\2&1\\6&6\end{bmatrix}\)
⇒A’=\(\begin{bmatrix}4&6\\1&2\end{bmatrix}\) and B’=\(\begin{bmatrix}-1&2&6\\3&1&6\end{bmatrix}\)
⇒A’ B’=\(\begin{bmatrix}4&6\\1&2\end{bmatrix}\)\(\begin{bmatrix}-1&2&6\\3&1&6\end{bmatrix}\)=\(\begin{bmatrix}14&14&60\\5&4&18\end{bmatrix}\).
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8. If P=\(\begin{bmatrix}-1&5\\8&3\end{bmatrix}\) and Q=\(\begin{bmatrix}4&2\\8&5\end{bmatrix}\). Find (2P+3Q’)’.
a) \(\begin{bmatrix}10&22\\34&21\end{bmatrix}\)
b) \(\begin{bmatrix}10&21\\34&22\end{bmatrix}\)
c) \(\begin{bmatrix}10&34\\22&21\end{bmatrix}\)
d) \(\begin{bmatrix}10&22\\21&34\end{bmatrix}\)
View Answer

Answer: a
Explanation: Given that, P=\(\begin{bmatrix}-1&5\\8&3\end{bmatrix}\) and Q=\(\begin{bmatrix}4&2\\8&5\end{bmatrix}\)
⇒Q’=\(\begin{bmatrix}4&8\\2&5\end{bmatrix}\)
⇒2P+3Q’=2\(\begin{bmatrix}-1&5\\8&3\end{bmatrix}\)+3\(\begin{bmatrix}4&8\\2&5\end{bmatrix}\)=\(\begin{bmatrix}-2+12&10+24\\16+6&6+15\end{bmatrix}\)=\(\begin{bmatrix}10&34\\22&21\end{bmatrix}\)
∴(2P+3Q’ )’=\(\begin{bmatrix}10&22\\34&21\end{bmatrix}\).

9. Which of the following is the reversal law of transposes?
a) (A-B)’=B’-A’
b) (AB)’=B’A’
c) (AB)’=(BA)’
d) (A+B)’=B’+A’
View Answer

Answer: b
Explanation: According to the reverse law of transposes the transpose of the product is the product of the transposes taken in the reverse order i.e. (AB)’=B’ A’.
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10. If A=\(\begin{bmatrix}i&1\\0&i\end{bmatrix}\), then the correct relation is ___________
a) A+A’=\(\begin{bmatrix}1&0\\-1&0\end{bmatrix}\)
b) A-A’=\(\begin{bmatrix}1&0\\-1&0\end{bmatrix}\)
c) A+A’=\(\begin{bmatrix}0&1\\-1&0\end{bmatrix}\)
d) A-A’=\(\begin{bmatrix}0&1\\-1&0\end{bmatrix}\)
View Answer

Answer: d
Explanation: Given that, A=\(\begin{bmatrix}i&1\\0&i\end{bmatrix}\)
⇒A’=\(\begin{bmatrix}i&0\\1&i\end{bmatrix}\)
∴A-A’=\(\begin{bmatrix}i&1\\0&i\end{bmatrix}\)–\(\begin{bmatrix}i&0\\1&i\end{bmatrix}\)=\(\begin{bmatrix}i-i&1-0\\0-1&i-i\end{bmatrix}\)=\(\begin{bmatrix}0&1\\-1&0\end{bmatrix}\).

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

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Manish Bhojasia - Founder & CTO at Sanfoundry
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