Transpose of a Matrix - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 3 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’

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)’=\(\begin{bmatrix}-6&-21&-24\\8&28&32\\2&7&8\end{bmatrix}\)

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’=\(\begin{bmatrix}4&6\\1&2\end{bmatrix}\)\(\begin{bmatrix}-1&2&6\\3&1&6\end{bmatrix}\)=\(\begin{bmatrix}14&14&60\\5&4&18\end{bmatrix}\).

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’.

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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