# Class 12 Maths MCQ – Transpose of a Matrix

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’

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=$$\begin{bmatrix}1&-1\\-2&5\end{bmatrix}$$

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

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) $$\begin{bmatrix}26&20\\12&10\end{bmatrix}$$

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) $$\begin{bmatrix}1&0&0\\1&1&0\\1&1&1\end{bmatrix}$$

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) $$\begin{bmatrix}1&4&7\\-2&5&2\\1&8&9\end{bmatrix}$$

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) $$\begin{bmatrix}14&14&18\\5&4&60\end{bmatrix}$$

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) $$\begin{bmatrix}10&22\\21&34\end{bmatrix}$$

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’

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’=$$\begin{bmatrix}0&1\\-1&0\end{bmatrix}$$

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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