Mathematics Multiple Choice Questions 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’
Answer: c
Clarification: (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&5end{bmatrix}).
a) A=(begin{bmatrix}-1&-2\-1&-5end{bmatrix})
b) A=(begin{bmatrix}1&2\1&5end{bmatrix})
c) A=(begin{bmatrix}-1&2\-1&5end{bmatrix})
d) A=(begin{bmatrix}1&-1\-2&5end{bmatrix})
Answer: d
Clarification: A=(begin{bmatrix}1&-2\-1&5end{bmatrix}). To find the transpose of the matrix, interchange the rows with columns and columns with rows.
Hence, A’=(begin{bmatrix}1&-1\-2&5end{bmatrix}).
3. If A=(begin{bmatrix}2\7\8end{bmatrix}), B=(begin{bmatrix}-3&4&1end{bmatrix}), find (AB)’.
a) (AB)’=(begin{bmatrix}-6&-21&-24\8&28&32\2&7&8end{bmatrix})
b) (AB)’=(begin{bmatrix}-6&8&2\-21&-28&7\-24&32&8end{bmatrix})
c) (AB)’=(begin{bmatrix}6&21&24\-8&28&7\-2&7&-8end{bmatrix})
d) (AB)’=(begin{bmatrix}-6&8&-21\8&2&7\-24&8&2end{bmatrix})
Answer: a
Clarification: Given that, A=(begin{bmatrix}2\7\8end{bmatrix}), B=(begin{bmatrix}-3&4&1end{bmatrix})
AB=(begin{bmatrix}2\7\8end{bmatrix})(begin{bmatrix}-3&4&1end{bmatrix})=(begin{bmatrix}-6&8&2\-21&28&7\-24&32&8end{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&8end{bmatrix})
4. If A’=(begin{bmatrix}8&2\6&4end{bmatrix}) and B’=(begin{bmatrix}9&5\7&3end{bmatrix}). Find (A+2B)’.
a) (begin{bmatrix}26&20\10&12end{bmatrix})
b) (begin{bmatrix}26&12\20&10end{bmatrix})
c) (begin{bmatrix}26&10\20&12end{bmatrix})
d) (begin{bmatrix}26&20\12&10end{bmatrix})
Answer: b
Clarification: Given that A’=(begin{bmatrix}8&2\6&4end{bmatrix}) and B’=(begin{bmatrix}9&5\7&3end{bmatrix})
Calculating the transpose of A’ and B’, we get
A=(begin{bmatrix}8&6\2&4end{bmatrix}) and B=(begin{bmatrix}9&7\5&3end{bmatrix})
⇒(A+2B)=(begin{bmatrix}8&6\2&4end{bmatrix})+2(begin{bmatrix}9&7\5&3end{bmatrix})
=(begin{bmatrix}8+18&6+14\2+10&4+6end{bmatrix})=(begin{bmatrix}26&20\12&10end{bmatrix})
Hence, (A+2B)’=(begin{bmatrix}26&12\20&10end{bmatrix}).
5. If A=(begin{bmatrix}cosx&-sinx&-cosx\sinx&-cosx&sinx end{bmatrix}). Find A’A.
a) (begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
b) (begin{bmatrix}1&0&1\1&0&1\1&0&1end{bmatrix})
c) (begin{bmatrix}1&0&1\0&1&0\1&0&1end{bmatrix})
d) (begin{bmatrix}1&0&0\1&1&0\1&1&1end{bmatrix})
Answer: d
Clarification: Given that, A=(begin{bmatrix}cosx&-sinx&cosx\sinx&cosx&sinx end{bmatrix})
∴ A’=(begin{bmatrix}cosx&sinx\-sinx&cosx\cosx&sinx end{bmatrix})
⇒A’ A=(begin{bmatrix}cosx&sinx\-sinx&cosx\cosx&sinx end{bmatrix})(begin{bmatrix}cos x &-sinx&cosx\sinx&cosx &sinx end{bmatrix})
=(begin{bmatrix}cos^2x+sin^2x&-sinx cosx+sinx cosx&cos^2x+sin^2x\-sinx cosx+sinx cosx&sin^2x+cos^2x&-sinx cosx+cosx sinx\cos^2x+sin^2 x&-cosx sinx+sinx cosx &cos^2x+sin^2x end{bmatrix})
=(begin{bmatrix}1&0&1\0&1&0\1&0&1end{bmatrix})
6. Find the transpose of the matrix A=(begin{bmatrix}-1&2&sqrt{3}\-4&5&sqrt{6}\-7&8&-9end{bmatrix})
a) (begin{bmatrix}1&-2&-sqrt{3}\4&-5&-sqrt{6}\7&-8&9end{bmatrix})
b) (begin{bmatrix}-1&-4&-7\2&5&8\sqrt{3}&sqrt{6}&-9end{bmatrix})
c) (begin{bmatrix}1&4&7\-2&-5&-8\-sqrt{3}&-sqrt{6}&9end{bmatrix})
d) (begin{bmatrix}1&4&7\-2&5&2\1&8&9end{bmatrix})
Answer: b
Clarification: 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}&-9end{bmatrix})
7. If matrix A=(begin{bmatrix}4&1\6&2end{bmatrix}) and B=(begin{bmatrix}-1&3\2&1\6&6end{bmatrix}), then find A’ B’.
a) (begin{bmatrix}14&14\5&4\6&18end{bmatrix})
b) (begin{bmatrix}14&5\14&4\6&18end{bmatrix})
c) (begin{bmatrix}14&14&60\5&4&18end{bmatrix})
d) (begin{bmatrix}14&14&18\5&4&60end{bmatrix})
Answer: c
Clarification: Given that, A=(begin{bmatrix}4&1\6&2end{bmatrix}) and B=(begin{bmatrix}-1&3\2&1\6&6end{bmatrix})
⇒A’=(begin{bmatrix}4&6\1&2end{bmatrix}) and B’=(begin{bmatrix}-1&2&6\3&1&6end{bmatrix})
⇒A’ B’=(begin{bmatrix}4&6\1&2end{bmatrix})(begin{bmatrix}-1&2&6\3&1&6end{bmatrix})=(begin{bmatrix}14&14&60\5&4&18end{bmatrix}).
8. If P=(begin{bmatrix}-1&5\8&3end{bmatrix}) and Q=(begin{bmatrix}4&2\8&5end{bmatrix}). Find (2P+3Q’)’.
a) (begin{bmatrix}10&22\34&21end{bmatrix})
b) (begin{bmatrix}10&21\34&22end{bmatrix})
c) (begin{bmatrix}10&34\22&21end{bmatrix})
d) (begin{bmatrix}10&22\21&34end{bmatrix})
Answer: a
Clarification: Given that, P=(begin{bmatrix}-1&5\8&3end{bmatrix}) and Q=(begin{bmatrix}4&2\8&5end{bmatrix})
⇒Q’=(begin{bmatrix}4&8\2&5end{bmatrix})
⇒2P+3Q’=2(begin{bmatrix}-1&5\8&3end{bmatrix})+3(begin{bmatrix}4&8\2&5end{bmatrix})=(begin{bmatrix}-2+12&10+24\16+6&6+15end{bmatrix})=(begin{bmatrix}10&34\22&21end{bmatrix})
∴(2P+3Q’ )’=(begin{bmatrix}10&22\34&21end{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’
Answer: b
Clarification: 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&iend{bmatrix}), then the correct relation is ___________
a) A+A’=(begin{bmatrix}1&0\-1&0end{bmatrix})
b) A-A’=(begin{bmatrix}1&0\-1&0end{bmatrix})
c) A+A’=(begin{bmatrix}0&1\-1&0end{bmatrix})
d) A-A’=(begin{bmatrix}0&1\-1&0end{bmatrix})
Answer: d
Clarification: Given that, A=(begin{bmatrix}i&1\0&iend{bmatrix})
⇒A’=(begin{bmatrix}i&0\1&iend{bmatrix})
∴A-A’=(begin{bmatrix}i&1\0&iend{bmatrix})–(begin{bmatrix}i&0\1&iend{bmatrix})=(begin{bmatrix}i-i&1-0\0-1&i-iend{bmatrix})=(begin{bmatrix}0&1\-1&0end{bmatrix}).