Mathematics Questions and Answers – Operations on Matrices

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

1. The addition of matrices is only possible if they are of the same order.
a) True
b) False
View Answer

Answer: a
Explanation: The given statement is true. Addition of matrices is possible only if the matrices are of the same order. If there are two matrices of different order, then A+B is not defined.
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2. If A = \(\begin{bmatrix}1&2&3\\9&10&11\end{bmatrix}\) and B = \(\begin{bmatrix}0&5&0\\5&0&5\end{bmatrix}\), then find A+B.
a) A+B = \(\begin{bmatrix}1&7&3\\11&10&16\end{bmatrix}\)
b) A+B = \(\begin{bmatrix}1&7&3\\14&11&13\end{bmatrix}\)
c) A+B = \(\begin{bmatrix}1&7&3\\14&10&16\end{bmatrix}\)
d) A+B = \(\begin{bmatrix}1&5&3\\14&10&16\end{bmatrix}\)
View Answer

Answer: c
Explanation: Given that, A = \(\begin{bmatrix}1&2&3\\9&10&11\end{bmatrix}\) and B = \(\begin{bmatrix}0&5&0\\5&0&5\end{bmatrix}\)
Then A+B = \(\begin{bmatrix}1+0&2+5&3+0\\9+5&10+0&11+5\end{bmatrix}\) = \(\begin{bmatrix}1&7&3\\14&10&16\end{bmatrix}\).

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

Answer: d
Explanation: Given that, A = \(\begin{bmatrix}3&4\\1&2\end{bmatrix}\) and B = \(\begin{bmatrix}1&5\\2&3\end{bmatrix}\)
⇒2A=2\(\begin{bmatrix}3&4\\1&2\end{bmatrix}\)=\(\begin{bmatrix}6&8\\2&4\end{bmatrix}\) and 3B=3\(\begin{bmatrix}1&5\\2&3\end{bmatrix}\)=\(\begin{bmatrix}3&15\\6&9\end{bmatrix}\)
∴2A-3B = \(\begin{bmatrix}6&8\\2&4\end{bmatrix}\)–\(\begin{bmatrix}3&15\\6&9\end{bmatrix}\)=\(\begin{bmatrix}3&-7\\-4&-5\end{bmatrix}\).
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4. If A+B = \(\begin{bmatrix}6&7\\5&0\end{bmatrix}\)and A = \(\begin{bmatrix}2&5\\1&-1\end{bmatrix}\). Find the matrix B.
a) B = \(\begin{bmatrix}4&1\\2&4\end{bmatrix}\)
b) B = \(\begin{bmatrix}4&2\\4&1\end{bmatrix}\)
c) B = \(\begin{bmatrix}4&1\\4&2\end{bmatrix}\)
d) B = \(\begin{bmatrix}4&4\\4&2\end{bmatrix}\)
View Answer

Answer: b
Explanation: Given that, A+B = \(\begin{bmatrix}6&7\\5&0\end{bmatrix}\)and A = \(\begin{bmatrix}2&5\\1&-1\end{bmatrix}\)
⇒B=(A+B)-A = \(\begin{bmatrix}6&7\\5&0\end{bmatrix}\)–\(\begin{bmatrix}2&5\\1&-1\end{bmatrix}\)
B = \(\begin{bmatrix}4&2\\4&1\end{bmatrix}\)

5. Find the matrix M and N, if M+N = \(\begin{bmatrix}5&6\\7&8\end{bmatrix}\),M-N = \(\begin{bmatrix}4&5\\6&8\end{bmatrix}\).
a) M=1/2 \(\begin{bmatrix}9&11\\13&16\end{bmatrix}\), N=1/2 \(\begin{bmatrix}1&1\\1&0\end{bmatrix}\)
b) M=\(\begin{bmatrix}5&6\\7&8\end{bmatrix}\), N=\(\begin{bmatrix}4&5\\8&6\end{bmatrix}\)
c) M=1/2 \(\begin{bmatrix}9&2\\13&16\end{bmatrix}\), N=1/2 \(\begin{bmatrix}1&1\\2&5\end{bmatrix}\)
d) M=1/2 \(\begin{bmatrix}4&5\\1&2\end{bmatrix}\), N=1/2 \(\begin{bmatrix}1&2\\1&2\end{bmatrix}\)
View Answer

Answer: a
Explanation:M+N = \(\begin{bmatrix}5&6\\7&8\end{bmatrix}\)-(1) and M-N = \(\begin{bmatrix}4&5\\6&8\end{bmatrix}\)-(2)
Adding equation (1) and equation (2), (M+N)+(M-N)=2M=\(\begin{bmatrix}5&6\\7&8\end{bmatrix}\)+\(\begin{bmatrix}4&5\\6&8\end{bmatrix}\)
M=1/2 \(\begin{bmatrix}9&11\\13&16\end{bmatrix}\).
Subtracting equation (1) and equation (2), (M+N)-(M-N)=2N=\(\begin{bmatrix}5&6\\7&8\end{bmatrix}\)–\(\begin{bmatrix}4&5\\6&8\end{bmatrix}\)
N=1/2 \(\begin{bmatrix}1&1\\1&0\end{bmatrix}\).
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6. Find the value of x and y if 2\(\begin{bmatrix}5&x\\y-4&6\end{bmatrix}\)+\(\begin{bmatrix}-4&1\\3&2\end{bmatrix}\)=\(\begin{bmatrix}6&3\\10&14\end{bmatrix}\)?
a) x=-1, y=9
b) x=-1, y=-9
c) x=1, y=-9
d) x=1, y=9
View Answer

Answer: d
Explanation: Given that, 2\(\begin{bmatrix}5&x\\y-4&6\end{bmatrix}\)+\(\begin{bmatrix}-4&1\\3&2\end{bmatrix}\)=\(\begin{bmatrix}6&3\\10&14\end{bmatrix}\)
⇒\(\begin{bmatrix}2(5)-4&2x+1\\2(y-4)+3&2(6)+2\end{bmatrix}\)=\(\begin{bmatrix}6&3\\10&14\end{bmatrix}\)
Comparing the two matrices, 2x+1=3, 2y-8=10
Solving the two equations we get, x=1, y=9.

7. Find AB if A = \(\begin{bmatrix}1&2\\3&4\end{bmatrix}\) and B = \(\begin{bmatrix}1&5\\3&2\end{bmatrix}\).
a) AB = \(\begin{bmatrix}15&23\\9&7\end{bmatrix}\)
b) AB = \(\begin{bmatrix}9&7\\23&15\end{bmatrix}\)
c) AB = \(\begin{bmatrix}7&9\\15&23\end{bmatrix}\)
d) AB = \(\begin{bmatrix}7&9\\23&15\end{bmatrix}\)
View Answer

Answer: c
Explanation: Given that, A = \(\begin{bmatrix}1&2\\3&4\end{bmatrix}\) and B = \(\begin{bmatrix}1&5\\3&2\end{bmatrix}\)
Then, AB = \(\begin{bmatrix}1&2\\3&4\end{bmatrix}\)\(\begin{bmatrix}1&5\\3&2\end{bmatrix}\)
=\(\begin{bmatrix}1×1+2×3&1×5+2×2\\3×1+4×3&3×5+4×2\end{bmatrix}\)=\(\begin{bmatrix}7&9\\15&23\end{bmatrix}\).
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8. Matrix addition and matrix multiplication both are commutative.
a) True
b) False
View Answer

Answer: b
Explanation: The given statement is false. Matrix addition is commutative i.e. A+B=B+A. But matrix multiplication is not commutative i.e.AB≠BA.

9. Let A=\(\begin{bmatrix}3&-5&2\\-4&-6&2\\7&1&5\end{bmatrix}\). Find the additive inverse of A.
a) \(\begin{bmatrix}-3&5&-2\\-4&6&2\\7&1&5\end{bmatrix}\)
b) \(\begin{bmatrix}3&-5&2\\-4&-6&2\\7&1&5\end{bmatrix}\)
c) \(\begin{bmatrix}-3&5&-2\\4&6&-2\\-7&-1&-5\end{bmatrix}\)
d) \(\begin{bmatrix}-3&5&2\\-4&6&-2\\-7&-1&5\end{bmatrix}\)
View Answer

Answer: c
Explanation: Additive inverse of matrix A is the negative of A i.e. -A.
Therefore, -A=\(\begin{bmatrix}-3&5&-2\\4&6&-2\\-7&-1&-5\end{bmatrix}\)
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10. Which of the following condition is incorrect for matrix multiplication?
a) A(BC)=(AB)C
b) A(B+C)=AB+AC
c) AB=0 if either A or B is 0
d) AB=BA
View Answer

Answer: d
Explanation: Matrix multiplication is never commutative i.e. AB≠BA. Therefore, the condition AB=BA is incorrect.

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

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