Mathematics Multiple Choice Questions & Answers (MCQs) on “Symmetric and Skew Symmetric Matrices”.
1. The matrix A=(begin{bmatrix}1&2\2&1end{bmatrix}) is a ____________
a) symmetric matrix
b) skew-symmetric matrix
c) null matrix
d) diagonal matrix
Answer: a
Clarification: Given that, A=(begin{bmatrix}1&2\2&1end{bmatrix})
⇒ A’=(begin{bmatrix}1&2\2&1end{bmatrix})
i.e.A=A’. Hence, it is a symmetric matrix.
2. Which of the following conditions holds true for a symmetric matrix?
a) A=-A’
b) A=A’
c) A=IA
d) A=|A|
Answer: b
Clarification: A matrix is A said to be a symmetric matrix if it is equal to its transpose i.e. A=A’.
3. Which of the following conditions holds true for a skew-symmetric matrix?
a) A=IA
b) A=|A|
c) A=A’
d) A=-A’
Answer: a
Clarification: A matrix is said to be skew-symmetric if it is equal to the negative of its transpose i.e. A=-A’.
4. Any square matrix can be expressed as a sum of symmetric and skew-symmetric matrix.
a) True
b) False
Answer: a
Clarification: The given statement is true. Every square matrix can be expressed as a sum of sum of symmetric and skew-symmetric matrix.
If A is a square matrix then it can be expressed as
A = (frac{1}{2})(A+A’)+(frac{1}{2})(A-A’), where (A+A’) is symmetric and (A-A’) is skew-symmetric.
5. The matrix A=(begin{bmatrix}0&1&-1\-1&0&1\1&-1&0end{bmatrix}) is __________
a) scalar matrix
b) identity matrix
c) symmetric matrix
d) skew-symmetric matrix
Answer: d
Clarification: The given matrix A=(begin{bmatrix}0&1&-1\-1&0&1\1&-1&0end{bmatrix}) is skew symmetric.
⇒A’=(begin{bmatrix}0&-1&1\1&0&-1\-1&1&0end{bmatrix})=A
∴A=-A’. Hence, it is a skew-symmetric matrix.
6. Which of the following matrices is both symmetric and skew symmetric?
a) A=(begin{bmatrix}1&0\1&0end{bmatrix})
b) A=(begin{bmatrix}0&0&0\0&0&0\0&0&0end{bmatrix})
c) A=(begin{bmatrix}1&0&1\1&0&1end{bmatrix})
d) A=(begin{bmatrix}0&0&-2\1&0&-1\2&0&0end{bmatrix})
Answer: b
Clarification: The matrix A=(begin{bmatrix}0&0&0\0&0&0\0&0&0end{bmatrix})=A’=-A’.Hence, a null matrix is both symmetric and skew-symmetric.
7. The matrix A=(begin{bmatrix}0&1&1\1&0&-1\-1&1&0end{bmatrix}) is symmetric.
a) True
b) False
Answer: b
Clarification: Given that, A=(begin{bmatrix}0&1&1\1&0&-1\-1&1&0end{bmatrix})
⇒A’=(begin{bmatrix}0&1&-1\1&0&1\1&-1&0end{bmatrix}). ∴A ≠ A’. Hence, it is not symmetric.
8. The matrix A=(begin{bmatrix}2&9\2&6end{bmatrix}) as a sum of symmetric and skew-symmetric matrix is ______
a) ( frac{1}{4} begin{bmatrix}4&11\11&12end{bmatrix} – frac{1}{2} begin{bmatrix}0&7\-7&0end{bmatrix})
b) ( frac{1}{4} begin{bmatrix}4&11\11&12end{bmatrix} + frac{1}{2} begin{bmatrix}0&7\7&0end{bmatrix})
c) ( frac{1}{2} begin{bmatrix}4&11\11&12end{bmatrix} + frac{1}{2} begin{bmatrix}0&7\-7&0end{bmatrix})
d) ( frac{1}{2} begin{bmatrix}4&11\11&12end{bmatrix} – frac{1}{2} begin{bmatrix}0&7\-7&0end{bmatrix})
Answer: c
Clarification: Given that A=(begin{bmatrix}2&9\2&6end{bmatrix}).
A’=(begin{bmatrix}2&2\9&6end{bmatrix})
⇒A+A’=(begin{bmatrix}2&9\2&6end{bmatrix})+(begin{bmatrix}2&2\9&6end{bmatrix})=(begin{bmatrix}4&11\11&12end{bmatrix})
⇒A-A’=(begin{bmatrix}2&9\2&6end{bmatrix})–(begin{bmatrix}2&2\9&6end{bmatrix})=(begin{bmatrix}0&7\-7&0end{bmatrix})
The given square matrix can be written as
⇒A = ( frac{1}{2}) (A+A’) + ( frac{1}{2}) (A-A’)=( frac{1}{2} begin{bmatrix}4&11\11&12end{bmatrix} + frac{1}{2} begin{bmatrix}0&7\-7&0end{bmatrix}).
9. If A=(begin{bmatrix}1&0\0&1end{bmatrix}), then which of the following statement is incorrect?
a) A is a skew-symmetric matrix
b) A is a square matrix
c) A is a symmetric
d) A is an identity matrix
Answer: a
Clarification: Given that, A=(begin{bmatrix}1&0\0&1end{bmatrix})
∴A’=(begin{bmatrix}1&0\0&1end{bmatrix})
⇒-A’=(begin{bmatrix}-1&0\0&-1end{bmatrix})≠A. Hence, it is not a skew symmetric matrix.
10. If A=(begin{bmatrix}a&b\c&dend{bmatrix}), then which of the following is skew-symmetric?
a) AA’
b) A+A’
c) 2(A+A’)
d) A-A’
Answer: c
Clarification: Given that, A=(begin{bmatrix}a&b\c&dend{bmatrix})
⇒A’=(begin{bmatrix}a&c\b&dend{bmatrix})
Let B=A-A’=(begin{bmatrix}a&b\c&dend{bmatrix})–(begin{bmatrix}a&c\b&dend{bmatrix})=(begin{bmatrix}a-a&b-c\c-b&d-dend{bmatrix})=(begin{bmatrix}0&b-c\c-b&0end{bmatrix})
B’=(begin{bmatrix}0&c-b\b-c&0end{bmatrix})=B’
Thus, B=A-A’ is a skew – symmetric.