# Class 12 Maths MCQ – Symmetric and Skew Symmetric Matrices

This set of Class 12 Maths Chapter 3 Multiple Choice Questions & Answers (MCQs) focuses on “Symmetric and Skew Symmetric Matrices”.

1. The matrix A=$$\begin{bmatrix}1&2\\2&1\end{bmatrix}$$ is a ____________
a) symmetric matrix
b) skew-symmetric matrix
c) null matrix
d) diagonal matrix

Explanation: Given that, A=$$\begin{bmatrix}1&2\\2&1\end{bmatrix}$$
⇒ A’=$$\begin{bmatrix}1&2\\2&1\end{bmatrix}$$
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|

Explanation: 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’

Explanation: 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

Explanation: 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=$$\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}$$ is __________
a) scalar matrix
b) identity matrix
c) symmetric matrix
d) skew-symmetric matrix

Explanation: The given matrix A=$$\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}$$ is skew symmetric.
⇒A’=$$\begin{bmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{bmatrix}$$=A
∴A=-A’. Hence, it is a skew-symmetric matrix.
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6. Which of the following matrices is both symmetric and skew symmetric?
a) A=$$\begin{bmatrix}1&0\\1&0\end{bmatrix}$$
b) A=$$\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$
c) A=$$\begin{bmatrix}1&0&1\\1&0&1\end{bmatrix}$$
d) A=$$\begin{bmatrix}0&0&-2\\1&0&-1\\2&0&0\end{bmatrix}$$

Explanation: The matrix A=$$\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$=A’=-A’.Hence, a null matrix is both symmetric and skew-symmetric.

7. The matrix A=$$\begin{bmatrix}0&1&1\\1&0&-1\\-1&1&0\end{bmatrix}$$ is symmetric.
a) True
b) False

Explanation: Given that, A=$$\begin{bmatrix}0&1&1\\1&0&-1\\-1&1&0\end{bmatrix}$$
⇒A’=$$\begin{bmatrix}0&1&-1\\1&0&1\\1&-1&0\end{bmatrix}$$. ∴A ≠ A’. Hence, it is not symmetric.

8. The matrix A=$$\begin{bmatrix}2&9\\2&6\end{bmatrix}$$ as a sum of symmetric and skew-symmetric matrix is ______
a) $$\frac{1}{4} \begin{bmatrix}4&11\\11&12\end{bmatrix} – \frac{1}{2} \begin{bmatrix}0&7\\-7&0\end{bmatrix}$$
b) $$\frac{1}{4} \begin{bmatrix}4&11\\11&12\end{bmatrix} + \frac{1}{2} \begin{bmatrix}0&7\\7&0\end{bmatrix}$$
c) $$\frac{1}{2} \begin{bmatrix}4&11\\11&12\end{bmatrix} + \frac{1}{2} \begin{bmatrix}0&7\\-7&0\end{bmatrix}$$
d) $$\frac{1}{2} \begin{bmatrix}4&11\\11&12\end{bmatrix} – \frac{1}{2} \begin{bmatrix}0&7\\-7&0\end{bmatrix}$$

Explanation: Given that A=$$\begin{bmatrix}2&9\\2&6\end{bmatrix}$$.
A’=$$\begin{bmatrix}2&2\\9&6\end{bmatrix}$$
⇒A+A’=$$\begin{bmatrix}2&9\\2&6\end{bmatrix}$$+$$\begin{bmatrix}2&2\\9&6\end{bmatrix}$$=$$\begin{bmatrix}4&11\\11&12\end{bmatrix}$$
⇒A-A’=$$\begin{bmatrix}2&9\\2&6\end{bmatrix}$$–$$\begin{bmatrix}2&2\\9&6\end{bmatrix}$$=$$\begin{bmatrix}0&7\\-7&0\end{bmatrix}$$
The given square matrix can be written as
⇒A = $$\frac{1}{2}$$ (A+A’) + $$\frac{1}{2}$$ (A-A’)=$$\frac{1}{2} \begin{bmatrix}4&11\\11&12\end{bmatrix} + \frac{1}{2} \begin{bmatrix}0&7\\-7&0\end{bmatrix}$$.

9. If A=$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$, 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

Explanation: Given that, A=$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$
∴A’=$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$
⇒-A’=$$\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$$≠A. Hence, it is not a skew symmetric matrix.

10. If A=$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$, then which of the following is skew-symmetric?
a) AA’
b) A+A’
c) 2(A+A’)
d) A-A’

Explanation: Given that, A=$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$
⇒A’=$$\begin{bmatrix}a&c\\b&d\end{bmatrix}$$
Let B=A-A’=$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$–$$\begin{bmatrix}a&c\\b&d\end{bmatrix}$$=$$\begin{bmatrix}a-a&b-c\\c-b&d-d\end{bmatrix}$$=$$\begin{bmatrix}0&b-c\\c-b&0\end{bmatrix}$$
B’=$$\begin{bmatrix}0&c-b\\b-c&0\end{bmatrix}$$=B’
Thus, B=A-A’ is a skew – symmetric.

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

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