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
View Answer
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|
View Answer
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’
View Answer
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
View Answer
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
View Answer
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.
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=\(\begin{bmatrix}0&0&-2\\1&0&-1\\2&0&0\end{bmatrix}\)
View Answer
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
View Answer
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) \( \frac{1}{2} \begin{bmatrix}4&11\\11&12\end{bmatrix} – \frac{1}{2} \begin{bmatrix}0&7\\-7&0\end{bmatrix}\)
View Answer
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
View Answer
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’
View Answer
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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