Matrices: Chapter 3 Links
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Matrices 3.3 of Class 12 Maths Chapter 3 focuses on the transpose of matrices, properties of transpose, symmetric matrices and skew symmetric matrices. These questions will help you understand important matrix identities and verification-based problems.
In this exercise, you will learn how to find the transpose of a matrix, verify transpose properties like (AB)' = B'A' and express matrices in symmetric and skew symmetric forms. These concepts are important for board exams and CUET.
Key Concepts
- Transpose of a Matrix: The transpose of a matrix is obtained by interchanging its rows and columns.
- If A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} , then A' = \begin{bmatrix} a & c \\ b & d \end{bmatrix} .
- Order of Transpose: If a matrix is of order m \times n , then its transpose is of order n \times m .
- Transpose of a Transpose: (A')' = A
- Transpose of Sum: (A + B)' = A' + B'
- Transpose of Difference: (A - B)' = A' - B'
- Reversal Law for Transpose: (AB)' = B'A'
- Transpose of Scalar Multiplication: (kA)' = kA' , where k is a scalar.
- Symmetric Matrix: A square matrix A is called symmetric if A' = A .
- In a symmetric matrix, the elements on opposite sides of the principal diagonal are equal.
- Skew Symmetric Matrix: A square matrix A is called skew symmetric if A' = -A .
- In a skew symmetric matrix, all diagonal elements are zero.
- Orthogonal Matrix Property: If A'A = I , then the matrix is orthogonal. (This term is not explicitly mentioned in NCERT, but Question 6 uses this concept. A square matrix whose transpose is equal to its inverse is called an orthogonal matrix.)
Decomposition of a Square Matrix
- Decomposition of a Square Matrix: Every square matrix can be expressed as the sum of a symmetric matrix and a skew symmetric matrix. (The word “decomposition” is not used in NCERT. It simply means expressing a square matrix uniquely as the sum of a symmetric matrix and a skew symmetric matrix)
- The decomposition formula is:
A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') - \frac{1}{2}(A + A') is always a symmetric matrix.
- \frac{1}{2}(A - A') is always a skew symmetric matrix.
- Important Result: If A and B are symmetric matrices of same order, then AB + BA is symmetric and AB - BA is skew symmetric.
- For any square matrix A , A + A' is symmetric and A - A' is skew symmetric.
- Zero or null matrix is both symmetric as well as skew symmetric.
Now, let us begin Matrices 3.3 exercise with Question 1.
Question 1: Matrices 3.3
1. Find the transpose of each of the following matrices.
(i) A = \begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}
Solution
The transpose of a matrix is obtained by interchanging rows and columns.
Therefore, A' = \begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}
Hence, the transpose of the matrix is \begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}
(ii) A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}
Solution
Interchanging rows and columns, we get A' = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}
Hence, the transpose of the matrix is \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}
(iii) A = \begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}
Solution
On interchanging rows and columns, we get A' = \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}
Hence, the transpose of the matrix is \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}
Question 2: Properties of Transpose
2. If A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} and B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix} then verify that
(i) (A+B)' = A' + B'
Solution
First, find A+B .
A+B = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} + \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}
= \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}
Taking transpose on both sides, we get
(A+B)' = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}
Now, find A' and B' .
A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}
B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}
Therefore, A' + B'
= \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix} + \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}
= \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}
Hence,
(A+B)' = A' + B'
(ii) (A-B)' = A'-B'
Solution
First, find A-B .
A-B = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} - \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}
= \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}
Taking transpose, we get
(A-B)' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}
Now, A' - B'
= \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix} - \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}
= \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}
Hence,
(A-B)' = A'-B'
You may already be following Maths Better for important concepts and exam-oriented questions, ranging from integration tricks to solving linear equations using the Matrix Method. Similarly, this NCERT Solutions series for Class 12 Maths will help strengthen your concepts and improve problem-solving skills.
Now, let’s move on to the next question of Matrices 3.3.
Question 3: Ex 3.3 Matrices
3. If A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} and B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} then verify that
(i) (A+B)' = A'+B'
Solution
Since transpose of A is given, first find A .
A = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}
Now,
A+B = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}
= \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix}
Taking transpose, we get
(A+B)' = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}
Now, A'+B'
= \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}
= \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}
Hence verified,
(A+B)' = A'+B'
(ii) (A-B)' = A'-B'
Since transpose of A is given, first find A .
A = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}
Now,
A-B = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}
= \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix}
Taking transpose,
(A-B)' = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}
Now, A'-B'
= \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}
= \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}
Hence,
(A-B)' = A'-B'
Question 4: Matrices 3.3
4. If A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} and B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} then find (A+2B)' .
Solution
Since transpose of A is given, first find A .
By taking transpose of A' , we get
A = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}
Now,
2B = 2 \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}
Therefore,
A+2B = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}
= \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}
Taking transpose,
(A+2B)' = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}
Question 5: Reversal Law
5. For the matrices A \text{ and } B , verify that (AB)' = B'A' , where
(i) A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}
Solution
First, find AB .
AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}
= \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}
Taking transpose,
(AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}
Now, find B' and A' .
B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}, \quad A' = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}
Therefore,
B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}
= \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}
Hence verified that,
(AB)' = B'A'
(ii) A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}
Solution
First, find AB .
AB = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}
= \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}
Taking transpose,
(AB)' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}
Now,
B' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}, \quad A' = \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}
Therefore,
B'A' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}
= \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}
Hence verified that,
(AB)' = B'A'
Try solving the next question on your own first, and then compare your approach with the solution given below.
Question 6: Exercise 3.3 Matrices
6. (i) If A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} then verify that A'A = I .
Solution
First, find transpose of A .
A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}
Now,
A'A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}
So, A'A =
\begin{bmatrix} \cos^2\alpha + \sin^2\alpha & \cos\alpha\sin\alpha - \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha - \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha \end{bmatrix}
Using \sin^2\alpha + \cos^2\alpha = 1 , we get
A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I
Hence verified.
(ii) If A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} then verify that A'A = I .
Solution
First, find transpose of A .
A' = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}
Now,
A'A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}
So, A'A =
\begin{bmatrix} \sin^2\alpha + \cos^2\alpha & \sin\alpha\cos\alpha - \cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha - \sin\alpha\cos\alpha & \cos^2\alpha + \sin^2\alpha \end{bmatrix}
Using \sin^2\alpha + \cos^2\alpha = 1 , we get
A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I
Hence verified.
Question 7: Matrices 3.3
7. (i) Show that the matrix A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} is a symmetric matrix.
Solution
To show that a matrix is symmetric, we verify whether
A' = A
Taking transpose of A , we get
A' = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}
Since
A' = A
therefore, the given matrix is a symmetric matrix.
(ii) Show that the matrix A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} is a skew symmetric matrix.
Solution
To show that a matrix is skew symmetric, we verify whether
A' = -A
Taking transpose of A , we get
A' = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}
Now,
-A = - \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}
= \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}
Since
A' = -A
therefore, the given matrix is a skew symmetric matrix.
Matrix-based questions are generally direct, but presentation and correct steps play an important role in scoring full marks.
Question 8: Exercise 3.3
8. For the matrix A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} verify that
(i) (A + A') is a symmetric matrix.
Solution
First, find transpose of A .
A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}
Now,
A + A' = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}
= \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}
Taking transpose,
(A + A')' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}
Since
(A + A')' = A + A'
therefore, (A + A') is a symmetric matrix.
(ii) (A - A') is a skew symmetric matrix.
Solution
Using
A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}, \quad A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}
Now,
A - A' = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} - \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}
= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}
Taking transpose,
(A - A')' = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
Also,
-(A - A') = - \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
Since
(A - A')' = -(A - A')
therefore, (A - A') is a skew symmetric matrix.
NCERT Class 12 Maths has a total of 61 exercises across Part 1 and Part 2. I’ll be covering each exercise one by one with clear explanations and exam-oriented solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter.
Here’s the next question.
Question 9: Matrices 3.3
9. Find \frac{1}{2}(A + A') \text{ and } \frac{1}{2}(A - A') , when A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}
Solution
First, find transpose of A .
A' = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}
Now,
A + A' = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} + \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}
= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
Therefore,
\frac{1}{2}(A + A') = \frac{1}{2} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
Also,
A - A' = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} - \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}
= \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix}
Therefore,
\frac{1}{2}(A - A') = \frac{1}{2} \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix}
= \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}
Question 10: Symmetric and Skew Symmetric Matrices
10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix.
(i) A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}
Solution
We know that
A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A')
First, find transpose of A .
A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}
Now, A + A' = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}
= \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix}
\frac{1}{2}(A + A') = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}
Also,
A - A' = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}
= \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}
i.e. \frac{1}{2}(A - A') = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}
Therefore,
\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}
(ii) A = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}
Solution
Since
A' = A
the matrix is symmetric.
Hence,
A = A + 0
Therefore,
\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
💡Every Square Matrix can be uniquely expressed as the sum of a symmetric and a skew symmetric matrix.
(iii) A = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}
Solution
First, find transpose of A .
A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}
Now, A + A' = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} + \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}
= \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix}
\frac{1}{2}(A + A') = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix}
Also, A - A' = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}
= \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix}
\frac{1}{2}(A - A') = \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix}
Therefore,
\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix}
(iv) A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}
Solution
First, find transpose of A .
A' = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}
Now, A + A' = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} + \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}
= \begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix}
\frac{1}{2}(A + A') = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}
Also, A - A' = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} - \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}
= \begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix}
\frac{1}{2}(A - A') = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}
Therefore,
\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}
Detailed video explanations of many NCERT questions are also available on the Mathsbetter YouTube channel.
Question 11: Matrices 3.3 – MCQ
11. If A \text{ and } B are symmetric matrices of same order, then AB - BA is a
- (A) Skew symmetric matrix
- (B) Symmetric matrix
- (C) Zero matrix
- (D) Identity matrix
Solution
Since A \text{ and } B are symmetric matrices,
A' = A \text{ and } B' = B
Now,
(AB - BA)' = (AB)' - (BA)'
Now, by transversal law, we get it
= B'A' - A'B'
= BA - AB
i.e. (AB - BA)' = -(AB - BA)
Hence, AB - BA is a skew symmetric matrix.
Correct option: (A) Skew symmetric matrix
Question 12: Matrices Ex 3.3 – MCQ
12. If A= \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} and A+A'=I , then the value of \alpha is
- (A) \frac{\pi}{6}
- (B) \frac{\pi}{3}
- (C) \pi
- (D) \frac{3\pi}{2}
Solution
First, find transpose of A .
A'= \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}
Now,
A+A'= \begin{bmatrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{bmatrix}
Given that
A+A'=I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
Comparing corresponding elements,
2\cos \alpha = 1
\cos \alpha = \frac{1}{2}
Therefore,
\alpha=\frac{\pi}{3}
Correct option: (B) \frac{\pi}{3}
Common Mistakes to Avoid
- Not interchanging rows and columns properly while finding transpose: Students often rewrite the same matrix instead of converting rows into columns correctly.
- Confusing transpose with inverse: Transpose only changes rows into columns. It does not involve determinants or reciprocals.
- Wrong order in reversal law:
(AB)' \neq A'B'
The correct formula is
(AB)' = B'A' - Using transpose properties for matrices of incompatible orders: Matrix addition and subtraction are possible only when matrices have the same order.
- Forgetting that symmetric and skew symmetric matrices must be square matrices: Non-square matrices cannot be symmetric or skew symmetric.
- Incorrect condition for symmetric matrix: A matrix is symmetric only if A' = A
- Incorrect condition for skew symmetric matrix: A matrix is skew symmetric only if
A' = -A - Forgetting negative sign in skew symmetric matrices: Students often compare A' directly with A instead of -A
- Not checking diagonal elements in skew symmetric matrices: All diagonal elements of a skew symmetric matrix are always zero.
- Sign errors while finding A - A' : Negative signs are commonly missed during subtraction.
- Errors in decomposition formula: Some students forget the factor \frac{1}{2} in A=\frac{1}{2}(A+A')+\frac{1}{2}(A-A')
- Mixing symmetric and skew symmetric parts:
\frac{1}{2}(A+A') gives the symmetric part, while
\frac{1}{2}(A-A') gives the skew symmetric part. - Arithmetic mistakes during matrix operations: Addition, subtraction, and scalar multiplication errors are very common in long calculations.
- Not comparing corresponding elements carefully: In MCQs and proof-based questions, one wrong element comparison may lead to an incorrect conclusion.
- Assuming every symmetric matrix is diagonal: Symmetric matrices may have non-zero elements outside the principal diagonal as long as opposite elements are equal.
Continue Learning
In this exercise, we learned how to find the transpose of matrices and verified important properties such as:
- (A+B)' = A' + B'
- (A-B)' = A' - B'
- (AB)' = B'A'
We also studied symmetric and skew symmetric matrices and learned how every square matrix can be expressed as the sum of a symmetric matrix and a skew symmetric matrix.
These concepts are important for board exams, MCQs, and CUET topics involving matrices.
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All the best and keep learning 👍
