GSEB Class 12 Maths Solutions Chapter 3 Matrices Exercise 3.3

Question 1. Find the transpose of each of the following matrices:
(i) \( \left[ \begin{array}{c} 5 \\ 1 \\ -1 \end{array} \right] \)
(ii) \( \left[ \begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array} \right] \)
(iii) \( \left[ \begin{array}{ccc} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{array} \right] \)
Answer:
(i) Transpose of \( \left[ \begin{array}{c} 5 \\ 1 \\ -1 \end{array} \right] \) is \( \left[ \begin{array}{ccc} 5 & 1 & -1 \end{array} \right] \).
(ii) Transpose of \( \left[ \begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array} \right] \) is \( \left[ \begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array} \right] \).
(iii) Transpose of \( \left[ \begin{array}{ccc} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{array} \right] \) is \( \left[ \begin{array}{ccc} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{array} \right] \).
In simple words: To find the transpose of a matrix, you switch its rows with its columns. The first row becomes the first column, the second row becomes the second column, and so on.

Exam Tip: Remember that the order of rows and columns changes. If a matrix is \( m \times n \), its transpose will be \( n \times m \).

Question 2. If \( A = \left[ \begin{array}{ccc} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{array} \right] \) and \( B = \left[ \begin{array}{ccc} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{array} \right] \), then verify that:
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
Answer:
(i) To verify \( (A+B)' = A' + B' \):
First, we find \( A+B \):
\( A+B = \left[ \begin{array}{ccc} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{array} \right] + \left[ \begin{array}{ccc} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{array} \right] = \left[ \begin{array}{ccc} -1+(-4) & 2+1 & 3+(-5) \\ 5+1 & 7+2 & 9+0 \\ -2+1 & 1+3 & 1+1 \end{array} \right] = \left[ \begin{array}{ccc} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{array} \right] \)
Now, we find the transpose of \( A+B \):
\( (A+B)' = \left[ \begin{array}{ccc} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{array} \right] \)
Next, we find \( A' \) and \( B' \):
\( A' = \left[ \begin{array}{ccc} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{array} \right] \)
\( B' = \left[ \begin{array}{ccc} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{array} \right] \)
Now, we find \( A' + B' \):
\( A' + B' = \left[ \begin{array}{ccc} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{array} \right] + \left[ \begin{array}{ccc} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{array} \right] = \left[ \begin{array}{ccc} -1+(-4) & 5+1 & -2+1 \\ 2+1 & 7+2 & 1+3 \\ 3+(-5) & 9+0 & 1+1 \end{array} \right] = \left[ \begin{array}{ccc} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{array} \right] \)
Since \( (A+B)' = A' + B' \), the identity is verified.

(ii) To verify \( (A-B)' = A' - B' \):
First, we find \( A-B \):
\( A-B = \left[ \begin{array}{ccc} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{array} \right] - \left[ \begin{array}{ccc} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{array} \right] = \left[ \begin{array}{ccc} -1-(-4) & 2-1 & 3-(-5) \\ 5-1 & 7-2 & 9-0 \\ -2-1 & 1-3 & 1-1 \end{array} \right] = \left[ \begin{array}{ccc} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{array} \right] \)
Now, we find the transpose of \( A-B \):
\( (A-B)' = \left[ \begin{array}{ccc} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{array} \right] \)
Next, we find \( A' - B' \):
\( A' - B' = \left[ \begin{array}{ccc} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{array} \right] - \left[ \begin{array}{ccc} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{array} \right] = \left[ \begin{array}{ccc} -1-(-4) & 5-1 & -2-1 \\ 2-1 & 7-2 & 1-3 \\ 3-(-5) & 9-0 & 1-1 \end{array} \right] = \left[ \begin{array}{ccc} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{array} \right] \)
Since \( (A-B)' = A' - B' \), the identity is verified.
In simple words: This problem shows that when you add or subtract two matrices and then flip them (transpose), the answer is the same as flipping each matrix first and then adding or subtracting them. We calculate both sides of the equations and confirm they are equal.

Exam Tip: Remember that for matrix operations like addition, subtraction, and transposition, the order of operations can often be swapped, as shown by these verification problems. Always calculate each side of the equation separately to avoid errors.

Question 3. If \( A = \left[ \begin{array}{ccc} 3 & -1 & 0 \\ 4 & 2 & 1 \end{array} \right] \) and \( B = \left[ \begin{array}{ccc} -1 & 2 & 1 \\ 1 & 2 & 3 \end{array} \right] \), then verify that:
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
Answer:
Given:
\( A = \left[ \begin{array}{ccc} 3 & -1 & 0 \\ 4 & 2 & 1 \end{array} \right] \implies A' = \left[ \begin{array}{cc} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{array} \right] \)
\( B = \left[ \begin{array}{ccc} -1 & 2 & 1 \\ 1 & 2 & 3 \end{array} \right] \implies B' = \left[ \begin{array}{cc} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{array} \right] \)

(i) To verify \( (A+B)' = A' + B' \):
First, find \( A+B \):
\( A+B = \left[ \begin{array}{ccc} 3 & -1 & 0 \\ 4 & 2 & 1 \end{array} \right] + \left[ \begin{array}{ccc} -1 & 2 & 1 \\ 1 & 2 & 3 \end{array} \right] = \left[ \begin{array}{ccc} 3-1 & -1+2 & 0+1 \\ 4+1 & 2+2 & 1+3 \end{array} \right] = \left[ \begin{array}{ccc} 2 & 1 & 1 \\ 5 & 4 & 4 \end{array} \right] \)
Now, find the transpose of \( A+B \):
L.H.S. \( = (A+B)' = \left[ \begin{array}{cc} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{array} \right] \)
Next, find \( A' + B' \):
R.H.S. \( = A' + B' = \left[ \begin{array}{cc} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{array} \right] + \left[ \begin{array}{cc} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{array} \right] = \left[ \begin{array}{cc} 3-1 & 4+1 \\ -1+2 & 2+2 \\ 0+1 & 1+3 \end{array} \right] = \left[ \begin{array}{cc} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{array} \right] \)
Since L.H.S. \( = \) R.H.S., the identity \( (A+B)' = A' + B' \) is verified.

(ii) To verify \( (A-B)' = A' - B' \):
First, find \( A-B \):
\( A-B = \left[ \begin{array}{ccc} 3 & -1 & 0 \\ 4 & 2 & 1 \end{array} \right] - \left[ \begin{array}{ccc} -1 & 2 & 1 \\ 1 & 2 & 3 \end{array} \right] = \left[ \begin{array}{ccc} 3-(-1) & -1-2 & 0-1 \\ 4-1 & 2-2 & 1-3 \end{array} \right] = \left[ \begin{array}{ccc} 4 & -3 & -1 \\ 3 & 0 & -2 \end{array} \right] \)
Now, find the transpose of \( A-B \):
L.H.S. \( = (A-B)' = \left[ \begin{array}{cc} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{array} \right] \)
Next, find \( A' - B' \):
R.H.S. \( = A' - B' = \left[ \begin{array}{cc} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{array} \right] - \left[ \begin{array}{cc} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{array} \right] = \left[ \begin{array}{cc} 3-(-1) & 4-1 \\ -1-2 & 2-2 \\ 0-1 & 1-3 \end{array} \right] = \left[ \begin{array}{cc} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{array} \right] \)
Since L.H.S. \( = \) R.H.S., the identity \( (A-B)' = A' - B' \) is verified.
In simple words: This question confirms that the transpose of the sum or difference of two matrices is equal to the sum or difference of their individual transposes. This rule works for both adding and subtracting matrices.

Exam Tip: Be careful with signs, especially when subtracting negative numbers within matrix elements. Each element must be handled separately and correctly.

Question 4. If \( A' = \left[ \begin{array}{cc} -2 & 3 \\ 1 & 2 \end{array} \right] \) and \( B = \left[ \begin{array}{cc} -1 & 0 \\ 1 & 2 \end{array} \right] \), find \( (A + 2B') \).
Answer:
Given:
\( A' = \left[ \begin{array}{cc} -2 & 3 \\ 1 & 2 \end{array} \right] \)
To find \( A \), we take the transpose of \( A' \):
\( A = (A')' = \left[ \begin{array}{cc} -2 & 1 \\ 3 & 2 \end{array} \right] \)
Given:
\( B = \left[ \begin{array}{cc} -1 & 0 \\ 1 & 2 \end{array} \right] \)
To find \( B' \), we take the transpose of \( B \):
\( B' = \left[ \begin{array}{cc} -1 & 1 \\ 0 & 2 \end{array} \right] \)
Now, we calculate \( 2B' \):
\( 2B' = 2 \left[ \begin{array}{cc} -1 & 1 \\ 0 & 2 \end{array} \right] = \left[ \begin{array}{cc} 2(-1) & 2(1) \\ 2(0) & 2(2) \end{array} \right] = \left[ \begin{array}{cc} -2 & 2 \\ 0 & 4 \end{array} \right] \)
Finally, we calculate \( A + 2B' \):
\( A + 2B' = \left[ \begin{array}{cc} -2 & 1 \\ 3 & 2 \end{array} \right] + \left[ \begin{array}{cc} -2 & 2 \\ 0 & 4 \end{array} \right] = \left[ \begin{array}{cc} -2+(-2) & 1+2 \\ 3+0 & 2+4 \end{array} \right] = \left[ \begin{array}{cc} -4 & 3 \\ 3 & 6 \end{array} \right] \)
So, \( A + 2B' = \left[ \begin{array}{cc} -4 & 3 \\ 3 & 6 \end{array} \right] \).
In simple words: First, we find matrix A by flipping A'. Then, we find matrix B' by flipping B. After that, we multiply B' by 2. Finally, we add matrix A to the result of 2B' to get the final answer.

Exam Tip: Always be careful to take the correct transpose (of A' to get A, and of B to get B') and to perform scalar multiplication correctly on each element of the matrix.

Question 5. For the following matrices A and B, verify that \( (AB)' = B'A' \):
(i) \( A = \left[ \begin{array}{c} 1 \\ -4 \\ 3 \end{array} \right], B = \left[ \begin{array}{ccc} -1 & 2 & 1 \end{array} \right] \)
(ii) \( A = \left[ \begin{array}{c} 0 \\ 1 \\ 2 \end{array} \right], B = \left[ \begin{array}{ccc} 1 & 5 & 7 \end{array} \right] \)
Answer:
(i) Given:
\( A = \left[ \begin{array}{c} 1 \\ -4 \\ 3 \end{array} \right] \implies A' = \left[ \begin{array}{ccc} 1 & -4 & 3 \end{array} \right] \)
\( B = \left[ \begin{array}{ccc} -1 & 2 & 1 \end{array} \right] \implies B' = \left[ \begin{array}{c} -1 \\ 2 \\ 1 \end{array} \right] \)
First, find \( AB \):
\( AB = \left[ \begin{array}{c} 1 \\ -4 \\ 3 \end{array} \right] \left[ \begin{array}{ccc} -1 & 2 & 1 \end{array} \right] = \left[ \begin{array}{ccc} 1 \times (-1) & 1 \times 2 & 1 \times 1 \\ -4 \times (-1) & -4 \times 2 & -4 \times 1 \\ 3 \times (-1) & 3 \times 2 & 3 \times 1 \end{array} \right] = \left[ \begin{array}{ccc} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{array} \right] \)
Now, find the transpose of \( AB \):
L.H.S. \( = (AB)' = \left[ \begin{array}{ccc} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{array} \right] \)
Next, find \( B'A' \):
R.H.S. \( = B'A' = \left[ \begin{array}{c} -1 \\ 2 \\ 1 \end{array} \right] \left[ \begin{array}{ccc} 1 & -4 & 3 \end{array} \right] = \left[ \begin{array}{ccc} -1 \times 1 & -1 \times (-4) & -1 \times 3 \\ 2 \times 1 & 2 \times (-4) & 2 \times 3 \\ 1 \times 1 & 1 \times (-4) & 1 \times 3 \end{array} \right] = \left[ \begin{array}{ccc} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{array} \right] \)
Since L.H.S. \( = \) R.H.S., the identity \( (AB)' = B'A' \) is verified.

(ii) Given:
\( A = \left[ \begin{array}{c} 0 \\ 1 \\ 2 \end{array} \right] \implies A' = \left[ \begin{array}{ccc} 0 & 1 & 2 \end{array} \right] \)
\( B = \left[ \begin{array}{ccc} 1 & 5 & 7 \end{array} \right] \implies B' = \left[ \begin{array}{c} 1 \\ 5 \\ 7 \end{array} \right] \)
First, find \( AB \):
\( AB = \left[ \begin{array}{c} 0 \\ 1 \\ 2 \end{array} \right] \left[ \begin{array}{ccc} 1 & 5 & 7 \end{array} \right] = \left[ \begin{array}{ccc} 0 \times 1 & 0 \times 5 & 0 \times 7 \\ 1 \times 1 & 1 \times 5 & 1 \times 7 \\ 2 \times 1 & 2 \times 5 & 2 \times 7 \end{array} \right] = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{array} \right] \)
Now, find the transpose of \( AB \):
L.H.S. \( = (AB)' = \left[ \begin{array}{ccc} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{array} \right] \)
Next, find \( B'A' \):
R.H.S. \( = B'A' = \left[ \begin{array}{c} 1 \\ 5 \\ 7 \end{array} \right] \left[ \begin{array}{ccc} 0 & 1 & 2 \end{array} \right] = \left[ \begin{array}{ccc} 1 \times 0 & 1 \times 1 & 1 \times 2 \\ 5 \times 0 & 5 \times 1 & 5 \times 2 \\ 7 \times 0 & 7 \times 1 & 7 \times 2 \end{array} \right] = \left[ \begin{array}{ccc} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{array} \right] \)
Since L.H.S. \( = \) R.H.S., the identity \( (AB)' = B'A' \) is verified.
In simple words: This problem shows that if you multiply two matrices and then take the transpose, it is the same as taking the transpose of each matrix individually and then multiplying them in reverse order (B'A' instead of A'B'). This is a key property of matrix transposes.

Exam Tip: Remember the order change for matrix multiplication and transpose: \( (AB)' \) is always \( B'A' \), not \( A'B' \). Matrix multiplication is not commutative, so order matters greatly.

Question 6. If:
(i) \( A = \left[ \begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array} \right] \), then verify that \( A'A = I \).
(ii) \( A = \left[ \begin{array}{cc} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array} \right] \), then verify that \( A'A = I \).
Answer:
(i) Given \( A = \left[ \begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array} \right] \).
Then \( A' = \left[ \begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array} \right] \).
Now, calculate \( A'A \):
\( A'A = \left[ \begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array} \right] \left[ \begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array} \right] \)
\( = \left[ \begin{array}{cc} \cos^2 \alpha + (-\sin \alpha)(-\sin \alpha) & \cos \alpha \sin \alpha + (-\sin \alpha)\cos \alpha \\ \sin \alpha \cos \alpha + \cos \alpha (-\sin \alpha) & \sin^2 \alpha + \cos^2 \alpha \end{array} \right] \)
\( = \left[ \begin{array}{cc} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{array} \right] \)
Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\( = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = I \).
Thus, \( A'A = I \) is verified.

(ii) Given \( A = \left[ \begin{array}{cc} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array} \right] \).
Then \( A' = \left[ \begin{array}{cc} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array} \right] \).
Now, calculate \( A'A \):
\( A'A = \left[ \begin{array}{cc} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array} \right] \left[ \begin{array}{cc} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array} \right] \)
\( = \left[ \begin{array}{cc} \sin^2 \alpha + (-\cos \alpha)(-\cos \alpha) & \sin \alpha \cos \alpha + (-\cos \alpha)\sin \alpha \\ \cos \alpha \sin \alpha + \sin \alpha (-\cos \alpha) & \cos^2 \alpha + \sin^2 \alpha \end{array} \right] \)
\( = \left[ \begin{array}{cc} \sin^2 \alpha + \cos^2 \alpha & \sin \alpha \cos \alpha - \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha - \sin \alpha \cos \alpha & \cos^2 \alpha + \sin^2 \alpha \end{array} \right] \)
Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\( = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = I \).
Thus, \( A'A = I \) is verified.
In simple words: These problems check if a matrix, when multiplied by its transpose, results in an identity matrix. This means the matrix is orthogonal. We perform the matrix multiplication and use the trigonometric identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \) to show that the result is indeed the identity matrix.

Exam Tip: Matrices where \( A'A = I \) (or \( AA' = I \)) are called orthogonal matrices. They are very important in linear algebra and transformations. Always remember the fundamental trigonometric identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \).

Question 7.
(i) Show that the matrix \( A = \left[ \begin{array}{ccc} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{array} \right] \) is a symmetric matrix.
(ii) Show that the matrix \( A = \left[ \begin{array}{ccc} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{array} \right] \) is a skew-symmetric matrix.
Answer:
(i) For a symmetric matrix, we must have \( A' = A \) (or \( a_{ij} = a_{ji} \) for all i, j).
Given \( A = \left[ \begin{array}{ccc} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{array} \right] \).
Let's find the transpose of \( A \):
\( A' = \left[ \begin{array}{ccc} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{array} \right]' = \left[ \begin{array}{ccc} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{array} \right] \)
Since \( A' = A \), the matrix \( A \) is a symmetric matrix.

(ii) For a skew-symmetric matrix, we must have \( A' = -A \) (or \( a_{ij} = -a_{ji} \) and \( a_{ii} = 0 \)).
Given \( A = \left[ \begin{array}{ccc} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{array} \right] \).
Let's find the transpose of \( A \):
\( A' = \left[ \begin{array}{ccc} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{array} \right]' = \left[ \begin{array}{ccc} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{array} \right] \)
Now, let's find \( -A \):
\( -A = - \left[ \begin{array}{ccc} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{array} \right] = \left[ \begin{array}{ccc} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{array} \right] \)
Since \( A' = -A \), the matrix \( A \) is a skew-symmetric matrix.
In simple words: Part (i) asks to confirm if a matrix is symmetric, which means it stays the same when you flip its rows and columns. We show that A and A' are identical. Part (ii) asks to confirm if a matrix is skew-symmetric, which means that flipping its rows and columns gives you the negative of the original matrix. We find A' and show it equals -A.

Exam Tip: To prove a matrix is symmetric, show \( A' = A \). To prove it's skew-symmetric, show \( A' = -A \). Also, remember that all diagonal elements of a skew-symmetric matrix must be zero.

Question 8. For a matrix \( A = \left[ \begin{array}{cc} 1 & 5 \\ 6 & 7 \end{array} \right] \), verify that:
(i) \( (A + A') \) is a symmetric matrix.
(ii) \( (A - A') \) is a skew-symmetric matrix.
Answer:
Given \( A = \left[ \begin{array}{cc} 1 & 5 \\ 6 & 7 \end{array} \right] \).
First, find \( A' \):
\( A' = \left[ \begin{array}{cc} 1 & 6 \\ 5 & 7 \end{array} \right] \)

(i) To verify \( (A + A') \) is a symmetric matrix:
Calculate \( A + A' \):
\( A + A' = \left[ \begin{array}{cc} 1 & 5 \\ 6 & 7 \end{array} \right] + \left[ \begin{array}{cc} 1 & 6 \\ 5 & 7 \end{array} \right] = \left[ \begin{array}{cc} 1+1 & 5+6 \\ 6+5 & 7+7 \end{array} \right] = \left[ \begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array} \right] \)
Let \( B = A + A' = \left[ \begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array} \right] \).
Now, find the transpose of \( B \):
\( B' = \left[ \begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array} \right]' = \left[ \begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array} \right] \)
Since \( B' = B \), \( (A + A') \) is a symmetric matrix. This is verified.

(ii) To verify \( (A - A') \) is a skew-symmetric matrix:
Calculate \( A - A' \):
\( A - A' = \left[ \begin{array}{cc} 1 & 5 \\ 6 & 7 \array} \right] - \left[ \begin{array}{cc} 1 & 6 \\ 5 & 7 \array} \right] = \left[ \begin{array}{cc} 1-1 & 5-6 \\ 6-5 & 7-7 \end{array} \right] = \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right] \)
Let \( C = A - A' = \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right] \).
Now, find the transpose of \( C \):
\( C' = \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right]' = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] \)
Now, find \( -C \):
\( -C = - \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] \)
Since \( C' = -C \), \( (A - A') \) is a skew-symmetric matrix. This is verified.
In simple words: This problem shows that if you add a matrix to its own transpose, the result is always a symmetric matrix. Also, if you subtract a matrix from its transpose, the result is always a skew-symmetric matrix. We do the calculations for both cases and then check the conditions for symmetric and skew-symmetric matrices.

Exam Tip: Remember these fundamental properties: \( (A+A') \) is always symmetric, and \( (A-A') \) is always skew-symmetric. These properties are useful for decomposing any square matrix into the sum of a symmetric and a skew-symmetric matrix.

Question 8. For a matrix \(A = \left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right]\), verify that
(i) \((A + A')\) is a symmetric matrix.
(ii) \((A - A')\) is a skew symmetric matrix.

Answer:
(i) Given matrix \(A = \left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right]\).
The transpose of matrix A is \(A' = \left[\begin{array}{ll} 1 & 6 \\ 5 & 7 \end{array}\right]\).
Now, we calculate \(A + A'\):
\(A + A' = \left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right] + \left[\begin{array}{ll} 1 & 6 \\ 5 & 7 \end{array}\right] = \left[\begin{array}{ll} 1+1 & 5+6 \\ 6+5 & 7+7 \end{array}\right] = \left[\begin{array}{ll} 2 & 11 \\ 11 & 14 \end{array}\right]\)
Let \(M = A + A'\). To check if M is symmetric, we need to verify if \(M = M'\).
The elements are \(m_{21} = 11\) and \(m_{12} = 11\), so \(m_{21} = m_{12}\). The diagonal elements are \(m_{11} = 2\) and \(m_{22} = 14\).
Since \(M' = \left[\begin{array}{ll} 2 & 11 \\ 11 & 14 \end{array}\right]\) which is equal to M, therefore, \((A + A')\) is a symmetric matrix.
(ii) Now, we calculate \(A - A'\):
\(A - A' = \left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right] - \left[\begin{array}{ll} 1 & 6 \\ 5 & 7 \end{array}\right] = \left[\begin{array}{ll} 1-1 & 5-6 \\ 6-5 & 7-7 \end{array}\right] = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]\)
Let \(N = A - A'\). To check if N is skew-symmetric, we need to verify if \(N' = -N\).
The transpose of N is \(N' = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]\).
Now, let's find \(-N\):
\(-N = -\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & -(-1) \\ -1 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]\)
Since \(N' = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]\) and \(-N = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]\), we can see that \(N' = -N\).
Therefore, \((A - A')\) is a skew-symmetric matrix.
In simple words: To check if a matrix is symmetric, compare it to its transpose; they should be exactly the same. For a skew-symmetric matrix, its transpose should be the negative of the original matrix. We performed these checks for both parts of the problem.

Exam Tip: Remember that for a symmetric matrix \(M\), \(M = M'\), and for a skew-symmetric matrix \(N\), \(N = -N'\) (or \(N' = -N\)). The diagonal elements of a skew-symmetric matrix are always zero.

Question 9. Find \(\frac { 1 }{ 2 }(A +A')\) and \(\frac {1}{ 2 }(A-A')\), when \(A = \left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\).
Answer:
Given matrix \(A = \left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\).
First, find the transpose of A, denoted as \(A'\):
\(A' = \left[\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right]\)
Next, calculate the sum \(A + A'\):
\(A+A' = \left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right] + \left[\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right]\)
\(A+A' = \left[\begin{array}{ccc} 0+0 & a+(-a) & b+(-b) \\ -a+a & 0+0 & c+(-c) \\ -b+b & -c+c & 0+0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
Now, find \(\frac{1}{2}(A + A')\):
\(\frac{1}{2}(A + A') = \frac{1}{2} \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
Next, calculate the difference \(A - A'\):
\(A-A' = \left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right] - \left[\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right]\)
\(A-A' = \left[\begin{array}{ccc} 0-0 & a-(-a) & b-(-b) \\ -a-a & 0-0 & c-(-c) \\ -b-b & -c-c & 0-0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{array}\right]\)
Finally, find \(\frac{1}{2}(A - A')\):
\(\frac{1}{2}(A-A') = \frac{1}{2} \left[\begin{array}{ccc} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\)
In simple words: We first found the transpose of matrix A. Then, we calculated half of the sum of A and its transpose, which resulted in a zero matrix. After that, we found half of the difference between A and its transpose, which brought us back to the original matrix A.

Exam Tip: Always remember that any square matrix A can be expressed as the sum of a symmetric and a skew-symmetric matrix: \(A = \frac{1}{2}(A+A') + \frac{1}{2}(A-A')\).

Question 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix.
(i) \(\left[\begin{array}{ll} 3 & 5 \\ 1 & -1 \end{array}\right]\)
(ii) \(\left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\)
(iii) \(\left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right]\)
(iv) \(\left[\begin{array}{ll} 1 & 5 \\ -1 & 2 \end{array}\right]\)
Answer:
For any square matrix A, we can write it as \(A = P + Q\), where \(P\) is a symmetric matrix and \(Q\) is a skew-symmetric matrix. The formulas are:
\(P = \frac{1}{2}(A + A')\)
\(Q = \frac{1}{2}(A - A')\)

(i) Let \(A = \left[\begin{array}{ll} 3 & 5 \\ 1 & -1 \end{array}\right]\).
The transpose is \(A' = \left[\begin{array}{cc} 3 & 1 \\ 5 & -1 \end{array}\right]\).
Calculate \(A + A'\):
\(A + A' = \left[\begin{array}{ll} 3 & 5 \\ 1 & -1 \end{array}\right] + \left[\begin{array}{cc} 3 & 1 \\ 5 & -1 \end{array}\right] = \left[\begin{array}{cc} 3+3 & 5+1 \\ 1+5 & -1-1 \end{array}\right] = \left[\begin{array}{cc} 6 & 6 \\ 6 & -2 \end{array}\right]\)
Now, find \(P = \frac{1}{2}(A + A')\):
\(P = \frac{1}{2} \left[\begin{array}{cc} 6 & 6 \\ 6 & -2 \end{array}\right] = \left[\begin{array}{cc} 3 & 3 \\ 3 & -1 \end{array}\right]\). This is a symmetric matrix.
Calculate \(A - A'\):
\(A - A' = \left[\begin{array}{ll} 3 & 5 \\ 1 & -1 \end{array}\right] - \left[\begin{array}{cc} 3 & 1 \\ 5 & -1 \end{array}\right] = \left[\begin{array}{cc} 3-3 & 5-1 \\ 1-5 & -1-(-1) \end{array}\right] = \left[\begin{array}{cc} 0 & 4 \\ -4 & 0 \end{array}\right]\)
Now, find \(Q = \frac{1}{2}(A - A')\):
\(Q = \frac{1}{2} \left[\begin{array}{cc} 0 & 4 \\ -4 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & 2 \\ -2 & 0 \end{array}\right]\). This is a skew-symmetric matrix.
Thus, \(A = P + Q = \left[\begin{array}{cc} 3 & 3 \\ 3 & -1 \end{array}\right] + \left[\begin{array}{cc} 0 & 2 \\ -2 & 0 \end{array}\right]\).
(ii) Let \(A = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\).
The transpose is \(A' = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\). (Note: A is already symmetric)
Calculate \(A + A'\):
\(A + A' = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right] + \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right] = \left[\begin{array}{ccc} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{array}\right]\)
Now, find \(P = \frac{1}{2}(A + A')\):
\(P = \frac{1}{2} \left[\begin{array}{ccc} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{array}\right] = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\). This is a symmetric matrix.
Calculate \(A - A'\):
\(A - A' = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right] - \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
Now, find \(Q = \frac{1}{2}(A - A')\):
\(Q = \frac{1}{2} \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\). This is a skew-symmetric matrix.
Thus, \(A = P + Q = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right] + \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\).
(iii) Let \(A = \left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right]\).
The transpose is \(A' = \left[\begin{array}{ccc} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right]\).
Calculate \(A + A'\):
\(A + A' = \left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right] + \left[\begin{array}{ccc} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right] = \left[\begin{array}{ccc} 3+3 & 3-2 & -1-4 \\ -2+3 & -2-2 & 1-5 \\ -4-1 & -5+1 & 2+2 \end{array}\right] = \left[\begin{array}{ccc} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{array}\right]\)
Now, find \(P = \frac{1}{2}(A + A')\):
\(P = \frac{1}{2} \left[\begin{array}{ccc} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{array}\right] = \left[\begin{array}{ccc} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{array}\right]\). This is a symmetric matrix.
Calculate \(A - A'\):
\(A - A' = \left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right] - \left[\begin{array}{ccc} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right] = \left[\begin{array}{ccc} 3-3 & 3-(-2) & -1-(-4) \\ -2-3 & -2-(-2) & 1-(-5) \\ -4-(-1) & -5-1 & 2-2 \end{array}\right] = \left[\begin{array}{ccc} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{array}\right]\)
Now, find \(Q = \frac{1}{2}(A - A')\):
\(Q = \frac{1}{2} \left[\begin{array}{ccc} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{array}\right]\). This is a skew-symmetric matrix.
Thus, \(A = P + Q = \left[\begin{array}{ccc} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{array}\right] + \left[\begin{array}{ccc} 0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{array}\right]\).
(iv) Let \(A = \left[\begin{array}{ll} 1 & 5 \\ -1 & 2 \end{array}\right]\).
The transpose is \(A' = \left[\begin{array}{cc} 1 & -1 \\ 5 & 2 \end{array}\right]\).
Calculate \(A + A'\):
\(A + A' = \left[\begin{array}{ll} 1 & 5 \\ -1 & 2 \end{array}\right] + \left[\begin{array}{cc} 1 & -1 \\ 5 & 2 \end{array}\right] = \left[\begin{array}{cc} 1+1 & 5-1 \\ -1+5 & 2+2 \end{array}\right] = \left[\begin{array}{cc} 2 & 4 \\ 4 & 4 \end{array}\right]\)
Now, find \(P = \frac{1}{2}(A + A')\):
\(P = \frac{1}{2} \left[\begin{array}{cc} 2 & 4 \\ 4 & 4 \end{array}\right] = \left[\begin{array}{cc} 1 & 2 \\ 2 & 2 \end{array}\right]\). This is a symmetric matrix.
Calculate \(A - A'\):
\(A - A' = \left[\begin{array}{ll} 1 & 5 \\ -1 & 2 \end{array}\right] - \left[\begin{array}{cc} 1 & -1 \\ 5 & 2 \end{array}\right] = \left[\begin{array}{cc} 1-1 & 5-(-1) \\ -1-5 & 2-2 \end{array}\right] = \left[\begin{array}{cc} 0 & 6 \\ -6 & 0 \end{array}\right]\)
Now, find \(Q = \frac{1}{2}(A - A')\):
\(Q = \frac{1}{2} \left[\begin{array}{cc} 0 & 6 \\ -6 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & 3 \\ -3 & 0 \end{array}\right]\). This is a skew-symmetric matrix.
Thus, \(A = P + Q = \left[\begin{array}{cc} 1 & 2 \\ 2 & 2 \end{array}\right] + \left[\begin{array}{cc} 0 & 3 \\ -3 & 0 \end{array}\right]\).
In simple words: For each matrix, we found its symmetric part by adding the matrix to its transpose and then dividing by two. We found its skew-symmetric part by subtracting its transpose and dividing by two. The original matrix can then be shown as the sum of these two new matrices.

Exam Tip: Always double-check your calculations for \(A+A'\) and \(A-A'\), especially when dealing with negative numbers. Verify that P is indeed symmetric (\(P=P'\)) and Q is skew-symmetric (\(Q=-Q'\)) before writing the final sum.

Question 11. If A and B are symmetric matrices of the same order, then AB – BA is a ____ matrix.
(a) Skew symmetric matrix
(b) Symmetric matrix
(c) Zero matrix
(d) Identity matrix
Answer: (a) Skew symmetric matrix
In simple words: When you have two matrices, A and B, that are symmetric (meaning they equal their own transposes), and you subtract BA from AB, the result is always a skew-symmetric matrix.

Exam Tip: This is a standard property of symmetric matrices. The proof relies on properties of transpose: \((AB-BA)' = (AB)' - (BA)' = B'A' - A'B'\). Since A and B are symmetric, \(A'=A\) and \(B'=B\), so \((AB-BA)' = BA - AB = -(AB-BA)\), which proves it is skew-symmetric.

Question 12. If \(A = \left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\) then \(A + A'= I\), if the value of \(\alpha\) is
(A) \(\frac{\pi}{2}\)
(B) \(\frac{\pi}{3}\)
(C) \(\pi\)
(D) \(\frac{3\pi}{2}\)
Answer: (B) \(\frac{\pi}{3}\)
In simple words: When you add matrix A to its transpose and the result is the identity matrix, it means that twice the cosine of alpha must be 1. This helps us find the specific angle alpha.

Exam Tip: For problems involving trigonometric functions in matrices, always recall the values of common angles for sine and cosine. Also, carefully perform matrix addition and equality checks.