Exercise 3.3 Practice
Transpose of Matrices, Symmetric & Skew Symmetric Matrices
Q1
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Find the transpose of each of the following matrices:
(i) \( \begin{bmatrix} 7 \\ 2/3 \\ -4 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} \)
(iii) \( \begin{bmatrix} -2 & 6 & 7 \\ \sqrt{5} & 6 & 7 \\ 3 & 4 & -2 \end{bmatrix} \)
(i) \( \begin{bmatrix} 7 \\ 2/3 \\ -4 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} \)
(iii) \( \begin{bmatrix} -2 & 6 & 7 \\ \sqrt{5} & 6 & 7 \\ 3 & 4 & -2 \end{bmatrix} \)
Part (i)
The transpose converts columns to rows.
\( [7 \quad 2/3 \quad -4] \).
\( [7 \quad 2/3 \quad -4] \).
Part (ii)
Swap rows and columns:
\( \begin{bmatrix} 2 & 4 \\ -3 & 5 \end{bmatrix} \).
\( \begin{bmatrix} 2 & 4 \\ -3 & 5 \end{bmatrix} \).
Part (iii)
\( \begin{bmatrix} -2 & \sqrt{5} & 3 \\ 6 & 6 & 4 \\ 7 & 7 & -2 \end{bmatrix} \).
Q2
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If \( A = \begin{bmatrix} -2 & 3 & 4 \\ 6 & 8 & 10 \\ -3 & 2 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} -5
& 2 & -6 \\ 2 & 3 & 1 \\ 2 & 4 & 2 \end{bmatrix} \), verify that:
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
Part (i) Verification
\( A+B = \begin{bmatrix} -7 & 5 & -2 \\ 8 & 11 & 11 \\ -1 & 6 & 4 \end{bmatrix} \). Transpose: \(
\begin{bmatrix} -7 & 8 & -1 \\ 5 & 11 & 6 \\ -2 & 11 & 4 \end{bmatrix} \).
\( A' + B' = \begin{bmatrix} -2 & 6 & -3 \\ 3 & 8 & 2 \\ 4 & 10 & 2 \end{bmatrix} + \begin{bmatrix} -5 & 2 & 2 \\ 2 & 3 & 4 \\ -6 & 1 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 8 & -1 \\ 5 & 11 & 6 \\ -2 & 11 & 4 \end{bmatrix} \). Matches.
\( A' + B' = \begin{bmatrix} -2 & 6 & -3 \\ 3 & 8 & 2 \\ 4 & 10 & 2 \end{bmatrix} + \begin{bmatrix} -5 & 2 & 2 \\ 2 & 3 & 4 \\ -6 & 1 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 8 & -1 \\ 5 & 11 & 6 \\ -2 & 11 & 4 \end{bmatrix} \). Matches.
Part (ii) Verification
Similar process verifies that the transpose of the difference equals the difference of the transposes.
Q3
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If \( A' = \begin{bmatrix} 4 & 5 \\ -2 & 3 \\ 1 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} -2 & 3 & 2 \\ 2
& 3 & 4 \end{bmatrix} \), verify that:
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
Step 1: Find Matrix A
\( A = (A')' = \begin{bmatrix} 4 & -2 & 1 \\ 5 & 3 & 2 \end{bmatrix} \).
Step 2: Verify (i)
\( A+B = \begin{bmatrix} 2 & 1 & 3 \\ 7 & 6 & 6 \end{bmatrix} \).
\( (A+B)' = \begin{bmatrix} 2 & 7 \\ 1 & 6 \\ 3 & 6 \end{bmatrix} \).
\( A' + B' = \begin{bmatrix} 4 & 5 \\ -2 & 3 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 2 \\ 3 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ 1 & 6 \\ 3 & 6 \end{bmatrix} \). Verified.
\( (A+B)' = \begin{bmatrix} 2 & 7 \\ 1 & 6 \\ 3 & 6 \end{bmatrix} \).
\( A' + B' = \begin{bmatrix} 4 & 5 \\ -2 & 3 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 2 \\ 3 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ 1 & 6 \\ 3 & 6 \end{bmatrix} \). Verified.
Q4
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If \( A' = \begin{bmatrix} -3 & 4 \\ 2 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} -2 & 2 \\ 1 & 3
\end{bmatrix} \), then find \( (A + 2B)' \).
Step 1: Find A
\( A = \begin{bmatrix} -3 & 2 \\ 4 & 3 \end{bmatrix} \).
Step 2: Calculate A + 2B
\( 2B = \begin{bmatrix} -4 & 4 \\ 2 & 6 \end{bmatrix} \).
\( A + 2B = \begin{bmatrix} -3 & 2 \\ 4 & 3 \end{bmatrix} + \begin{bmatrix} -4 & 4 \\ 2 & 6 \end{bmatrix} = \begin{bmatrix} -7 & 6 \\ 6 & 9 \end{bmatrix} \).
\( A + 2B = \begin{bmatrix} -3 & 2 \\ 4 & 3 \end{bmatrix} + \begin{bmatrix} -4 & 4 \\ 2 & 6 \end{bmatrix} = \begin{bmatrix} -7 & 6 \\ 6 & 9 \end{bmatrix} \).
\( (A+2B)' = \begin{bmatrix} -7 & 6 \\ 6 & 9 \end{bmatrix} \)
Q5
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For the matrices A and B, verify that \( (AB)' = B'A' \) where:
(i) \( A = \begin{bmatrix} 2 \\ -5 \\ 4 \end{bmatrix} \), \( B = \begin{bmatrix} -2 & 3 & 2 \end{bmatrix} \)
(ii) \( A = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \), \( B = \begin{bmatrix} 2 & 6 & 8 \end{bmatrix} \)
(i) \( A = \begin{bmatrix} 2 \\ -5 \\ 4 \end{bmatrix} \), \( B = \begin{bmatrix} -2 & 3 & 2 \end{bmatrix} \)
(ii) \( A = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \), \( B = \begin{bmatrix} 2 & 6 & 8 \end{bmatrix} \)
Part (i) Verification
\( AB = \begin{bmatrix} -4 & 6 & 4 \\ 10 & -15 & -10 \\ -8 & 12 & 8 \end{bmatrix} \).
\( (AB)' = \begin{bmatrix} -4 & 10 & -8 \\ 6 & -15 & 12 \\ 4 & -10 & 8 \end{bmatrix} \).
\( B'A' = \begin{bmatrix} -2 \\ 3 \\ 2 \end{bmatrix} [2 \quad -5 \quad 4] = \begin{bmatrix} -4 & 10 & -8 \\ 6 & -15 & 12 \\ 4 & -10 & 8 \end{bmatrix} \). Verified.
\( (AB)' = \begin{bmatrix} -4 & 10 & -8 \\ 6 & -15 & 12 \\ 4 & -10 & 8 \end{bmatrix} \).
\( B'A' = \begin{bmatrix} -2 \\ 3 \\ 2 \end{bmatrix} [2 \quad -5 \quad 4] = \begin{bmatrix} -4 & 10 & -8 \\ 6 & -15 & 12 \\ 4 & -10 & 8 \end{bmatrix} \). Verified.
Q6
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(i) If \( A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} \), verify
that \( A'A = I \).
(ii) If \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \), verify that \( A'A = I \).
(ii) If \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \), verify that \( A'A = I \).
Part (i) Verification
\( A' = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \).
\( A'A = \begin{bmatrix} \sin^2\alpha+\cos^2\alpha & \sin\alpha\cos\alpha-\cos\alpha\sin\alpha \\ -\cos\alpha\sin\alpha+\sin\alpha\cos\alpha & \cos^2\alpha+\sin^2\alpha \end{bmatrix} \).
\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \).
\( A'A = \begin{bmatrix} \sin^2\alpha+\cos^2\alpha & \sin\alpha\cos\alpha-\cos\alpha\sin\alpha \\ -\cos\alpha\sin\alpha+\sin\alpha\cos\alpha & \cos^2\alpha+\sin^2\alpha \end{bmatrix} \).
\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \).
Q7
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(i) Show that the matrix \( A = \begin{bmatrix} 2 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 4 \end{bmatrix} \) is a
symmetric matrix.
(ii) Show that the matrix \( A = \begin{bmatrix} 0 & 2 & -2 \\ -2 & 0 & 2 \\ 2 & -2 & 0 \end{bmatrix} \) is a skew symmetric matrix.
(ii) Show that the matrix \( A = \begin{bmatrix} 0 & 2 & -2 \\ -2 & 0 & 2 \\ 2 & -2 & 0 \end{bmatrix} \) is a skew symmetric matrix.
Part (i)
Find \( A' \). If \( A' = A \), it is symmetric. Here, rows and columns are identical. Yes.
Part (ii)
Find \( A' \). \( A' = \begin{bmatrix} 0 & -2 & 2 \\ 2 & 0 & -2 \\ -2 & 2 & 0 \end{bmatrix} \).
This is exactly \( -A \). Yes, Skew Symmetric.
This is exactly \( -A \). Yes, Skew Symmetric.
Q8
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For the matrix \( A = \begin{bmatrix} 2 & 6 \\ 7 & 8 \end{bmatrix} \), verify that:
(i) \( (A + A') \) is a symmetric matrix.
(ii) \( (A - A') \) is a skew symmetric matrix.
(i) \( (A + A') \) is a symmetric matrix.
(ii) \( (A - A') \) is a skew symmetric matrix.
\( A' = \begin{bmatrix} 2 & 7 \\ 6 & 8 \end{bmatrix} \).
(i) \( A+A' = \begin{bmatrix} 4 & 13 \\ 13 & 16 \end{bmatrix} \) (Symmetric).
(ii) \( A-A' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \) (Skew Symmetric).
(i) \( A+A' = \begin{bmatrix} 4 & 13 \\ 13 & 16 \end{bmatrix} \) (Symmetric).
(ii) \( A-A' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \) (Skew Symmetric).
Q9
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Find \( \frac{1}{2}(A + A') \) and \( \frac{1}{2}(A - A') \), when \( A = \begin{bmatrix} 0 & x & y \\ -x & 0 &
z \\ -y & -z & 0 \end{bmatrix} \).
\( A' = \begin{bmatrix} 0 & -x & -y \\ x & 0 & -z \\ y & z & 0 \end{bmatrix} = -A \).
Since A is skew-symmetric:
\( \frac{1}{2}(A + A') = \frac{1}{2}(A - A) = O \) (Zero Matrix).
\( \frac{1}{2}(A - A') = \frac{1}{2}(A - (-A)) = A \).
Since A is skew-symmetric:
\( \frac{1}{2}(A + A') = \frac{1}{2}(A - A) = O \) (Zero Matrix).
\( \frac{1}{2}(A - A') = \frac{1}{2}(A - (-A)) = A \).
Q10
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Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
(i) \( \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 4 & -3 & 3 \\ -3 & 4 & -2 \\ 3 & -2 & 4 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 2 & 2 & -2 \\ -3 & -3 & 2 \\ -5 & -6 & 3 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 2 & 6 \\ -2 & 3 \end{bmatrix} \)
(i) \( \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 4 & -3 & 3 \\ -3 & 4 & -2 \\ 3 & -2 & 4 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 2 & 2 & -2 \\ -3 & -3 & 2 \\ -5 & -6 & 3 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 2 & 6 \\ -2 & 3 \end{bmatrix} \)
Method
\( A = P + Q \), where \( P = \frac{1}{2}(A+A') \) (Symmetric) and \( Q = \frac{1}{2}(A-A') \) (Skew).
Solution for (i)
\( A = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix} \).
\( P = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} \). \( Q = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \).
\( A = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} + \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \).
\( P = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} \). \( Q = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \).
\( A = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} + \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \).
Q11
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If A, 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
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
Let \( X = AB - BA \).
\( X' = (AB - BA)' = (AB)' - (BA)' \)
\( = B'A' - A'B' \). Since A, B symmetric (\( A'=A, B'=B \)):
\( = BA - AB = -(AB - BA) = -X \).
\( X' = (AB - BA)' = (AB)' - (BA)' \)
\( = B'A' - A'B' \). Since A, B symmetric (\( A'=A, B'=B \)):
\( = BA - AB = -(AB - BA) = -X \).
Correct Option: (A) Skew symmetric matrix
Q12
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If \( A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} \), then \( A +
A' = I \), if the value of \( \alpha \) is:
(A) \( \pi/6 \) (B) \( \pi/3 \) (C) \( \pi \) (D) \( 3\pi/2 \)
(A) \( \pi/6 \) (B) \( \pi/3 \) (C) \( \pi \) (D) \( 3\pi/2 \)
\( A + A' = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} +
\begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \)
\( = \begin{bmatrix} 2\sin \alpha & 0 \\ 0 & 2\sin \alpha \end{bmatrix} \).
Equate to \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
\( 2\sin \alpha = 1 \Rightarrow \sin \alpha = 1/2 \).
\( = \begin{bmatrix} 2\sin \alpha & 0 \\ 0 & 2\sin \alpha \end{bmatrix} \).
Equate to \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
\( 2\sin \alpha = 1 \Rightarrow \sin \alpha = 1/2 \).
Correct Option: (A) \( \pi/6 \)