Question 1: Find the transpose of each of the following matrices:
(i) \begin{pmatrix}5\\[4pt]\tfrac12\\[4pt]-1\end{pmatrix}
(ii) \begin{pmatrix}1 & -1\\[4pt]2 & 3\end{pmatrix}
(iii) \begin{pmatrix}-1 & 5 & 6\\[4pt]\sqrt{3} & 5 & 6\\[4pt]2 & 3 & -1\end{pmatrix}
Solution 1:
(i) Let A=\begin{pmatrix}5\\\tfrac12\\-1\end{pmatrix}.
Then
A^T = \begin{pmatrix}5 & \tfrac12 & -1\end{pmatrix}
(ii) Let B=\begin{pmatrix}1 & -1\\2 & 3\end{pmatrix}.
Then
B^T = \begin{pmatrix}1 & 2\\-1 & 3\end{pmatrix}
(iii) Let C=\begin{pmatrix}-1 & 5 & 6\\\sqrt{3} & 5 & 6\\2 & 3 & -1\end{pmatrix}.
Then
C^T = \begin{pmatrix}-1 & \sqrt{3} & 2\\5 & 5 & 3\\6 & 6 & -1\end{pmatrix}
Question 2 If
A=\begin{pmatrix}-1 & 2 & 3\\5 & 7 & 9\\-2 & 1 & 1\end{pmatrix} and B=\begin{pmatrix}-4 & 1 & -5\\1 & 2 & 0\\1 & 3 & 1\end{pmatrix}
verify that
(i) (A+B)^T = A^T + B^T
(ii) (A-B)^T = A^T - B^T
Solution 2
(i) Compute A+B:
A+B=\begin{pmatrix}-5 & 3 & -2\\6 & 9 & 9\\-1 & 4 & 2\end{pmatrix}
Then
(A+B)^T = \begin{pmatrix}-5 & 6 & -1\\3 & 9 & 4\\-2 & 9 & 2\end{pmatrix}
Now compute A^T + B^T:
A^T=\begin{pmatrix}-1 & 5 & -2\\2 & 7 & 1\\3 & 9 & 1\end{pmatrix}, B^T=\begin{pmatrix}-4 & 1 & 1\\1 & 2 & 3\\-5 & 0 & 1\end{pmatrix}
A^T+B^T = \begin{pmatrix}-5 & 6 & -1\\3 & 9 & 4\\-2 & 9 & 2\end{pmatrix}
Thus (i) is verified.
(ii) Compute A-B:
A-B=\begin{pmatrix}3 & 1 & 8\\4 & 5 & 9\\-3 & -2 & 0\end{pmatrix}
Then
(A-B)^T = \begin{pmatrix}3 & 4 & -3\\1 & 5 & -2\\8 & 9 & 0\end{pmatrix}
A^T-B^T = \begin{pmatrix}3 & 4 & -3\\1 & 5 & -2\\8 & 9 & 0\end{pmatrix}
Thus (ii) is verified.
Question 3
If A^T = \begin{pmatrix}3 & 4\\-1 & 2\\0 & 1\end{pmatrix} and B = \begin{pmatrix}-1 & 2 & 1\\1 & 2 & 3\end{pmatrix}
verify that
(i) (A+B)^T = A^T + B^T
(ii) (A-B)^T = A^T - B^T
Solution 3
First recover A by transposing A^T:
A = \begin{pmatrix}3 & -1 & 0\\4 & 2 & 1\end{pmatrix}
(i). Compute A+B:
A+B = \begin{pmatrix}2 & 1 & 1\\5 & 4 & 4\end{pmatrix}
Then
(A+B)^T = \begin{pmatrix}2 & 5\\1 & 4\\1 & 4\end{pmatrix}
B^T = \begin{pmatrix}-1 & 1\\2 & 2\\1 & 3\end{pmatrix}
A^T + B^T = \begin{pmatrix}2 & 5\\1 & 4\\1 & 4\end{pmatrix}
Therefore. (A+B)^T = A^T + B^T
Thus (i) is verified.
(ii). A-B = \begin{pmatrix}4 & -3 & -1\\3 & 0 & -2\end{pmatrix}
(A-B)^T = \begin{pmatrix}4 & 3\\-3 & 0\\-1 & -2\end{pmatrix}
Compute A^T - B^T
A^T - B^T = \begin{pmatrix}4 & 3\\-3 & 0\\-1 & -2\end{pmatrix}
Therefore (A-B)^T = A^T - B^T
Thus (ii) is verified.
Question 4: Given
A' = \begin{pmatrix}-2 & 3 \\ 1 & 2\end{pmatrix} and B = \begin{pmatrix}-1 & 0 \\ 1 & 2\end{pmatrix}
Find (A + 2B)^{\prime}
Solution 4
Compute 2B:
2B = 2 \begin{pmatrix}-1 & 0 \\ 1 & 2\end{pmatrix} = \begin{pmatrix}-2 & 0 \\ 2 & 4\end{pmatrix}
Then
A + 2B = \begin{pmatrix}-2 & 3 \\ 1 & 2\end{pmatrix} + \begin{pmatrix}-2 & 0 \\ 2 & 4\end{pmatrix} = \begin{pmatrix}-4 & 3 \\ 3 & 6\end{pmatrix}
Finally, take transpose:
\bigl(A + 2B\bigr)^{\prime} = \boxed{\begin{pmatrix}-4 & 3 \\ 3 & 6\end{pmatrix}}
Question 5: Verify that (AB)^{\prime} = B^{\prime} A^{\prime} for each of the following:
(i) A = \begin{pmatrix}1 \\ -4 \\ 3\end{pmatrix},\quad B = \begin{pmatrix}-1 & 2 & 1\end{pmatrix}
(ii) A = \begin{pmatrix}0 \\ 1 \\ 2\end{pmatrix},\quad B = \begin{pmatrix}1 & 5 & 7\end{pmatrix}
Solution 5
(i) Compute AB:
AB = \begin{pmatrix}1 \\ -4 \\ 3\end{pmatrix}\begin{pmatrix}-1 & 2 & 1\end{pmatrix} = \begin{pmatrix}-1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3\end{pmatrix}
Transpose:
(AB)^{\prime} = \begin{pmatrix}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{pmatrix}
Also compute B^{\prime}A^{\prime}:
B^{\prime} = \begin{pmatrix}-1 \\ 2 \\ 1\end{pmatrix},\quad A^{\prime} = \begin{pmatrix}1 & -4 & 3\end{pmatrix}
B^{\prime}A^{\prime} = \begin{pmatrix}-1 \\ 2 \\ 1\end{pmatrix}\begin{pmatrix}1 & -4 & 3\end{pmatrix} = \begin{pmatrix}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{pmatrix}
Thus (AB)′ = B′A′.
(ii) Compute AB:
AB = \begin{pmatrix}0 \\ 1 \\ 2\end{pmatrix}\begin{pmatrix}1 & 5 & 7\end{pmatrix} = \begin{pmatrix}0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14\end{pmatrix}
Transpose:
(AB)^{\prime} = \begin{pmatrix}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{pmatrix}
Also compute B^{\prime}A^{\prime}:
B^{\prime} = \begin{pmatrix}1 \\ 5 \\ 7\end{pmatrix},\quad A^{\prime} = \begin{pmatrix}0 & 1 & 2\end{pmatrix}
B^{\prime}A^{\prime} = \begin{pmatrix}1 \\ 5 \\ 7\end{pmatrix}\begin{pmatrix}0 & 1 & 2\end{pmatrix} = \begin{pmatrix}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{pmatrix}
Thus (AB)′ = B′A′.
Question 6: Show that A^{\prime}A = I_{2} for each of the following:
(i) A = \begin{pmatrix}\cos x & \sin x \\ -\sin x & \cos x\end{pmatrix}
(ii) A = \begin{pmatrix}\sin x & \cos x \\ -\cos x & \sin x\end{pmatrix}
Solution 6
(i) Compute A^{\prime}:
A^{\prime} = \begin{pmatrix}\cos x & -\sin x \\ \sin x & \cos x\end{pmatrix}
Then multiply:
A^{\prime}A = \begin{pmatrix}\cos x & -\sin x \\ \sin x & \cos x\end{pmatrix}\begin{pmatrix}\cos x & \sin x \\ -\sin x & \cos x\end{pmatrix}
A^{\prime}A = \begin{pmatrix}\cos^{2}x+\sin^{2}x & 0 \\ 0 & \sin^{2}x+\cos^{2}x\end{pmatrix}
A^{\prime}A= I_{2}
(ii) Compute A^{\prime}:
A^{\prime} = \begin{pmatrix}\sin x & -\cos x \\ \cos x & \sin x\end{pmatrix}
Then multiply:
A^{\prime}A = \begin{pmatrix}\sin x & -\cos x \\ \cos x & \sin x\end{pmatrix}\begin{pmatrix}\sin x & \cos x \\ -\cos x & \sin x\end{pmatrix}
A^{\prime}A = \begin{pmatrix}\sin^{2}x+\cos^{2}x & 0 \\ 0 & \cos^{2}x+\sin^{2}x\end{pmatrix}
A^{\prime}A= I_{2}
Question 7
(i) Show that the matrix
A = \begin{pmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{pmatrix} is symmetric.
(ii) Show that the matrix
A = \begin{pmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{pmatrix} is skew-symmetric.
Solution 7
(i) A matrix is symmetric if A = A^T.
Given A = \begin{pmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{pmatrix}
Compute transpose:
A^T = \begin{pmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{pmatrix}
Since A^T = A,
A is symmetric.
(ii) A matrix is skew-symmetric if A^T = -A.
Given A = \begin{pmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{pmatrix}
Compute transpose:
A^T = \begin{pmatrix}0 & -1 & 1\\1 & 0 & -1\\-1 & 1 & 0\end{pmatrix}
Compute -A:
-A = \begin{pmatrix}0 & -1 & 1\\1 & 0 & -1\\-1 & 1 & 0\end{pmatrix}
Since A^T = -A,
A is skew-symmetric.
Question 8: For the matrix
A = \begin{pmatrix}1 & 5\\6 & 7\end{pmatrix}
verify that:
(i) A + A^T is symmetric.
(ii) A - A^T is skew-symmetric.
Solution 8
(i) Compute A^T:
A^T = \begin{pmatrix}1 & 6\\5 & 7\end{pmatrix}
Then A + A^T = \begin{pmatrix}2 & 11\\11 & 14\end{pmatrix}
Transpose of this sum is: \begin{pmatrix}2 & 11\\11 & 14\end{pmatrix}^T = \begin{pmatrix}2 & 11\\11 & 14\end{pmatrix}
Thus it is symmetric.
(ii) Compute A - A^T = \begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}
Transpose of this difference is:
\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}^T = \begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}
Observe that \begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix} = -\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}
Thus it is skew-symmetric.
Question 9: If
A = \begin{pmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{pmatrix}
find: \tfrac12(A + A^T)\quad\text{and}\quad\tfrac12(A - A^T)
Solution 9:
Compute A^T:
A^T = \begin{pmatrix}0 & -a & -b\\a & 0 & -c\\b & c & 0\end{pmatrix}
Then:
A + A^T = \begin{pmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{pmatrix}
Hence \tfrac12(A + A^T) = \begin{pmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{pmatrix}
Also:
A - A^T = \begin{pmatrix}0 & 2a & 2b\\-2a & 0 & 2c\\-2b & -2c & 0\end{pmatrix}
Therefore \tfrac12(A - A^T) = \begin{pmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{pmatrix} = A
10. Express the following matrices as the sum of a symmetric and a skew‑symmetric matrix:
(i) \begin{pmatrix}3 & 5 \ 1 & -1\end{pmatrix}
(ii) \begin{pmatrix}6 & -2 & 2 \ -2 & 3 & -1 \ 2 & -1 & 3\end{pmatrix}
(iii) \begin{pmatrix}3 & 3 & -1 \ -2 & -2 & 1 \ -4 & -5 & 2\end{pmatrix}
(iv) \begin{pmatrix}1 & 5 \ -1 & 2\end{pmatrix}
Solution to 10
(i) Let A = \begin{pmatrix}3 & 5 \\ 1 & -1\end{pmatrix}.
Compute the
symmetric part S = \frac{1}{2}(A + A^T) and the
skew-symmetric part K = \frac{1}{2}(A - A^T).
We have A^T = \begin{pmatrix}3 & 1 \\ 5 & -1\end{pmatrix}.
So A + A^T = \begin{pmatrix}6 & 6 \\ 6 & -2\end{pmatrix}, hence S = \frac{1}{2}(A + A^T) = \begin{pmatrix}3 & 3 \\ 3 & -1\end{pmatrix}.
Also A - A^T = \begin{pmatrix}0 & 4 \\ -4 & 0\end{pmatrix}, so K = \frac{1}{2}(A - A^T) = \begin{pmatrix}0 & 2 \\ -2 & 0\end{pmatrix}.
\boxed {A = \begin{pmatrix}3 & 5 \\ 1 & -1\end{pmatrix} = \begin{pmatrix}3 & 3 \\ 3 & -1\end{pmatrix} + \begin{pmatrix}0 & 2 \\ -2 & 0\end{pmatrix}}
(ii) Let A = \begin{pmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{pmatrix}.
Compute the
symmetric part S = \frac{1}{2}(A + A^T) and the
skew-symmetric part K = \frac{1}{2}(A - A^T).
Then A^T = \begin{pmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{pmatrix} (same as A since A is already symmetric).
Thus S = \frac{1}{2}(A + A^T) = A = \begin{pmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{pmatrix},
and K = \frac{1}{2}(A - A^T) = \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}.
\boxed{A = \begin{pmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{pmatrix} = \begin{pmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{pmatrix} + \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix} }.
(iii) Let A = \begin{pmatrix}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{pmatrix}.
Compute the
symmetric part S = \frac{1}{2}(A + A^T) and the
skew-symmetric part K = \frac{1}{2}(A - A^T).
Compute A^T = \begin{pmatrix}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{pmatrix}.
Then A + A^T = \begin{pmatrix}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{pmatrix}, so
S = \frac{1}{2}(A + A^T) = \begin{pmatrix}3 & \tfrac{1}{2} & -\tfrac{5}{2} \\ \tfrac{1}{2} & -2 & -2 \\ -\tfrac{5}{2} & -2 & 2\end{pmatrix}.
Also A - A^T = \begin{pmatrix}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{pmatrix}, thus
K = \frac{1}{2}(A - A^T) = \begin{pmatrix}0 & \tfrac{5}{2} & \tfrac{3}{2} \\ -\tfrac{5}{2} & 0 & 3 \\ -\tfrac{3}{2} & -3 & 0\end{pmatrix}.
\boxed{ A = \begin{pmatrix}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{pmatrix} = \begin{pmatrix}3 & \tfrac{1}{2} & -\tfrac{5}{2} \\ \tfrac{1}{2} & -2 & -2 \\ -\tfrac{5}{2} & -2 & 2\end{pmatrix} + \begin{pmatrix}0 & \tfrac{5}{2} & \tfrac{3}{2} \\ -\tfrac{5}{2} & 0 & 3 \\ -\tfrac{3}{2} & -3 & 0\end{pmatrix}}
(iv) Let A = \begin{pmatrix}1 & 5 \\ -1 & 2\end{pmatrix}.
Compute the
symmetric part S = \frac{1}{2}(A + A^T) and the
skew-symmetric part K = \frac{1}{2}(A - A^T).
Then A^T = \begin{pmatrix}1 & -1 \\ 5 & 2\end{pmatrix}.
So A + A^T = \begin{pmatrix}2 & 4 \\ 4 & 4\end{pmatrix}, giving
S = \frac{1}{2}(A + A^T) = \begin{pmatrix}1 & 2 \\ 2 & 2\end{pmatrix}.
And A - A^T = \begin{pmatrix}0 & 6 \\ -6 & 0\end{pmatrix}, so
K = \frac{1}{2}(A - A^T) = \begin{pmatrix}0 & 3 \\ -3 & 0\end{pmatrix}.
\boxed{A = \begin{pmatrix}1 & 5 \\ -1 & 2\end{pmatrix} = \begin{pmatrix}1 & 2 \\ 2 & 2\end{pmatrix} + \begin{pmatrix}0 & 3 \\ -3 & 0\end{pmatrix} }
Question 11: If A, B are symmetric matrices of the same order, then AB – BA is a
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
Solution:
Since for any matrices X and Y,
(XY)^T = Y^T X^T, and
if A and B are symmetric then
A^T = A, B^T = B.
Therefore
(AB - BA)^T = (AB)^T - (BA)^T
(AB - BA)^T = B^T A^T - A^T B^T,
(AB - BA)^T = BA - AB,
(AB - BA)^T = - (AB - BA),
which means AB – BA is a skew-symmetric matrix.
Answer: Option A
12. If A = \begin{pmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix} and A + A^T = I, then the value of α is
(A) \frac{\pi}{6}
(B) \frac{\pi}{3}
(C) \pi
(D) \frac{3\pi}{2}
Solution:
A = \begin{pmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix}
Compute A^T = \begin{pmatrix}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{pmatrix}.
Then A + A^T = \begin{pmatrix} \cos\alpha + \cos\alpha & - \sin\alpha + \sin\alpha\\ \sin\alpha - \sin\alpha & \cos\alpha + \cos\alpha\end{pmatrix}
A + A^T = \begin{pmatrix}2\cos\alpha & 0 \\ 0 & 2\cos\alpha\end{pmatrix}
Given A + A^T = I
\begin{pmatrix}2\cos\alpha & 0 \\ 0 & 2\cos\alpha\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}.
Hence 2\cos\alpha = 1\implies \cos\alpha = \tfrac{1}{2}.
So \alpha = \tfrac{\pi}{3}\quad(\text{principal value}),
Answer: Option B