NCERT Solutions for Class 12 Maths Chapter 3 Matrix - Exercise 3.3
Access Exercises of Class 12 Maths Chapter 3 – Matrix
Exercise 3.1 Solutions: 10 Questions (7 Short Answers, 3 MCQs)
Exercise 3.2 Solutions: 22 Questions (14 Long, 6 Short, 2 MCQs)
Exercise 3.3 Solutions: 12 Questions (10 Short Answers, 2 MCQs)
Exercise 3.4 Solutions: 18 Questions (4 Long, 13 Short, 1 MCQ)
Miscellaneous Exercise Solutions: 15 Questions (7 Long, 5 Short, 3 MCQs)
Exercise 3.3
1. Find the transpose of each of the following matrices.
$$\textbf{(i)\space}\begin{bmatrix}\textbf{5}\\\frac{\textbf{1}}{\textbf{2}}\\\textbf{\normalsize-1}\end{bmatrix}\\\textbf{(ii)\space}\begin{bmatrix}\textbf{1} &\normalsize-\textbf{1}\\\textbf{2} &\textbf{3} \end{bmatrix}\\\textbf{(iii)\space}\begin{bmatrix}\normalsize-\textbf{1} &\textbf{5} &\textbf{6}\\\sqrt{\textbf{3}} &\textbf{5} &\textbf{6}\\\textbf{2} &\textbf{3} &\normalsize-\textbf{1}\end{bmatrix}\\\textbf{Sol.\space} (i)\space\text{Let A} = \begin{bmatrix}5\\\frac{1}{2}\\ \normalsize-1\end{bmatrix}_{3×1}\\\text{then A' = }\begin{bmatrix}5 &\frac{1}{2} &\normalsize-1\end{bmatrix}_{1×3}\\\text{(ii)\space Let A = }\begin{bmatrix}1 &\normalsize-1 \\2 &3\end{bmatrix},$$
$$\text{then A' =}\begin{bmatrix}1 &2 \\\normalsize-1 &3\end{bmatrix}\\\text{(iii)\space Let}\\\text{A =}\begin{bmatrix}\normalsize-1 &5 &6\\\sqrt{3} &5 &6\\ 2 &3 &\normalsize-1\end{bmatrix},\\\text{then A'} = \begin{bmatrix}\normalsize-1 &\sqrt{3} &2\\5 &5 &3\\ 6 &6 &\normalsize-1\end{bmatrix}$$
$$\textbf{2.\space If A =}\begin{bmatrix}\normalsize-\textbf{1} &\textbf{2} &\textbf{3}\\\textbf{5} &\textbf{7} &\textbf{9}\\\normalsize-\textbf{2} &\textbf{1} &\textbf{1}\end{bmatrix}\space\textbf{and}\\\textbf{B = }\begin{bmatrix}\normalsize-\textbf{4} &\textbf{1} &\textbf{\normalsize-5}\\\textbf{1} &\textbf{2} &\textbf{0}\\\textbf{1} &\textbf{3} &\textbf{1}\end{bmatrix}\textbf{,}\space\textbf{then}$$
verify that
(i) (A + B)′ = A′ + B′
(ii) (A – B)′ = A′ – B′
Sol. (i) Here, A + B =
$$\begin{bmatrix}\normalsize-1 &2 &3\\5 &7 &9\\\normalsize-2 &1 &1\end{bmatrix}+\begin{bmatrix}\normalsize-4 &1 &\normalsize-5\\1 &2 &0\\ 1 &3 &1\end{bmatrix}\\=\begin{bmatrix}\normalsize-5 &3 &\normalsize-2\\6 &9 &9\\\normalsize-1 &4 &2\end{bmatrix}\\\Rarr\space\text{(A + B)'} = \begin{bmatrix}\normalsize-5 &6 &\normalsize-1\\3 &9 &4\\\normalsize-2 &9 &2\end{bmatrix}\\\text{...(i)}\\\text{Also\space A' + B'} = \\\begin{bmatrix}\normalsize-1 &2 &3\\5 &7 &9\\\normalsize-2 &1 &1\end{bmatrix}' + \begin{bmatrix}\normalsize-4 &1 &5\\1 &2 &0\\1 &3 &1\end{bmatrix}'$$
$$=\begin{bmatrix}\normalsize-1 &5 &\normalsize-2\\2 &7 &1\\3 &9 &1\end{bmatrix} + \begin{bmatrix}\normalsize-4 &1 &1\\1 &2 &3\\-5 &0 &1\end{bmatrix}\\=\begin{bmatrix}\normalsize-5 & 6 & \normalsize-1\\3 &9 &4\\ \normalsize-2 &9 &2\end{bmatrix}\\\text{...(ii)}$$
From equations (i) and (ii), it is verified that
(A + B)′ = A′ + B′.
$$\text{(ii)\space Here, A-B = }$$
$$\begin{bmatrix}\normalsize-1 &2 &3\\5 &7 &9\\\normalsize-2 &1 &1\end{bmatrix}-\begin{bmatrix}\normalsize-4 &1 &\normalsize-5\\1 &2 &0\\\normalsize1 &3 &1\end{bmatrix}\\\begin{bmatrix}\normalsize-1+4 &2-1 &3+5 \\5-1 &7-2 &9-0\\\normalsize-2-1 &1-3 &1-1\end{bmatrix}\\=\begin{bmatrix}3 &1 &8\\4 &5 &9\\ \normalsize-3 &\normalsize-2 &0\end{bmatrix}\\\Rarr\space\text{(A-B)'} =\begin{bmatrix}3 &4 &\normalsize-3\\1 &5 &\normalsize-2\\ 8 &9 &0\end{bmatrix}\\\text{...(i)}\\\text{Also, A' - B'}=$$
$$\begin{bmatrix}\normalsize-1 &5 &\normalsize-2\\2 &7 &1\\ 3 &9 &1\end{bmatrix} - \begin{bmatrix}-4 &1 &1\\1 &2 &3\\\normalsize-5 &0 &1\end{bmatrix}\\=\begin{bmatrix}\normalsize-1+4 &5-1 &-2-1\\2-1 &7-2 &1-3\\ 3-(\normalsize-5) &9-0 &1-1\end{bmatrix}\\=\begin{bmatrix}3 &4 &\normalsize-3\\1 &5 &\normalsize-2\\ 8 &9 &0\end{bmatrix}\space\text{...(ii)}$$
From Eqs. (i) and (ii), it is verified that
(A – B)′ = A′ – B′.
$$\textbf{3. If A'\space=\space}\begin{bmatrix}\textbf{3} &\textbf{4} \\\normalsize-\textbf{1} &\textbf{2}\\\textbf{0} &\textbf{1}\end{bmatrix}\space\\\textbf{and B = }\begin{bmatrix}\normalsize-\textbf{1} &\textbf{2} &\textbf{1} \\\textbf{1} &\textbf{2} &\textbf{3}\end{bmatrix}\textbf{,}$$
then verify that
(i) (A + B)′ = A′ + B′
(ii) (A – B)′ = A′ – B′
Sol. ∴ A = (A′)′ =
$$\begin{bmatrix}3 &4 \\\normalsize-1 &2\\0 &\normalsize1\end{bmatrix}' =\begin{bmatrix}3 &\normalsize-1 &0\\\normalsize4 &2 &1\end{bmatrix}\\\text{(i) Here, A + B =}\\\begin{bmatrix}3 &\normalsize-1 &0\\4 &2 &1\end{bmatrix} + \begin{bmatrix}\normalsize-1 &2 &1\\1 &2 &3\end{bmatrix}\\=\begin{bmatrix}2 &1 &1\\5 &4 &4\end{bmatrix}\\\therefore\space\text{LHS = (A+B)'} =\begin{bmatrix}2 &1 &1\\5 &4 &4\end{bmatrix}\\=\begin{bmatrix}2 &5 \\1 &4\\ 1 &4\end{bmatrix}$$
$$\text{B' = }\begin{bmatrix}\normalsize-1 &2 &1\\1 &2 &3\end{bmatrix}'\\=\begin{bmatrix}\normalsize-1 &1 \\2 &2\\1 &3\end{bmatrix},\text{also A' = }\begin{bmatrix}3 &4\\\normalsize-1 &2\\0 &1\end{bmatrix}\\\text{Here, RHS = A' + B' =}\\\begin{bmatrix}3 &4\\\normalsize-1 &2\\ 0 &1\end{bmatrix} + \begin{bmatrix}\normalsize-1 &1\\2 &2\\1 &3\end{bmatrix}\\=\begin{bmatrix}2 &5\\1 &4\\1 &4 \end{bmatrix}$$
So, verified that (A + B)′ = A′ + B′.
(ii) Here,
RHS = A′ – B′ =
$$\begin{bmatrix}3 &4\\\normalsize-1 &2\\0 &\normalsize-1\end{bmatrix}-\begin{bmatrix}\normalsize-1 &1\\\normalsize2 &2\\1 &\normalsize3\end{bmatrix}\\=\begin{bmatrix}3+1 &4-1\\-1-2 &2-2\\0-1 &1-3\end{bmatrix}\\=\begin{bmatrix}4 &3\\\normalsize-3 &0\\-1 &\normalsize-2\end{bmatrix}\\\text{Also (A - B)'} =\\\begin{pmatrix}\begin{bmatrix}3 &\normalsize-1 &0\\4 &2 &1\end{bmatrix}-\begin{bmatrix}\normalsize-1 &2 &1\\1 &2 &3\end{bmatrix}\end{pmatrix}\\\lbrack\because\space\text{A = (A')'}\rbrack$$
$$= \begin{bmatrix}3+1 &-1-2 &0-1\\4-1 &2-2 &1-3\end{bmatrix}'\\=\begin{bmatrix}4 &\normalsize-3 &\normalsize-1\\3 &0 &\normalsize-2\end{bmatrix}'\\=\begin{bmatrix}4 &3\\-3 &0\\\normalsize-1 &\normalsize-2\end{bmatrix}$$
So, verified that (A – B)′ = A′ – B′.
$$\textbf{4.\space}\textbf{If A}^′\space\textbf{=}\space\begin{bmatrix}\normalsize-\textbf{2} &\textbf{3}\\\textbf{1} &\textbf{2}\end{bmatrix}\space\textbf{and B = }\begin{bmatrix}\normalsize-\textbf{1} &\textbf{0}\\\textbf{1} &\textbf{2}\end{bmatrix}\textbf{,}\\\textbf{then find (A + 2B)}^′\textbf{.}\\\textbf{Sol.\space} \text{Given, A' = }\begin{bmatrix}\normalsize-2 &3\\1 &2\end{bmatrix}\\\Rarr\space\text{A = (A')'} = \begin{bmatrix}\normalsize-2 &3\\1 &2\end{bmatrix}' =\begin{bmatrix}\normalsize-2 &1\\3 &2\end{bmatrix}\\\lbrack\because(\text{A}')' = \text{A}\rbrack\\\text{and 2B = 2}\begin{bmatrix}\normalsize-1 &0\\1 &2\end{bmatrix}=\begin{bmatrix}\normalsize-2 &0\\2 &4\end{bmatrix}\\\therefore\space\text{A + 2B = }\begin{bmatrix}\normalsize-2 &1\\3 & 2\end{bmatrix} + \begin{bmatrix}\normalsize-2 &0\\2 &4\end{bmatrix}\\=\begin{bmatrix}\normalsize-4 &1\\5 &6\end{bmatrix} $$
$$\Rarr\space (\text{A + 2B})' = \begin{bmatrix}\normalsize-4 &1\\5 &6\end{bmatrix}'\\=\begin{bmatrix}\normalsize-4 &5\\1 &6\end{bmatrix}$$
5. For the matrices A and B, verify that (AB)′= B′A′, where
$$\textbf{(i) A =}\begin{bmatrix}\textbf{1}\\\normalsize-\textbf{4}\\\textbf{3}\end{bmatrix}\textbf{,}\textbf{B = }\begin{Bmatrix}\normalsize-\textbf{1} &\textbf{2} &\textbf{1}\end{Bmatrix}\\\textbf{(ii)\space}\textbf{A = }\begin{bmatrix}\textbf{0}\\\textbf{1}\\\textbf{2}\end{bmatrix}\textbf{,}\textbf{B = }\begin{bmatrix}\textbf{1} &\textbf{5} &\textbf{7}\end{bmatrix}\\\textbf{Sol.\space}\text{(i)\space}\text{Here, AB} =\\\begin{bmatrix}1\\\normalsize-4\\ 3\end{bmatrix}\begin{bmatrix}\normalsize-1 &2 &1\end{bmatrix}\\=\begin{bmatrix}\normalsize-1 &2 &1\\4 &-8 &-4\\\normalsize-3 &6 &3\end{bmatrix}$$
$$\text{(AB)'} = \begin{bmatrix}\normalsize-1 &2 &1\\4 &\normalsize-8 &\normalsize-4\\\normalsize-3 &6 &3\end{bmatrix}\\= \begin{bmatrix}\normalsize-1 &4 &\normalsize-3\\2 &\normalsize-8 &6\\1 &\normalsize-4 &3\end{bmatrix}\\\text{...(i)}\\\text{Also,}\\\text{B'A' =}\begin{bmatrix}\normalsize-1 &2 &1\end{bmatrix}\begin{bmatrix}1\\\normalsize-4\\ 3\end{bmatrix}'\\=\begin{bmatrix}\normalsize-1\\2\\ 1\end{bmatrix}\begin{bmatrix}1 &\normalsize-4 &3\end{bmatrix}$$
$$= \begin{bmatrix}\normalsize-1 &4 &\normalsize-3\\2 &\normalsize-8 &6\\ 1 &\normalsize-4 &3\end{bmatrix}\\\text{...(ii)}$$
From equations (i) and (ii), we find that (AB)′ = B′A′.
$$\textbf{6. If (i) A =}\space \begin{bmatrix}\textbf{cos}\space\alpha &\textbf{sin}\space\alpha\\-\textbf{sin}\space\alpha &\textbf{cos}\space\alpha\end{bmatrix}\textbf{,}$$
then verify that A′A = I.
$$\textbf{(ii)\space}\textbf{If A =}\begin{bmatrix}\textbf{sin}\space\alpha &\textbf{cos}\space\alpha\\\textbf{- cos}\space\alpha &\textbf{cos}\space\alpha\end{bmatrix}\textbf{,}$$
then verify that A′A = I.
Sol. (i) Here,
$$\text{A = }\begin{bmatrix}\text{cos}\space\alpha &\text{sin}\space\alpha\\-\text{sin}\space\alpha &\text{cos}\space\alpha\end{bmatrix}\\\Rarr\space \text{A}' = \begin{bmatrix}\text{cos}\space\alpha &\text{sin}\space\alpha\\-\text{sin}\space\alpha &\text{cos}\space\alpha\end{bmatrix}'\\=\begin{bmatrix}\text{cos}\space\alpha &-\text{sin}\space\alpha\\\text{sin}\space\alpha &\text{cos}\space\alpha\end{bmatrix}\\\therefore\space\text{A'A} =\\\begin{bmatrix}\text{cos}\space\alpha &-\text{sin}\space\alpha\\\text{sin}\space\alpha &\text{cos}\space\alpha\end{bmatrix}\begin{bmatrix}\text{cos}\space\alpha &\text{sin}\space\alpha\\-\text{sin}\space\alpha &\text{cos}\space\alpha\end{bmatrix}\\=\\\begin{bmatrix}\text{(cos}\space\alpha)(\text{cos}\space\alpha) +(\text{sin}\space\alpha)\\(\text{sin}\space\alpha) &(\text{cos}\space\alpha)(\text{sin}\alpha) +(-\text{sin}\space\alpha)(\text{cos}\alpha)\\(\text{sin}\space\alpha)(\text{cos}\space\alpha) +(\text{cos}\space\alpha)\\(\normalsize-\text{sin}\space\alpha) &\text{sin}^{2}\alpha + \text{cos}^{2}\alpha\end{bmatrix}$$
=
$$\begin{bmatrix}\text{cos}^{2}\alpha + \text{sin}^{2}\alpha &\text{cos}\alpha\text{sin}\alpha -\text{sin}\alpha\text{cos}\alpha\\\text{sin}\alpha\text{cos}\alpha -\text{cos}\alpha\text{sin}\alpha & \text{sin}^{2}\alpha + \text{cos}^{2}\alpha\end{bmatrix}\\=\begin{bmatrix}1 &0 \\0 &1\end{bmatrix}= 1.\\\lbrack\because\space\text{sin}^{2}\alpha + \text{cos}^{2}\alpha=1\rbrack$$
(ii) Here,
$$\text{A = }\begin{bmatrix}\text{sin}\space\alpha &\text{cos}\space\alpha\\-\text{cos}\space\alpha &\text{sin}\space\alpha\end{bmatrix}\\\Rarr\space\text{A' =} \begin{bmatrix}\text{sin}\space\alpha&\text{cos}\space\alpha\\-\text{cos}\space\alpha &\text{sin}\space\alpha\end{bmatrix}\\=\begin{bmatrix}\text{sin}\space\alpha &-\text{cos}\space\alpha\\\text{cos}\space\alpha &\text{sin}\space\alpha\end{bmatrix}\\\therefore\space\text{A'A = }\begin{bmatrix}\text{sin}\space\alpha &-\text{cos}\space\alpha\\\text{cos}\space\alpha&\text{sin}\space\alpha\end{bmatrix}\\\begin{bmatrix}\text{sin}\space\alpha &\text{cos}\space\alpha\\-\text{cos}\space\alpha&\text{sin}\space\alpha\end{bmatrix}$$
=
$$\begin{bmatrix}(\text{sin}\space\alpha)(\text{sin}\space\alpha)+ (\normalsize-\text{cos}\alpha)(-\text{cos}\alpha) &(\text{sin}\alpha)(\text{cos}\alpha) + (-\text{cos}\alpha)(\text{sin}\alpha)\\(\text{sin}\space\alpha)(\text{cos}\space\alpha)+(\text{sin}\alpha)(\normalsize-\text{cos}\alpha) &(\text{cos}\space\alpha)(\text{cos}\space\alpha) + (\text{sin}\space\alpha)(\text{sin}\space\alpha)\end{bmatrix}\\=\\\begin{bmatrix}\text{sin}^{2}\alpha + \text{cos}^{2}\alpha &\text{sin}\alpha\text{cos}\alpha - \text{cos}\space\alpha\text{sin}\alpha\\\text{sin}\space\alpha\text{cos}\alpha -\text{sin}\alpha\text{cos}\alpha &\text{cos}^{2}\alpha + \text{sin}^{2}\alpha\end{bmatrix}\\=\begin{bmatrix}1 &0\\0 &1\end{bmatrix}=1.\\\lbrack\because\space \text{sin}^{2}\alpha + \text{cos}^{2}\alpha=1\rbrack$$
Hence, we have verified that A′A = I.
7. (i) Show that the matrix,
$$\textbf{A = }\begin{bmatrix} \textbf{1} &\normalsize-\textbf{1} &\textbf{5}\\\normalsize-\textbf{1} &\textbf{2} &\textbf{1}\\\textbf{5} &\textbf{1} &\textbf{3}\end{bmatrix}\space\\\textbf{is a symmetric matrix.}\\\textbf{(ii)\space}\textbf{Show that the matrix,}\space\\\textbf{A =}\begin{bmatrix}\textbf{0} &\textbf{1} &\textbf{-1}\\\textbf{-1} &\textbf{0} &\textbf{1}\\\textbf{1} &\textbf{\normalsize-1} &\textbf{0}\end{bmatrix}$$
is a skew-symmetric matrix.
Sol. (i) Here,
$$\text{A'} = \begin{bmatrix}1 &\normalsize-1 &5 \\\normalsize-1 &2 &1\\ 5 &1 &3\end{bmatrix}\\=\begin{bmatrix}1 &\normalsize-1 &5\\\normalsize-1 &2 &1\\ 5 &1 &3\end{bmatrix}= \text{A}\\\lbrack\because\space \text{A}' = \text{A}\rbrack$$
Hence, A is a symmetric matrix.
$$\text{(ii)\space}\text{Now, A'} = \begin{bmatrix}0 &1 &\normalsize-1\\\normalsize-1 &0 &1\\ 1 &\normalsize-1 &0\end{bmatrix}'\\=\begin{bmatrix}0 &\normalsize-1 &1\\1 &0 &\normalsize-1\\\normalsize-1 &1 &0\end{bmatrix}\\=-\begin{bmatrix}0 &1 &\normalsize-1\\\normalsize-1 &0 &1\\ 1 &\normalsize-1 &0\end{bmatrix} = -\text{A}$$
∴ A′ = – A. Hence, A is skew-symmetric matrix.
$$\textbf{8. For the matrix, A =}\begin{bmatrix}\textbf{1} &\textbf{5} \\\textbf{6} &\textbf{7}\end{bmatrix}\textbf{,}$$
verify that
(i) (A + A′) is a symmetric matrix.
(ii) (A – A′) is skew-symmetric matrix.
Sol. (i) Here, A + A′ =
$$\begin{bmatrix}1 &5 \\6 &7\end{bmatrix} + \begin{bmatrix}1 &5 \\6 &7\end{bmatrix}'\\=\begin{bmatrix}1 &5\\6 &7\end{bmatrix} + \begin{bmatrix}1 &6\\5 &7\end{bmatrix}\\= \begin{bmatrix}2 &11\\11 &14\end{bmatrix}\\\text{A + A'} = \begin{bmatrix}2 &11 \\11 &14\end{bmatrix}\\\text{and}\space(\text{A + A}') = \begin{bmatrix}2 &11\\11 &14\end{bmatrix}\\\lbrack\because\space \text{A' = A,}\space\text{then A is a symmetric matrix}\rbrack$$
Hence, (A + A′) is a symmetric matrix.
(ii) A – A′ =
$$\begin{bmatrix}1 &5 \\6 &7\end{bmatrix} - \begin{bmatrix}1 &5 \\6 &7\end{bmatrix}'\\=\begin{bmatrix}1 &5 \\6 &7\end{bmatrix} - \begin{bmatrix}1 &6 \\5 &7\end{bmatrix}\\=\begin{bmatrix}0 &\normalsize-1\\1 & 0\end{bmatrix}\\\Rarr\space\text{A - A'} = \begin{bmatrix}0 &\normalsize-1\\1 &0\end{bmatrix}\\\text{and}\space\text{(A - A') =}\begin{bmatrix}0 &1 \\\normalsize-1 &0\end{bmatrix}\\=-\begin{bmatrix}0 &\normalsize-1 \\1 &0\end{bmatrix}$$
= – (A – A′)
[∵ A′ = – A, then A is a skew-symmetric matrix]
Hence, (A – A′)′ is a skew-symmetric matrix.
$$\textbf{9.\space Find}\space\frac{\textbf{1}}{\textbf{2}}\textbf{(A + A')}\space\textbf{and}\space\frac{\textbf{1}}{\textbf{2}}\textbf{(A - A')},\\\textbf{when}\space \textbf{A =}\begin{bmatrix}\textbf{0}& \textbf{a} &\textbf{b}\\\normalsize-\textbf{a} &\textbf{0} &\textbf{c}\\\textbf{-b} &\textbf{-c} &\textbf{0}\end{bmatrix}\textbf{.}\\\textbf{Sol.\space}\text{Now,}\frac{1}{2}(\text{A + A}') =\frac{1}{2}\\\begin{pmatrix}\begin{bmatrix}0 &a &b\\-a &0 &c\\-b &-c &0\end{bmatrix} + \begin{bmatrix}0 &a &b\\-a &0 &c\\-b &-c &0\end{bmatrix}'\end{pmatrix}\\=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}0 &a &b\\-a &0 &c\\\normalsize-b &\normalsize-c &0\end{bmatrix} + \begin{bmatrix}0 &\normalsize-a &\normalsize-b\\\normalsize a &0 &-c\\ \normalsize b&\normalsize c &0\end{bmatrix}\end{pmatrix}\\=\frac{1}{2}\begin{bmatrix}0 &0 &0\\0 &0 &0\\0 &0 &0\end{bmatrix} \\=\begin{bmatrix}0 &0 &0\\0 &0 &0\\0 &0 &0\end{bmatrix}\text{and}\space\frac{1}{2}(A - A')$$
$$= \frac{1}{2}\begin{pmatrix}\begin{bmatrix}0 &a &b\\-a &0 &c\\-b &-c &0\end{bmatrix} - \begin{bmatrix}0 &a &b\\\normalsize-a &0 &c\\-b &-c &0\end{bmatrix}\end{pmatrix}\\=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}0 &a &b\\-a &0 &c\\-b &-c &0\end{bmatrix} - \begin{bmatrix}0 &-a &-b\\a &0 &-c\\ b &c &0\end{bmatrix}\end{pmatrix}\\=\frac{1}{2}\begin{bmatrix}0 &2a &2ab\\-2a &0 &2c\\-2b &-2c &0\end{bmatrix}\\=\begin{bmatrix}0 &a &b\\-a &0 &c\\-b &-c &0\end{bmatrix}$$
10. Express the following matrices as the sum of a symmetric and a skew-symmetric matrix :
$$\textbf{(i)\space}\begin{bmatrix}\textbf{3} &\textbf{5}\\\textbf{1} &\textbf{\normalsize-1}\end{bmatrix}\\\textbf{(ii)\space}\begin{bmatrix}\textbf{6} &\normalsize-\textbf{2} &\textbf{2}\\\normalsize-\textbf{2} &\textbf{3} &\normalsize-\textbf{1}\\\textbf{2} &\textbf{-1} &\textbf{3}\end{bmatrix}\\\textbf{(iii)\space}\begin{bmatrix}\textbf{3} &\textbf{3} &\textbf{-1}\\\normalsize-\textbf{2} & \normalsize-\textbf{2} &\textbf{1}\\ \normalsize-\textbf{4}&-\textbf{5} &-\textbf{2}\end{bmatrix}\\\textbf{(iv)\space}\begin{bmatrix}\textbf{1} &\textbf{5} \\\normalsize-\textbf{1} &\textbf{2}\end{bmatrix}$$
$$\textbf{Sol.\space}\text{Let A} = \begin{bmatrix}3 &5 \\1 &\normalsize-1\end{bmatrix}$$, then A = P + Q
$$\text{where,}\space \text{P} =\frac{1}{2}(\text{A + A}')\space\\\text{and Q =}\frac{1}{2}\text{(A - A')}\\\text{Now,\space P} = \frac{1}{2}(\text{A + A}')\\=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}3 & 5 \\1 &\normalsize-1\end{bmatrix} + \begin{bmatrix}3 &1 \\5 & \normalsize-1\end{bmatrix}\end{pmatrix}\\=\frac{1}{2}\begin{bmatrix}6 &6\\6 &\normalsize-2\end{bmatrix} = \begin{bmatrix}3 &3\\3 &\normalsize-1 \end{bmatrix}\\\therefore\space\text{P'} = \begin{bmatrix}3 &3\\3 &\normalsize-1\end{bmatrix}' = \begin{bmatrix}3 &3 \\3 &\normalsize-1\end{bmatrix} = \text{P}$$
(∵ P′ = P, then P is a symmetric matrix.)
$$\text{Thus, P} =\frac{1}{2}(\text{A + A}')\space$$
is a symmetric matrix.
$$\text{Now,\space Q = }\frac{1}{2}(\text{A - A'})\\=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}3 &5\\1 &\normalsize-1\end{bmatrix} - \begin{bmatrix}3 &1\\5 &\normalsize-1\end{bmatrix}\end{pmatrix}\\=\frac{1}{2}\begin{bmatrix}0 &4 \\\normalsize-4 &0\end{bmatrix} = \begin{bmatrix}0 &2 \\\normalsize-2 &0\end{bmatrix}\\\therefore\space Q' =\begin{bmatrix}0 &2 \\\normalsize-2 &0\end{bmatrix}'=\begin{bmatrix}0 & \normalsize-2 \\2 &0 \end{bmatrix}\\=-\text{Q}$$
[∵ Q′ = – Q, then Q is a skew-symmetric matrix]
$$\text{Thus, Q = }\frac{1}{2}\text{(A - A')}\space$$
is a skew-symmetric matrix.
Representing A as the sum of P and Q,
$$\text{P + Q} = \begin{bmatrix}3 &3 \\3 &\normalsize-1\end{bmatrix} +\begin{bmatrix}0 &2\\\normalsize-2 &0\end{bmatrix}\\\begin{bmatrix}3 &5 \\1 &\normalsize-1\end{bmatrix} = \text{A}.\\\text{(ii)\space}\text{Let A = }\begin{bmatrix}6 &\normalsize-2 &2 \\\normalsize-2 &3 &\normalsize-1\\ 2 &\normalsize-1 &3\end{bmatrix},\\\text{then A'} = \begin{bmatrix}6 &\normalsize-2 &2 \\\normalsize-2 &3 &\normalsize-1\\ 2 &\normalsize-1 &3\end{bmatrix}'\\=\begin{bmatrix}6 &\normalsize-2 &2\\\normalsize-2 &3 &\normalsize-1\\ 2 &\normalsize-1 &3\end{bmatrix} =\text{A}$$
Now,
$$\text{A + A'} =\\\begin{bmatrix}6 &\normalsize-2 &2 \\\normalsize-2 &3 &\normalsize-1\\2 &\normalsize-1 &3\end{bmatrix} + \begin{bmatrix}6 &\normalsize-2 &2\\\normalsize-2 &3 &\normalsize-1\\2&\normalsize-1 &3\end{bmatrix}\\=\begin{bmatrix}12 &-4 &4\\\normalsize-4 &6 &\normalsize-2\\ 4 &\normalsize-2 &6\end{bmatrix}\\\text{Let\space P} = \frac{1}{2}(\text{A + A'}) \\=\frac{1}{2}\begin{bmatrix}12 &\normalsize-4 &4\\\normalsize-4 &6 &\normalsize-2\\ 4&\normalsize-2 &6\end{bmatrix}\\=\begin{bmatrix}6 &\normalsize-2 &2\\\normalsize-2 &3 &\normalsize-1\\ 2 &\normalsize-1 &3\end{bmatrix}$$
$$\text{Now, P' = }\begin{bmatrix}6 &\normalsize-2 &2\\\normalsize-2 &3 &\normalsize-1\\ 2 &\normalsize-1 &3\end{bmatrix}'\\=\begin{bmatrix}6 &\normalsize-2 &2\\\normalsize-2 &3 &\normalsize-1\\2 &\normalsize-1 &3\end{bmatrix}' =\text{P}\\\text{Thus,\space P} = \frac{1}{2}(\text{A + A}')$$
is a symmetric matrix.
Now, A – A′ =
$$\begin{bmatrix}6 &\normalsize-2 &2\\\normalsize-2 &3 &\normalsize-1\\ 2 &\normalsize-1 &3\end{bmatrix} - \begin{bmatrix}6 &\normalsize-2 &2\\\normalsize-2 &3 &\normalsize-1\\ 2 &\normalsize-1&3\end{bmatrix}\\=\begin{bmatrix}0 &0 &0\\0 &0 &0\\0 &0 &0\end{bmatrix}\\\text{Let}\space Q =\frac{1}{2}(\text{A - A'})\\=\frac{1}{2} \begin{bmatrix} 0 &0 &0\\0 &0 &0\\0 &0 &0\end{bmatrix} =\begin{bmatrix}0 &0 &0\\0 &0 &0\\ 0 &0 &0\end{bmatrix}\\\text{Now, Q' =}\begin{bmatrix}0 &0 &0\\0 &0 &0\\0 &0 &0\end{bmatrix} =-\text{Q}$$
$$\text{Thus, \space Q} = \frac{1}{2}(\text{A - A}')$$
is a skew symmetric matrix.
Representing A as the sum of P and Q,
P + Q =
$$\begin{bmatrix}6 &-2 &2\\\normalsize-2 &3 &\normalsize-1\\2 &\normalsize-1 &3\end{bmatrix} + \begin{bmatrix}0 &0 &0\\0 &0 &0\\0 &0 &0\end{bmatrix}\\=\begin{bmatrix}6 &\normalsize-2 &2\\\normalsize-2 & 3 &\normalsize-1\\ 2 &\normalsize-1 &3\end{bmatrix} = \text{A}$$
$$\text{(iii)\space Let A =}\begin{bmatrix}3 &3 &\normalsize-1\\\normalsize-2 &\normalsize-2 &1\\\normalsize-4 &\normalsize-5 &2\end{bmatrix},\\\text{then A' = }\begin{bmatrix} 3 &\normalsize-2 &\normalsize-4\\3 &\normalsize-2 &\normalsize-5\\ \normalsize-1 &1 &2\end{bmatrix}$$
$$\text{Now, A + A'} = \\\begin{bmatrix}3 &3 &\normalsize-1\\\normalsize-2 &\normalsize-2 &1\\ \normalsize-4 &\normalsize-5 &2\end{bmatrix} + \begin{bmatrix}3 &\normalsize-2 &\normalsize-4\\3 &\normalsize-2 &\normalsize-5\\\normalsize-1 &1 &2\end{bmatrix}\\=\begin{bmatrix}6 &1 &\normalsize-5\\1 &\normalsize-4 &\normalsize-4\\\normalsize-5 &\normalsize-4 &4\end{bmatrix}\\\text{P} =\frac{1}{2}(\text{A + A'})\\=\frac{1}{2}\begin{bmatrix}6 &1 &\normalsize-5\\1 &\normalsize-4 &\normalsize-4\\\normalsize-5 &\normalsize-4 &4\end{bmatrix}\\=\begin{bmatrix}3 &\frac{1}{2} &\frac{\normalsize-5}{2}\\\frac{1}{2}&-2 &-2\\\frac{\normalsize-5}{2} &-2 &2\end{bmatrix}$$
$$\text{Now,\space P'} = \begin{bmatrix}3 &\frac{1}{2} &\frac{\normalsize-5}{2}\\\frac{1}{2} &\normalsize-2 &\normalsize-2\\\frac{\normalsize-5}{2} &\normalsize-2 &2\end{bmatrix}'\\=\begin{bmatrix}3 &\frac{1}{2} &\frac{-5}{2}\\\frac{1}{2} &\normalsize-2 &\normalsize-2\\-\frac{5}{2} &\normalsize-2 &2\end{bmatrix} =\text{P}\\\text{Thus,\space}\text{P} =\frac{1}{2}(\text{A + A'})$$
is a symmetric matrix.
Now, A – A′ =
$$= \begin{bmatrix} 3 &3 &\normalsize-1\\\normalsize-2 &\normalsize-2 &1\\\normalsize-4 &\normalsize-5 &2\end{bmatrix} - \begin{bmatrix}3 &\normalsize-2 &\normalsize-4\\3 &\normalsize-2 &\normalsize 5\\\normalsize-1 &1 &2\end{bmatrix}\\=\begin{bmatrix}0 &5 &3\\\normalsize-5 &0 &6\\\normalsize-3 &\normalsize-6 &0\end{bmatrix}\\\text{Let\space Q = }\frac{1}{2}(\text{A - A'})\\=\frac{1}{2}\begin{bmatrix}0 &5 &3\\\normalsize-5 &0 &6\\\normalsize-3 &\normalsize-6 &0\end{bmatrix}\\=\begin{bmatrix}0 &\frac{5}{2} &\frac{3}{2}\\-\frac{5}{2} &0 &3\\\frac{\normalsize-3}{2} &\normalsize-3 &0\end{bmatrix}$$
$$\text{Now\space Q' = }\begin{bmatrix}0 &\frac{5}{2} &\frac{3}{2}\\\frac{\normalsize-5}{2} &0 &3\\\frac{-3}{2} &3 &0\end{bmatrix}'\\=\begin{bmatrix}0 &\frac{\normalsize-5}{2} &\frac{\normalsize-3}{2}\\\frac{5}{2} &0 &\normalsize-3\\\frac{3}{2} &3 &0\end{bmatrix} = -\text{Q}\\\text{Thus,\space Q =}\frac{1}{2}(\text{A - A'})$$
is a skew-symmetric matrix.
Representing A as the sum of P and Q,
$$\text{P + Q} = \begin{bmatrix}3 &\frac{1}{2} &\frac{\normalsize-5}{2}\\\frac{1}{2} &\normalsize-2 &\normalsize-2\\\frac{\normalsize-5}{2} &\normalsize-2 &2\end{bmatrix}+\\\begin{bmatrix}0 &\frac{5}{2}&\frac{3}{2}\\\frac{\normalsize-5}{2} &0 &3\\\frac{\normalsize-3}{2} &-3 &0\end{bmatrix}\\=\begin{bmatrix}3 &3 &\normalsize-1\\\normalsize-2 &\normalsize-2 &1\\\normalsize-4 &\normalsize-5 &2\end{bmatrix} =\text{A}$$
$$\text{(iv) Let A =}\begin{bmatrix} 1 &5 \\\normalsize-1 &2\end{bmatrix}.\text{Then,}\\\text{A'} =\begin{bmatrix}1 &5 \\\normalsize-1 &2\end{bmatrix}' = \begin{bmatrix}1 &\normalsize-1 \\5 &2\end{bmatrix}\\\text{Now, A + A'} =\\\begin{bmatrix}1 &5 \\\normalsize-1 &2\end{bmatrix}' = \begin{bmatrix}1 &\normalsize-1\\5 &2\end{bmatrix}\\=\begin{bmatrix}2 &4 \\4 &4\end{bmatrix}$$
$$\text{Let P} = \frac{1}{2}(\text{A + A}')\\=\frac{1}{2}\begin{bmatrix}2 &4 \\4 &4\end{bmatrix} =\begin{bmatrix}1 &2 \\2 &2\end{bmatrix}.\\\text{Now,}\space\text{P'} = \begin{bmatrix}1 &2 \\2 &2 \end{bmatrix}' = \begin{bmatrix}1 &2 \\2 &2\end{bmatrix} =\text{P}\\\text{Thus, P} =\frac{1}{2}(\text{A + A'})\\\text{is a symmetric matrix.}\\\text{Now,\space A - A'} =\\\begin{bmatrix}1 &5\\\normalsize-1 &2\end{bmatrix}\begin{bmatrix}1 &\normalsize-1\\\normalsize5 &2\end{bmatrix}\\=\begin{bmatrix}0 &6 \\-6 &0\end{bmatrix}$$
$$\text{Let}\space \text{Q = }\frac{1}{2}(\text{A - A'})\\=\frac{1}{2}\begin{bmatrix}0 &6 \\-6 &0\end{bmatrix}\\=\begin{bmatrix}0 & 3\\\normalsize-3 &0\end{bmatrix}\\\text{Now,}\\\text{Q'} = \begin{bmatrix}0 &3' \\-3 &0\end{bmatrix}' = \begin{bmatrix}0 &\normalsize-3 \\3 &0\end{bmatrix}\\=-\text{Q}\\\text{Thus, Q} =\frac{1}{2}(\text{A - A'})\\\text{is a skew-symmetric matrix.}$$
Representing A as the sum of P and Q,
$$\text{P + Q} =\\ \begin{bmatrix}1 &2 \\2 &2\end{bmatrix} + \begin{bmatrix}0 &3 \\\normalsize-3 &0\end{bmatrix}\\=\begin{bmatrix}1 &5\\\normalsize-1 &2\end{bmatrix} =\text{A}$$
Choose the correct answer in the following questions :
11. 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
Sol. Given, A and B are symmetric matrices.
$$\Rarr\space \text{A' = A}\space\text{and\space B' = B}$$
(AB – BA)′ = (AB)′ – (BA)′
[∵ (A – B)′ = A′ – B′]
= B′A′ – A′B′
[∵ (AB)′ = B′A′]
= BA – AB
[∵ A′ = A and B′ = B]
= – (AB – BA)
$$\Rarr\space(\text{AB - BA})' =- (\text{AB - BA})$$
Thus, (AB – BA) is a skew-symmetric matrix.
$$\textbf{12.\space If A = }\begin{bmatrix}\textbf{cos}\space\alpha &\textbf{-sin}\space\alpha\\\textbf{sin}\space\alpha &\textbf{cos}\space\alpha\end{bmatrix}\textbf{,}$$
then A + A′ = I, then the value of α is :
(a) π/6
(b) π/3
(c) π
(d) 3π/2
Sol. (b) π/3
$$\text{Here,\space A = }\begin{bmatrix}\text{cos}\space\alpha &-\text{sin}\space\alpha\\\text{sin}\space\alpha &\text{cos}\space\alpha\end{bmatrix}\\\text{and\space} \text{A}' = \begin{bmatrix}\text{cos}\space\alpha &-\text{sin}\space\alpha\\\text{sin}\space\alpha &\text{cos}\space\alpha\end{bmatrix}\\= \begin{bmatrix}\text{cos}\space\alpha &\text{sin}\space\alpha\\-\text{sin}\space\alpha &\text{cos}\space\alpha\end{bmatrix}$$
Given, A + A′ = I
$$\therefore\space\begin{bmatrix}\text{cos}\space\alpha &-\text{sin}\space\alpha\\\text{sin}\space\alpha &\text{cos}\space\alpha\end{bmatrix} + \begin{bmatrix}\text{cos}\space\alpha &\text{sin}\space \alpha\\-\text{sin}\space\alpha &\text{cos}\space\alpha\end{bmatrix}\\=\begin{bmatrix}1 &0\\0 &1\end{bmatrix}\\\Rarr\space\begin{bmatrix}\text{2 cos}\space\alpha &0\\0 &\text{2 cos}\alpha\end{bmatrix}\\= \begin{bmatrix}1 &0 \\0 &1\end{bmatrix}$$
Comparing the corresponding elements of the above matrices, we have
$$\text{2 cos}\space\alpha = \text{I}\\\Rarr\space a =\frac{1}{2} =\text{cos}\frac{\pi}{3}\\\Rarr\space \alpha = \frac{\pi}{3}.$$
