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NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3
Exercise 3.3: Transpose; symmetric and skew symmetric matrices. \(A'\) swaps rows and columns; \((A \pm B)' = A' \pm B'\), \((kA)' = kA'\), \((AB)' = B'A'\). A is symmetric if \(A' = A\), skew symmetric if \(A' = -A\) (then the diagonal is zero). Every square matrix is \(\tfrac12(A + A') + \tfrac12(A - A')\): symmetric plus skew symmetric.
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Exercise 3.3 questions and solutions
Exercise 3.3, Question 1
Find the transpose.
(i) \(\begin{bmatrix}5 \\ \frac{1}{2} \\ -1\end{bmatrix}\)
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The column becomes a row. Answer: \[\begin{bmatrix}5 & \frac{1}{2} & -1\end{bmatrix}\]
(ii) \(\begin{bmatrix}1 & -1 \\ 2 & 3\end{bmatrix}\)
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Row i becomes column i. Answer: \[\begin{bmatrix}1 & 2 \\ -1 & 3\end{bmatrix}\]
(iii) \(\begin{bmatrix}-1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1\end{bmatrix}\)
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Row i becomes column i. Answer: \[\begin{bmatrix}-1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1\end{bmatrix}\]
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Exercise 3.3, Question 2
\(A = \begin{bmatrix}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{bmatrix},\ B = \begin{bmatrix}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{bmatrix}\). Verify:
(i) \((A + B)' = A' + B'\)
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\[A + B = \begin{bmatrix}-5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2\end{bmatrix}\]; its transpose and \(A' + B'\) are both the matrix below. Answer: Both sides equal \[\begin{bmatrix}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{bmatrix}\]
(ii) \((A - B)' = A' - B'\)
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\[A - B = \begin{bmatrix}3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0\end{bmatrix}\]; transpose it, and compare with \(A' - B'\). Answer: Both sides equal \[\begin{bmatrix}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{bmatrix}\]
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Exercise 3.3, Question 3
\(A' = \begin{bmatrix}3 & 4 \\ -1 & 2 \\ 0 & 1\end{bmatrix},\ B = \begin{bmatrix}-1 & 2 & 1 \\ 1 & 2 & 3\end{bmatrix}\). Verify:
(i) \((A + B)' = A' + B'\)
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\[\begin{aligned}A &= (A')' \\ &= \begin{bmatrix}3 & -1 & 0 \\ 4 & 2 & 1\end{bmatrix}\end{aligned}\], \[A + B = \begin{bmatrix}2 & 1 & 1 \\ 5 & 4 & 4\end{bmatrix}\] Its transpose equals \(A' + B'\). Answer: Both sides equal \[\begin{bmatrix}2 & 5 \\ 1 & 4 \\ 1 & 4\end{bmatrix}\]
(ii) \((A - B)' = A' - B'\)
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\[A - B = \begin{bmatrix}4 & -3 & -1 \\ 3 & 0 & -2\end{bmatrix}\]; its transpose equals \(A' - B'\). Answer: Both sides equal \[\begin{bmatrix}4 & 3 \\ -3 & 0 \\ -1 & -2\end{bmatrix}\]
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Exercise 3.3, Question 4
\(A' = \begin{bmatrix}-2 & 3 \\ 1 & 2\end{bmatrix}\), \(B = \begin{bmatrix}-1 & 0 \\ 1 & 2\end{bmatrix}\). Find \((A + 2B)'\).
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\[\begin{aligned}(A + 2B)' &= A' + 2B' \\ &= \begin{bmatrix}-2 & 3 \\ 1 & 2\end{bmatrix} + \begin{bmatrix}-2 & 2 \\ 0 & 4\end{bmatrix}\end{aligned}\] Answer: \[\begin{bmatrix}-4 & 5 \\ 1 & 6\end{bmatrix}\]
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Exercise 3.3, Question 5
Verify \((AB)' = B'A'\):
(i) \(A = \begin{bmatrix}1 \\ -4 \\ 3\end{bmatrix},\ B = \begin{bmatrix}-1 & 2 & 1\end{bmatrix}\)
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\[AB = \begin{bmatrix}-1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3\end{bmatrix}\]; its transpose equals \(B'A'\) computed as a \((3 \times 1)(1 \times 3)\) product. Answer: Both sides equal \[\begin{bmatrix}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{bmatrix}\]
(ii) \(A = \begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix},\ B = \begin{bmatrix}1 & 5 & 7\end{bmatrix}\)
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\[AB = \begin{bmatrix}0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14\end{bmatrix}\]; transpose it and compare with \(B'A'\). Answer: Both sides equal \[\begin{bmatrix}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{bmatrix}\]
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Exercise 3.3, Question 6
Verify \(A'A = I\):
(i) \(A = \begin{bmatrix}\cos{\left(\alpha \right)} & \sin{\left(\alpha \right)} \\ - \sin{\left(\alpha \right)} & \cos{\left(\alpha \right)}\end{bmatrix}\)
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\[A'A = \begin{bmatrix}\cos{\left(\alpha \right)} & - \sin{\left(\alpha \right)} \\ \sin{\left(\alpha \right)} & \cos{\left(\alpha \right)}\end{bmatrix}\begin{bmatrix}\cos{\left(\alpha \right)} & \sin{\left(\alpha \right)} \\ - \sin{\left(\alpha \right)} & \cos{\left(\alpha \right)}\end{bmatrix}\]: diagonal entries \(\cos^2\alpha + \sin^2\alpha = 1\), off-diagonal \[\cos\alpha\sin\alpha - \sin\alpha\cos\alpha = 0\] Answer: \(A'A = \) \[\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
(ii) \(A = \begin{bmatrix}\sin{\left(\alpha \right)} & \cos{\left(\alpha \right)} \\ - \cos{\left(\alpha \right)} & \sin{\left(\alpha \right)}\end{bmatrix}\)
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Diagonal entries \(\sin^2\alpha + \cos^2\alpha = 1\); off-diagonal \[\sin\alpha\cos\alpha - \cos\alpha\sin\alpha = 0\] Answer: \(A'A = \) \[\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
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Exercise 3.3, Question 7
(i) Show \(A = \begin{bmatrix}1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3\end{bmatrix}\) is symmetric.
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\(a_{12} = a_{21} = -1\), \(a_{13} = a_{31} = 5\), \(a_{23} = a_{32} = 1\), so \(A' = A\). Answer: Symmetric
(ii) Show \(A = \begin{bmatrix}0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0\end{bmatrix}\) is skew symmetric.
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Diagonal zero and \(a_{ji} = -a_{ij}\) for each pair, so \(A' = -A\). Answer: Skew symmetric
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Exercise 3.3, Question 8
\(A = \begin{bmatrix}1 & 5 \\ 6 & 7\end{bmatrix}\). Verify:
(i) \(A + A'\) is symmetric
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\[A + A' = \begin{bmatrix}1 & 5 \\ 6 & 7\end{bmatrix} + \begin{bmatrix}1 & 6 \\ 5 & 7\end{bmatrix}\], which equals its own transpose. Answer: \(A + A' = \) \[\begin{bmatrix}2 & 11 \\ 11 & 14\end{bmatrix}\] (symmetric)
(ii) \(A - A'\) is skew symmetric
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\(A - A'\) has zero diagonal and off-diagonal entries \(-1\) and \(1\). Answer: \(A - A' = \) \[\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}\] (skew symmetric)
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Exercise 3.3, Question 9
Find \(\tfrac12(A + A')\) and \(\tfrac12(A - A')\) for \(A = \begin{bmatrix}0 & a & b \\ - a & 0 & c \\ - b & - c & 0\end{bmatrix}\).
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\[\begin{aligned}A' &= \begin{bmatrix}0 & - a & - b \\ a & 0 & - c \\ b & c & 0\end{bmatrix} \\ &= -A\end{aligned}\]: A is skew symmetric. So \(A + A' = O\) and \(A - A' = 2A\). Answer: \(\tfrac12(A + A') = O\) (zero matrix), \(\tfrac12(A - A') = A\)
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Exercise 3.3, Question 10
Write as the sum of a symmetric matrix P and a skew symmetric matrix Q.
(i) \(\begin{bmatrix}3 & 5 \\ 1 & -1\end{bmatrix}\)
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\(P = \tfrac12(A + A')\), \(Q = \tfrac12(A - A')\). Answer: \[\begin{gathered}P = \begin{bmatrix}3 & 3 \\ 3 & -1\end{bmatrix}, \\ Q = \begin{bmatrix}0 & 2 \\ -2 & 0\end{bmatrix}\end{gathered}\]
(ii) \(\begin{bmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{bmatrix}\)
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\(A' = A\): A is already symmetric. Answer: \(P = A\), \(Q = O\)
(iii) \(\begin{bmatrix}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{bmatrix}\)
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\[A + A' = \begin{bmatrix}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{bmatrix}\], \[A - A' = \begin{bmatrix}0 & 5 & 3 \\ -5 & 0 & 3 \\ -3 & -3 & 0\end{bmatrix}\]; halve each. Answer: \[\begin{gathered}P = \begin{bmatrix}3 & \frac{1}{2} & - \frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ - \frac{5}{2} & -2 & 2\end{bmatrix}, \\ Q = \begin{bmatrix}0 & \frac{5}{2} & \frac{3}{2} \\ - \frac{5}{2} & 0 & 3 \\ - \frac{3}{2} & -3 & 0\end{bmatrix}\end{gathered}\]
(iv) \(\begin{bmatrix}1 & 5 \\ -1 & 2\end{bmatrix}\)
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\[P = \tfrac12\begin{bmatrix}2 & 4 \\ 4 & 4\end{bmatrix}\], \[Q = \tfrac12\begin{bmatrix}0 & 6 \\ -6 & 0\end{bmatrix}\] Answer: \[\begin{gathered}P = \begin{bmatrix}1 & 2 \\ 2 & 2\end{bmatrix}, \\ Q = \begin{bmatrix}0 & 3 \\ -3 & 0\end{bmatrix}\end{gathered}\]
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Exercise 3.3, Question 11
A, B symmetric of the same order. Then \(AB - BA\) is: (A) skew symmetric (B) symmetric (C) zero (D) identity
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\[\begin{aligned}(AB - BA)' &= B'A' - A'B' \\ &= BA - AB \\ &= -(AB - BA)\end{aligned}\] Answer: (A)
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Exercise 3.3, Question 12
\(A = \begin{bmatrix}\cos{\left(\alpha \right)} & - \sin{\left(\alpha \right)} \\ \sin{\left(\alpha \right)} & \cos{\left(\alpha \right)}\end{bmatrix}\) and \(A + A' = I\). Then \(\alpha\) is: (A) \(\tfrac{\pi}{6}\) (B) \(\tfrac{\pi}{3}\) (C) \(\pi\) (D) \(\tfrac{3\pi}{2}\)
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\[\begin{aligned}&A + A' = \begin{bmatrix}2 \cos{\left(\alpha \right)} & 0 \\ 0 & 2 \cos{\left(\alpha \right)}\end{bmatrix} = I \\ \Rightarrow\ &\cos\alpha = \tfrac12\end{aligned}\] Answer: (B) \(\tfrac{\pi}{3}\)
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← Exercise 3.2 All Chapter 3 exercises Exercise 3.4 → Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in . Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log .