Miscellaneous Exercise, Question 1
A and B are symmetric. Prove \(AB - BA\) is skew symmetric.
Show solution
- \((AB - BA)' = (AB)' - (BA)' = B'A' - A'B'\).
- With \(A' = A\), \(B' = B\): this is \(BA - AB = -(AB - BA)\).
Answer: Proved
NCERT Solutions for Class 12 Maths Chapter 3 Miscellaneous Exercise
The Miscellaneous Exercise on Matrices. Transpose rules (\((AB)' = B'A'\)) settle the symmetric / skew symmetric proofs; equations like \(A'A = I\) or a product equal to O become ordinary equations in the unknowns once the products are multiplied out.
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Miscellaneous Exercise questions and solutions
Miscellaneous Exercise, Question 2
Show \(B'AB\) is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Show solution
- \((B'AB)' = B'A'(B')' = B'A'B\).
- If \(A' = A\): this is \(B'AB\), so \(B'AB\) is symmetric. If \(A' = -A\): it is \(-B'AB\), so \(B'AB\) is skew symmetric.
Answer: Shown
Miscellaneous Exercise, Question 3
Find x, y, z if \(A = \begin{bmatrix}0 & 2 y & z \\ x & y & - z \\ x & - y & z\end{bmatrix}\) satisfies \(A'A = I\).
Show solution
- \(A'A = I\) says the columns of A are unit vectors that are mutually perpendicular.
- Column 1: \(0 + x^2 + x^2 = 1\). Column 2: \(4y^2 + y^2 + y^2 = 1\). Column 3: \(z^2 + z^2 + z^2 = 1\). (The dot products of different columns are 0 automatically.)
Answer: \[\begin{gathered}x = \pm\tfrac{1}{\sqrt2}, \\ y = \pm\tfrac{1}{\sqrt6}, \\ z = \pm\tfrac{1}{\sqrt3}\end{gathered}\]
Miscellaneous Exercise, Question 4
For what x is \(\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{bmatrix}\begin{bmatrix}0 \\ 2 \\ x\end{bmatrix} = O\)?
Show solution
- \[\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{bmatrix} = \begin{bmatrix}6 & 2 & 4\end{bmatrix}\]
- \[\begin{aligned}&\begin{bmatrix}6 & 2 & 4\end{bmatrix}\begin{bmatrix}0 \\ 2 \\ x\end{bmatrix} = [4 + 4x] = O \\ \Rightarrow\ &x = -1\end{aligned}\]
Answer: \(x = -1\)
Miscellaneous Exercise, Question 5
\(A = \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}\). Show \(A^2 - 5A + 7I = O\).
Show solution
- \[A^2 = \begin{bmatrix}8 & 5 \\ -5 & 3\end{bmatrix}\]
- \[A^2 - 5A + 7I = \begin{bmatrix}8 & 5 \\ -5 & 3\end{bmatrix} - \begin{bmatrix}15 & 5 \\ -5 & 10\end{bmatrix} + \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}\]
Answer: Shown: the result is \[\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\]
Miscellaneous Exercise, Question 6
Find x: \(\begin{bmatrix}x & -5 & -1\end{bmatrix}\begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix}\begin{bmatrix}x \\ 4 \\ 1\end{bmatrix} = O\)
Show solution
- \[\begin{bmatrix}x & -5 & -1\end{bmatrix}\begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix} = \begin{bmatrix}x - 2 & -10 & 2 x - 8\end{bmatrix}\]
- Times \(\begin{bmatrix}x \\ 4 \\ 1\end{bmatrix}\): \(x^2 - 2x - 40 + 2x - 8 = x^2 - 48 = 0\).
Answer: \(x = \pm4\sqrt3\)
Miscellaneous Exercise, Question 7
Annual sales of products x, y, z: Market I 10,000, 2,000, 18,000; Market II 6,000, 20,000, 8,000.
(a) Unit prices Rs 2.50, Rs 1.50, Rs 1.00: find each market's revenue with matrices.
Show solution
- \[\begin{bmatrix}10000 & 2000 & 18000 \\ 6000 & 20000 & 8000\end{bmatrix}\begin{bmatrix}2.50 \\ 1.50 \\ 1.00\end{bmatrix}\]
- Market I: \(25000 + 3000 + 18000\); Market II: \(15000 + 30000 + 8000\).
Answer: Market I: Rs 46,000; Market II: Rs 53,000
(b) Unit costs Rs 2.00, Rs 1.00, Rs 0.50: find the gross profit.
Show solution
- Costs: \[\begin{bmatrix}10000 & 2000 & 18000 \\ 6000 & 20000 & 8000\end{bmatrix}\begin{bmatrix}2 \\ 1 \\ 0.50\end{bmatrix} = \begin{bmatrix}31000 \\ 36000\end{bmatrix}\]
- Profit = revenue − cost: \(46000 - 31000\) and \(53000 - 36000\).
Answer: Market I: Rs 15,000; Market II: Rs 17,000 (Rs 32,000 in all)
Miscellaneous Exercise, Question 8
Find X with \(X\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6\end{bmatrix} = \begin{bmatrix}-7 & -8 & -9 \\ 2 & 4 & 6\end{bmatrix}\).
Show solution
- X times a \(2 \times 3\) gives \(2 \times 3\), so X is \(2 \times 2\): \[X = \begin{bmatrix}a & b \\ c & d\end{bmatrix}\]
- Row 1: \(a + 4b = -7\), \(2a + 5b = -8\), \(3a + 6b = -9\) give \(a = 1, b = -2\).
- Row 2: \(c + 4d = 2\), \(2c + 5d = 4\), \(3c + 6d = 6\) give \(c = 2, d = 0\).
Answer: \(X = \) \[\begin{bmatrix}1 & -2 \\ 2 & 0\end{bmatrix}\]
Miscellaneous Exercise, Question 9
\(A = \begin{bmatrix}\alpha & \beta \\ \gamma & - \alpha\end{bmatrix}\) and \(A^2 = I\). Then: (A) \(1 + \alpha^2 + \beta\gamma = 0\) (B) \(1 - \alpha^2 + \beta\gamma = 0\) (C) \(1 - \alpha^2 - \beta\gamma = 0\) (D) \(1 + \alpha^2 - \beta\gamma = 0\)
Show solution
- \[\begin{aligned}A^2 &= \begin{bmatrix}\alpha^{2} + \beta \gamma & 0 \\ 0 & \alpha^{2} + \beta \gamma\end{bmatrix} \\ &= I\end{aligned}\], so \(\alpha^2 + \beta\gamma = 1\).
Answer: (C)
Miscellaneous Exercise, Question 10
If A is both symmetric and skew symmetric, then: (A) A is diagonal (B) A is a zero matrix (C) A is square (D) none of these
Show solution
- \(A' = A\) and \(A' = -A\) give \(A = -A\), so \(A = O\).
Answer: (B)
Miscellaneous Exercise, Question 11
A is square with \(A^2 = A\). Then \((I + A)^3 - 7A\) equals: (A) A (B) \(I - A\) (C) I (D) 3A
Show solution
- I and A commute, so \((I + A)^3 = I + 3A + 3A^2 + A^3\).
- \(A^2 = A\) and \(A^3 = A^2 A = A\): \((I + A)^3 = I + 7A\).
Answer: (C) I
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