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NCERT Solutions for Class 12 Maths Chapter 4 Miscellaneous Exercise
The Miscellaneous Exercise on Determinants. Expand along the row or column that keeps the arithmetic simplest (the rationalised book does not use row and column operations on determinants);for inverses of products use \((AB)^{-1} = B^{-1}A^{-1}\); systems in \(\tfrac1x, \tfrac1y, \tfrac1z\) become linear after substituting \(u = \tfrac1x\) and so on.
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Miscellaneous Exercise questions and solutions
Miscellaneous Exercise, Question 1
Prove that \(\begin{vmatrix}x & \sin{\left(\theta \right)} & \cos{\left(\theta \right)} \\ - \sin{\left(\theta \right)} & - x & 1 \\ \cos{\left(\theta \right)} & 1 & x\end{vmatrix}\) is independent of \(\theta\).
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Along row 1: \[x(-x^2 - 1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)\] \[= -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^2\theta = -x^3\] Answer: \(-x^3\), which has no \(\theta\) in it
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Miscellaneous Exercise, Question 2
Evaluate \(\begin{vmatrix}\cos{\left(\alpha \right)} \cos{\left(\beta \right)} & \sin{\left(\beta \right)} \cos{\left(\alpha \right)} & - \sin{\left(\alpha \right)} \\ - \sin{\left(\beta \right)} & \cos{\left(\beta \right)} & 0 \\ \sin{\left(\alpha \right)} \cos{\left(\beta \right)} & \sin{\left(\alpha \right)} \sin{\left(\beta \right)} & \cos{\left(\alpha \right)}\end{vmatrix}\).
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Along column 3: \[-\sin\alpha(-\sin\alpha\sin^2\beta - \sin\alpha\cos^2\beta) + \cos\alpha(\cos\alpha\cos^2\beta + \cos\alpha\sin^2\beta)\] \(= \sin^2\alpha + \cos^2\alpha\). Answer: 1
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Miscellaneous Exercise, Question 3
\(A^{-1} = \begin{bmatrix}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{bmatrix}\), \(B = \begin{bmatrix}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{bmatrix}\). Find \((AB)^{-1}\).
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\((AB)^{-1} = B^{-1}A^{-1}\). \[\begin{aligned}|B| &= 1(3 - 0) - 2(-1 - 0) - 2(2 - 0) \\ &= 1\end{aligned}\], \[\begin{aligned}B^{-1} &= \operatorname{adj} B \\ &= \begin{bmatrix}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{bmatrix}\end{aligned}\] \[B^{-1}A^{-1} = \begin{bmatrix}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{bmatrix}\begin{bmatrix}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{bmatrix}\] Answer: \[(AB)^{-1} = \begin{bmatrix}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{bmatrix}\]
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Miscellaneous Exercise, Question 4
\(A = \begin{bmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5\end{bmatrix}\). Verify:
(i) \([\operatorname{adj} A]^{-1} = \operatorname{adj}(A^{-1})\)
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\[\begin{aligned}|A| &= 1(15 - 1) - 2(10 - 1) + 1(2 - 3) \\ &= -5\end{aligned}\]; \[\operatorname{adj} A = \begin{bmatrix}14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1\end{bmatrix}\], \(A^{-1} = -\tfrac15\operatorname{adj} A\). \(|\operatorname{adj} A| = |A|^2 = 25\); inverting adj A and taking the adjoint of \(A^{-1}\) both give the matrix below (each equals \(\tfrac{1}{|A|}A\)). Answer: Both equal \[-\dfrac15\begin{bmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5\end{bmatrix}\]
(ii) \((A^{-1})^{-1} = A\)
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\[A^{-1} = -\tfrac15\begin{bmatrix}14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1\end{bmatrix}\]; its determinant is \(\tfrac{1}{|A|} = -\tfrac15\) and its adjoint is \(\tfrac{1}{|A|}A\), so \[\begin{aligned}(A^{-1})^{-1} &= \dfrac{\operatorname{adj}(A^{-1})}{|A^{-1}|} \\ &= A\end{aligned}\] Answer: Verified: \((A^{-1})^{-1} = A\)
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Miscellaneous Exercise, Question 5
Evaluate \(\begin{vmatrix}x & y & x + y \\ y & x + y & x \\ x + y & x & y\end{vmatrix}\).
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Expand along row 1: \[x\left[(x + y)y - x^2\right] - y\left[y^2 - x(x + y)\right] + (x + y)\left[xy - (x + y)^2\right]\] \[= (x^2y + xy^2 - x^3) + (x^2y + xy^2 - y^3) - (x + y)(x^2 + xy + y^2)\] \[(x + y)(x^2 + xy + y^2) = x^3 + 2x^2y + 2xy^2 + y^3\], so the total is \[2x^2y + 2xy^2 - x^3 - y^3 - x^3 - 2x^2y - 2xy^2 - y^3 = -2x^3 - 2y^3\] Answer: \(-2(x^3 + y^3)\)
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Miscellaneous Exercise, Question 6
Evaluate \(\begin{vmatrix}1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y\end{vmatrix}\).
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Expand along row 1: \[1\left[(x + y)^2 - xy\right] - x\left[(x + y) - y\right] + y\left[x - (x + y)\right]\] \(= (x^2 + xy + y^2) - x^2 - y^2 = xy\). Answer: \(xy\)
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Miscellaneous Exercise, Question 7
Solve \(\dfrac2x + \dfrac3y + \dfrac{10}{z} = 4,\ \dfrac4x - \dfrac6y + \dfrac5z = 1,\ \dfrac6x + \dfrac9y - \dfrac{20}{z} = 2\).
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Put \(u = \tfrac1x, v = \tfrac1y, w = \tfrac1z\): \[A = \begin{bmatrix}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{bmatrix}\], \[B = \begin{bmatrix}4 \\ 1 \\ 2\end{bmatrix}\] \[\begin{aligned}|A| &= 2(120 - 45) - 3(-80 - 30) + 10(36 + 36) \\ &= 150 + 330 + 720 \\ &= 1200\end{aligned}\]; \[\operatorname{adj} A = \begin{bmatrix}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{bmatrix}\] \[\begin{aligned}\begin{bmatrix}u \\ v \\ w\end{bmatrix} &= \tfrac{1}{1200}\begin{bmatrix}600 \\ 400 \\ 240\end{bmatrix} \\ &= \begin{bmatrix}\frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5}\end{bmatrix}\end{aligned}\] Answer: \(x = 2,\ y = 3,\ z = 5\)
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Miscellaneous Exercise, Question 8
x, y, z non-zero. The inverse of \(A = \begin{bmatrix}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{bmatrix}\) is: (A) \(\begin{bmatrix}\frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z}\end{bmatrix}\) (B) \(xyz\) times that (C) \(\tfrac{1}{xyz}A\) (D) \(\tfrac{1}{xyz}I\)
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A diagonal matrix with non-zero diagonal is inverted entry by entry: \[\operatorname{diag}(x, y, z)\operatorname{diag}(x^{-1}, y^{-1}, z^{-1}) = I\] Answer: (A)
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Miscellaneous Exercise, Question 9
\(A = \begin{bmatrix}1 & \sin{\left(\theta \right)} & 1 \\ - \sin{\left(\theta \right)} & 1 & \sin{\left(\theta \right)} \\ -1 & - \sin{\left(\theta \right)} & 1\end{bmatrix}\), \(0 \le \theta \le 2\pi\). Then: (A) \(\det A = 0\) (B) \(\det A \in (2, \infty)\) (C) \(\det A \in (2, 4)\) (D) \(\det A \in [2, 4]\)
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Along row 1: \[1(1 + \sin^2\theta) - \sin\theta(-\sin\theta + \sin\theta) + 1(\sin^2\theta + 1) = 2 + 2\sin^2\theta\] \(0 \le \sin^2\theta \le 1\), both ends reached (\(\theta = 0, \tfrac{\pi}{2}\)), so \(\det A\) runs over \([2, 4]\). Answer: (D)
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Determinants has 40 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.
Determinants in our sample papers: Sample paper 1 (questions 3, 4, 7, 32) · Sample paper 2 (questions 4, 6, 7, 35) · Sample paper 3 (questions 4, 7, 32) · Sample paper 4 (questions 4, 5, 7, 35) · Sample paper 5 (questions 4, 6, 7, 32).
Also useful: free MCQs and case studies for Determinants · formulas for this chapter (Class 12 formula sheet, free PDF) · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan
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