NCERT Class 12 Mathematics — Determinants, Miscellaneous Exercise on Chapter 4. All 9 questions solved.

The Miscellaneous Exercise pulls the chapter together: determinant identities in several variables, the properties of the adjoint and inverse from Exercise 4.4, and one system of equations that does not look linear until you change what you are solving for.

Key insight. Before expanding a determinant full of letters, look for structure. In question 1 the $\theta$ terms cancel through $\sin^2\theta + \cos^2\theta = 1$; in question 5 a row operation collapses the whole thing. Grinding out a general $3\times3$ expansion works but takes three times as long and offers three times as many places to lose a sign.

A note on the numbering. The current reprint carries two instruction lines left over from the pre-rationalisation edition: “Using properties of determinants in Exercises 11 to 15, prove that:” appears immediately above question 7, and “Choose the correct answer in Exercise 17 to 19” above questions 8 and 9. Neither refers to any question that still exists — the questions they introduced were removed when the syllabus was rationalised, but the headings were not. There are nine questions in this exercise and nothing is missing.

Question 1

Prove that the determinant $\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$ is independent of $\theta$.

Solution. Expand along the first row and let the trigonometry take care of itself.

$$= x\big[(-x)(x) – (1)(1)\big] – \sin\theta\big[(-\sin\theta)(x) – (1)(\cos\theta)\big] + \cos\theta\big[(-\sin\theta)(1) – (-x)(\cos\theta)\big]$$

$$= x(-x^2 – 1) – \sin\theta(-x\sin\theta – \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)$$

$$= -x^3 – x + x\sin^2\theta + \sin\theta\cos\theta – \sin\theta\cos\theta + x\cos^2\theta$$

The two $\sin\theta\cos\theta$ terms cancel, and the remaining $\theta$ terms combine:

$$= -x^3 – x + x\big(\sin^2\theta + \cos^2\theta\big) = -x^3 – x + x = -x^3$$

Since the result is $-x^3$ with no $\theta$ in it, the determinant is independent of $\theta$.

Question 2

Evaluate $\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}$.

Solution. The second row has a zero, so expand along it — two terms instead of three. Its signs are $-\,+\,-$.

$$= -(-\sin\beta)\begin{vmatrix} \cos\alpha\sin\beta & -\sin\alpha \\ \sin\alpha\sin\beta & \cos\alpha \end{vmatrix} + \cos\beta\begin{vmatrix} \cos\alpha\cos\beta & -\sin\alpha \\ \sin\alpha\cos\beta & \cos\alpha \end{vmatrix}$$

Each $2\times2$ simplifies using $\sin^2\alpha + \cos^2\alpha = 1$:

$$\begin{vmatrix} \cos\alpha\sin\beta & -\sin\alpha \\ \sin\alpha\sin\beta & \cos\alpha \end{vmatrix} = \cos^2\alpha\sin\beta + \sin^2\alpha\sin\beta = \sin\beta$$

$$\begin{vmatrix} \cos\alpha\cos\beta & -\sin\alpha \\ \sin\alpha\cos\beta & \cos\alpha \end{vmatrix} = \cos^2\alpha\cos\beta + \sin^2\alpha\cos\beta = \cos\beta$$

So the determinant is

$$\sin\beta\cdot\sin\beta + \cos\beta\cdot\cos\beta = \sin^2\beta + \cos^2\beta = 1$$

Question 3

If $\mathrm{A}^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$ and $\mathrm{B} = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$, find $(\mathrm{AB})^{-1}$.

Solution. Use $(\mathrm{AB})^{-1} = \mathrm{B}^{-1}\mathrm{A}^{-1}$ — note the reversed order. $\mathrm{A}^{-1}$ is already given, so only $\mathrm{B}^{-1}$ has to be computed.

$$|\mathrm{B}| = 1(3 – 0) – 2(-1 – 0) + (-2)(2 – 0) = 3 + 2 – 4 = 1$$

Since $|\mathrm{B}| = 1$, the inverse is just the adjoint:

$$\mathrm{B}^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$$

$$(\mathrm{AB})^{-1} = \mathrm{B}^{-1}\mathrm{A}^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$$

$$= \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}$$

Nothing here requires knowing A itself, which is the point of giving you $\mathrm{A}^{-1}$ directly.

Question 4

Let $\mathrm{A} = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that (i) $[\mathrm{adj}\,\mathrm{A}]^{-1} = \mathrm{adj}(\mathrm{A}^{-1})$ and (ii) $(\mathrm{A}^{-1})^{-1} = \mathrm{A}$.

Solution. First the determinant and adjoint.

$$|\mathrm{A}| = 1(15 – 1) – 2(10 – 1) + 1(2 – 3) = 14 – 18 – 1 = -5$$

$$\mathrm{adj}\,\mathrm{A} = \begin{bmatrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \end{bmatrix}, \qquad \mathrm{A}^{-1} = \frac{-1}{5}\begin{bmatrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \end{bmatrix}$$

(i) Computing each side separately:

$$[\mathrm{adj}\,\mathrm{A}]^{-1} = \frac{-1}{5}\begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix} = \mathrm{adj}(\mathrm{A}^{-1})$$

Both come to $\begin{bmatrix} -1/5 & -2/5 & -1/5 \\ -2/5 & -3/5 & -1/5 \\ -1/5 & -1/5 & -1 \end{bmatrix}$, so the identity holds.

(ii) Inverting $\mathrm{A}^{-1}$ returns A. This is immediate from the definition — if $\mathrm{A}\mathrm{A}^{-1} = \mathrm{I}$, then A is by definition the inverse of $\mathrm{A}^{-1}$ — but it can also be checked by computing $|\mathrm{A}^{-1}| = -\tfrac{1}{5}$ and working through the adjoint again.

Question 5

Evaluate $\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$.

Solution. Every row adds to the same total, $2(x+y)$, which is the structure to exploit. Replace $\mathrm{C}_1$ by $\mathrm{C}_1 + \mathrm{C}_2 + \mathrm{C}_3$ — an operation that leaves the determinant unchanged:

$$= \begin{vmatrix} 2(x+y) & y & x+y \\ 2(x+y) & x+y & x \\ 2(x+y) & x & y \end{vmatrix} = 2(x+y)\begin{vmatrix} 1 & y & x+y \\ 1 & x+y & x \\ 1 & x & y \end{vmatrix}$$

taking the common factor out of the first column. Now subtract row 1 from rows 2 and 3 to create zeros:

$$= 2(x+y)\begin{vmatrix} 1 & y & x+y \\ 0 & x & -y \\ 0 & x-y & -x \end{vmatrix} = 2(x+y)\big[(x)(-x) – (-y)(x-y)\big]$$

$$= 2(x+y)\big[-x^2 + xy – y^2\big] = -2(x+y)\big(x^2 – xy + y^2\big)$$

The bracket is the second factor of the sum of cubes, so this is

$$= -2\big(x^3 + y^3\big)$$

Question 6

Evaluate $\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$.

Solution. The first column is all ones, so subtracting row 1 from rows 2 and 3 clears it and leaves a triangular determinant.

$$\mathrm{R}_2 \to \mathrm{R}_2 – \mathrm{R}_1, \qquad \mathrm{R}_3 \to \mathrm{R}_3 – \mathrm{R}_1$$

$$= \begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{vmatrix}$$

An upper triangular determinant is the product of its diagonal:

$$= 1 \times y \times x = xy$$

Question 7

Solve the system of equations

$$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \qquad \frac{4}{x} – \frac{6}{y} + \frac{5}{z} = 1, \qquad \frac{6}{x} + \frac{9}{y} – \frac{20}{z} = 2$$

Solution. These equations are not linear in $x$, $y$, $z$ — but they are perfectly linear in $\tfrac1x$, $\tfrac1y$ and $\tfrac1z$. Substituting

$$u = \frac{1}{x}, \qquad v = \frac{1}{y}, \qquad w = \frac{1}{z}$$

turns them into an ordinary system:

$$2u + 3v + 10w = 4, \qquad 4u – 6v + 5w = 1, \qquad 6u + 9v – 20w = 2$$

$$\mathrm{A} = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}$$

$$|\mathrm{A}| = 2(120 – 45) – 3(-80 – 30) + 10(36 + 36) = 150 + 330 + 720 = 1200$$

Since $|\mathrm{A}| \neq 0$ there is a unique solution, and $\mathrm{X} = \mathrm{A}^{-1}\mathrm{B}$ gives

$$u = \frac{1}{2}, \qquad v = \frac{1}{3}, \qquad w = \frac{1}{5}$$

Finally, undo the substitution — this is the step people forget, leaving the answer as $u, v, w$:

$$x = \frac{1}{u} = 2, \qquad y = \frac{1}{v} = 3, \qquad z = \frac{1}{w} = 5$$

Question 8

If $x, y, z$ are nonzero real numbers, then the inverse of matrix $\mathrm{A} = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ is

Solution. The quickest route is to test the candidate directly rather than compute an adjoint. For a diagonal matrix, multiplication is entry by entry along the diagonal:

$$\begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \begin{bmatrix} 1/x & 0 & 0 \\ 0 & 1/y & 0 \\ 0 & 0 & 1/z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \mathrm{I}$$

So the inverse of a diagonal matrix is the diagonal matrix of reciprocals, which is option (A). This is also why the question insists $x, y, z$ are non-zero.

Option (C) is what you get by writing $\tfrac{1}{|\mathrm{A}|}\mathrm{A}$ instead of $\tfrac{1}{|\mathrm{A}|}\mathrm{adj}\,\mathrm{A}$ — a common slip, since here $|\mathrm{A}| = xyz$.

Question 9

Let $\mathrm{A} = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}$, where $0 \leq \theta \leq 2\pi$. Then

Solution. Write $s = \sin\theta$ and expand along the first row.

$$\mathrm{Det}(\mathrm{A}) = 1\big(1 + s^2\big) – s\big(-s + s\big) + 1\big(s^2 + 1\big)$$

The middle bracket is zero, so

$$= (1 + s^2) + (1 + s^2) = 2 + 2s^2$$

Now find the range. Over $0 \leq \theta \leq 2\pi$ the sine takes every value in $[-1, 1]$, so $s^2$ takes every value in $[0, 1]$:

$$2 + 2s^2 \in [2, 4]$$

Both endpoints are attained — $2$ at $\theta = 0$ and $4$ at $\theta = \tfrac{\pi}{2}$ — so the interval is closed. That is the whole distinction between (C) and (D).

Common mistakes

• Expanding a lettered determinant before looking for structure. Questions 5 and 6 both collapse after one row or column operation. Question 5 in particular becomes a short calculation once you notice every row sums to $2(x+y)$.
• Reversing the order in question 3. $(\mathrm{AB})^{-1} = \mathrm{B}^{-1}\mathrm{A}^{-1}$. Computing $\mathrm{A}^{-1}\mathrm{B}^{-1}$ gives a different matrix.
• Solving for the substituted variables and stopping. Question 7 returns $u = \tfrac12$, $v = \tfrac13$, $w = \tfrac15$ — but the question asked for $x$, $y$, $z$, which are the reciprocals $2$, $3$, $5$.
• Writing $\mathrm{A}^{-1} = \tfrac{1}{|\mathrm{A}|}\mathrm{A}$. It is $\tfrac{1}{|\mathrm{A}|}\mathrm{adj}\,\mathrm{A}$. Option (C) in question 8 is built from exactly this error.
• Choosing an open interval where the endpoints are attained. In question 9 both $2$ and $4$ occur, at $\theta = 0$ and $\theta = \tfrac{\pi}{2}$, so the answer is $[2, 4]$ and not $(2, 4)$.
• Hunting for questions 10 to 19. They do not exist in this edition; only the headings that introduced them survive.

Practise next

• Chapter 3, Matrices — the operations this chapter has been leaning on throughout, including the inverse from the other direction.
• Exercise 4.5 — worth a second pass, since the matrix method in question 7 is the same technique applied to a disguised system.