NCERT Exemplar Solutions Class 12 Maths Determinants — Objective Type Q24–37

NCERT Exemplar Problems, Class 12 Mathematics — Determinants, Objective Type questions 24 to 37. All 14 questions solved.

Fourteen multiple-choice questions. Most are a determinant to evaluate in two lines with one property — add the columns and take out the common sum (25, 28, 36), spot a skew-symmetric determinant (32) or two equal columns (27) — and then read off the option. The rest test the inverse (33, 34), area (26) and a limit (30).

One answer in the printed key is wrong, Question 25. It is explained where it occurs.

A note on the syllabus. The rationalised NCERT textbook no longer teaches the properties of determinants; JEE still does, and the quick solutions below depend on them. Each property is named where it is used.

Key insight. Before expanding, look for a row or column operation that makes a common factor appear. In an objective question the factored form usually is the answer, or rules options out at once — and when no option matches the factored form, test the options with small numbers rather than trusting the closest-looking one.

Question 24

If $\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix}$, then the value of $x$ is

Solution. $2x^2 – 40 = 18 + 14 = 32$, so $x^2 = 36$ and $x = \pm 6$. Both signs work, because $x$ appears only as $x^2$.

Question 25

The value of the determinant $\begin{vmatrix} a-b & b+c & a \\ b-a & c+a & b \\ c-a & a+b & c \end{vmatrix}$ is

Solution. $C_2 \to C_2 + C_3$ makes every entry of the second column $a + b + c$. Take it out:

$$\Delta = (a+b+c)\begin{vmatrix} a-b & 1 & a \\ b-a & 1 & b \\ c-a & 1 & c \end{vmatrix}.$$

$R_2 \to R_2 – R_1$ and $R_3 \to R_3 – R_1$:

$$\Delta = (a+b+c)\begin{vmatrix} a-b & 1 & a \\ 2(b-a) & 0 & b-a \\ b+c-2a & 0 & c-a \end{vmatrix}.$$

Expand along the second column (its only non-zero entry is in position $(1,2)$, sign $-$):

$$\Delta = -(a+b+c)\big[2(b-a)(c-a) – (b-a)(b+c-2a)\big] = -(a+b+c)(b-a)(c-b).$$

So $\Delta = -(a+b+c)(a-b)(b-c)$. This is none of (A), (B), (C).

The printed key says (C), and that is wrong. Test it with $a = 1$, $b = 2$, $c = 3$: the determinant is $\begin{vmatrix} -1 & 5 & 1 \\ 1 & 4 & 2 \\ 2 & 3 & 3 \end{vmatrix} = -1(12-6) – 5(3-4) + 1(3-8) = -6$, while (C) gives $1 + 8 + 27 – 18 = 18$, (A) gives $36$ and (B) gives $18$. A second check: when $a = b$ the first column becomes $(0, 0, c-a)$ and the determinant has a factor $a – b$ that (C) does not.

Question 26

The area of a triangle with vertices $(-3, 0)$, $(3, 0)$ and $(0, k)$ is 9 sq. units. The value of $k$ will be

Solution.

$$\frac12\left|\begin{vmatrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{vmatrix}\right| = \frac12\big|-3(0-k) + 1(3k-0)\big| = \frac12|6k| = 3|k| = 9,$$

so $k = \pm 3$. Only $3$ is offered. (The base is 6 and the height $|k|$, which gives the same thing.)

Question 27

The determinant $\begin{vmatrix} b^2-ab & b-c & bc-ac \\ ab-a^2 & a-b & b^2-ab \\ bc-ac & c-a & ab-a^2 \end{vmatrix}$ equals

Solution. The first column is $(b-a)(b, a, c)$ and the third is $(b-a)(c, b, a)$. Take $b – a$ out of each:

$$\Delta = (b-a)^2\begin{vmatrix} b & b-c & c \\ a & a-b & b \\ c & c-a & a \end{vmatrix}.$$

The middle column is now the first minus the third. $C_2 \to C_2 – C_1 + C_3$ turns it into zeros, so $\Delta = 0$. None of (A) to (C) is identically zero.

Question 28

The number of distinct real roots of $\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$ in the interval $-\dfrac\pi4 \le x \le \dfrac\pi4$ is

Solution. Each row sums to $\sin x + 2\cos x$. $C_1 \to C_1 + C_2 + C_3$, take it out, then subtract $R_1$ from the other rows:

$$\Delta = (\sin x + 2\cos x)\begin{vmatrix} 1 & \cos x & \cos x \\ 0 & \sin x – \cos x & 0 \\ 0 & 0 & \sin x – \cos x \end{vmatrix} = (\sin x + 2\cos x)(\sin x – \cos x)^2.$$

$\sin x = \cos x$ gives $\tan x = 1$, so $x = \frac\pi4$, which is in the interval. $\sin x + 2\cos x = 0$ gives $\tan x = -2$; but on $\left[-\frac\pi4, \frac\pi4\right]$, $\tan x$ only runs from $-1$ to $1$. One root.

Question 29

If $A$, $B$ and $C$ are angles of a triangle, then the determinant $\begin{vmatrix} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{vmatrix}$ is equal to

Solution. Expand along the first row:

$$\Delta = -(1-\cos^2 A) – \cos C(-\cos C – \cos A\cos B) + \cos B(\cos A\cos C + \cos B)$$

$$= -1 + \cos^2 A + \cos^2 B + \cos^2 C + 2\cos A\cos B\cos C.$$

In a triangle $C = \pi – (A + B)$, so $\cos C = -\cos(A+B)$ and

$$\cos^2 C + 2\cos A\cos B\cos C = \cos(A+B)\big[\cos(A+B) – 2\cos A\cos B\big] = -\cos(A+B)\cos(A-B) = -\cos^2 A + \sin^2 B.$$

So $\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A\cos B\cos C = \cos^2 B + \sin^2 B = 1$, and $\Delta = -1 + 1 = 0$.

Question 30

Let $f(t) = \begin{vmatrix} \cos t & t & 1 \\ 2\sin t & t & 2t \\ \sin t & t & t \end{vmatrix}$. Then $\displaystyle\lim_{t\to0}\frac{f(t)}{t^2}$ is equal to

Solution. Take $t$ out of the second column, then expand along it:

$$f(t) = t\begin{vmatrix} \cos t & 1 & 1 \\ 2\sin t & 1 & 2t \\ \sin t & 1 & t \end{vmatrix} = t\big[\cos t\,(t – 2t) – (2t\sin t – 2t\sin t) + (2\sin t – \sin t)\big] = t(\sin t – t\cos t).$$

So $\dfrac{f(t)}{t^2} = \dfrac{\sin t}{t} – \cos t \to 1 – 1 = 0$.

Question 31

The maximum value of $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1+\cos\theta & 1 & 1 \end{vmatrix}$ is ($\theta$ is a real number)

Solution. $R_2 \to R_2 – R_1$, $R_3 \to R_3 – R_1$:

$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & \sin\theta & 0 \\ \cos\theta & 0 & 0 \end{vmatrix} = 1\,(0 – \sin\theta\cos\theta) = -\frac12\sin2\theta.$$

Its largest value is $\frac12$, at $\sin2\theta = -1$.

Question 32

If $f(x) = \begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{vmatrix}$, then

Solution. At $x = 0$ the determinant is $\begin{vmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{vmatrix}$, which is skew-symmetric. A skew-symmetric determinant of odd order is $0$: transposing it changes nothing, yet equals multiplying every row by $-1$, which multiplies a $3 \times 3$ determinant by $(-1)^3$; so $\Delta = -\Delta$. In full, $f(x) = 2x(x^2 – ab + ac – bc)$, which is not $0$ at $a$, $b$ or $1$ in general.

Question 33

If $A = \begin{bmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{bmatrix}$, then $A^{-1}$ exists if

Solution. $|A| = 2(6 – 5) – \lambda(0 – 5) – 3(0 – 2) = 5\lambda + 8$. The inverse exists exactly when $|A| \ne 0$, that is $\lambda \ne -\frac85$. None of the options says that.

Question 34

If $A$ and $B$ are invertible matrices, then which of the following is not correct?

Solution. (A) is the definition $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ rearranged; (B) follows from $|A|\,|A^{-1}| = |I| = 1$; (C) is the reversal law. (D) fails already for $A = B = I$: $(2I)^{-1} = \frac12 I$, but $I + I = 2I$.

Question 35

If $x, y, z$ are all different from zero and $\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix} = 0$, then the value of $x^{-1} + y^{-1} + z^{-1}$ is

Solution. $R_2 \to R_2 – R_1$, $R_3 \to R_3 – R_1$:

$$\begin{vmatrix} 1+x & 1 & 1 \\ -x & y & 0 \\ -x & 0 & z \end{vmatrix} = (1+x)yz – (-xz) + (xy) = xyz + xy + yz + zx.$$

Setting this to $0$ and dividing by $xyz$: $1 + \frac1z + \frac1x + \frac1y = 0$.

Question 36

The value of the determinant $\begin{vmatrix} x & x+y & x+2y \\ x+2y & x & x+y \\ x+y & x+2y & x \end{vmatrix}$ is

Solution. Each row sums to $3(x+y)$. $C_1 \to C_1 + C_2 + C_3$, take it out, then subtract $R_1$:

$$\Delta = 3(x+y)\begin{vmatrix} 1 & x+y & x+2y \\ 0 & -y & -y \\ 0 & y & -2y \end{vmatrix} = 3(x+y)\,(2y^2 + y^2) = 9y^2(x+y).$$

Question 37

There are two values of $a$ which make the determinant $\Delta = \begin{vmatrix} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = 86$. The sum of these numbers is

Solution. Expand along the first column:

$$\Delta = 1(2a^2 + 4) – 2(-4a – 20) = 2a^2 + 8a + 44 = 86 \implies a^2 + 4a – 21 = 0.$$

The roots are $3$ and $-7$; their sum is $-4$ (or, directly, $-\frac{4}{1}$ by the sum of roots).

Common mistakes

• Matching the nearest-looking option. In Question 25 the factored value $-(a+b+c)(a-b)(b-c)$ resembles none of the options, and the temptation is to pick the familiar $a^3+b^3+c^3-3abc$ (as the printed key does). One numerical check settles it.
• Dropping a root. In Question 24, $x^2 = 36$ has two roots; in Question 28, the factor $(\sin x – \cos x)^2$ is a single root counted once, not twice — the question asks for distinct roots.
• Assuming $(A+B)^{-1}$ splits. Inversion reverses products but does not distribute over sums (Question 34).
• Forgetting the domain in a trigonometric root count. $\tan x = -2$ has solutions, just none in $\left[-\frac\pi4, \frac\pi4\right]$ (Question 28).

Practise next

• Exemplar Short and Long Answer Q1–23, the full proofs behind these shortcuts, including the identity used in Question 29.
• Exemplar Fill in the Blanks and True or False Q38–58, the rest of the unit.
• NCERT textbook Exercise 4.1 for evaluating determinants by expansion, the method behind Questions 24 and 37.
• NCERT textbook Exercise 4.4 for the adjoint and inverse, as in Questions 33 and 34.