NCERT Solutions Class 12 Maths Chapter 4 Ex 4.1 | LearnCBSEHub

Determinants mastery

NCERT EXERCISE 4.1 • Q1 — Q8 COMPLETE

💡 Key Mastery Rule

A determinant of order 2 is evaluated by subtracting the product of elements: $\Delta = (a_{11}a_{22}) – (a_{12}a_{21})$.

For order 3, we expand along any row (usually Row 1) using the sign pattern:

Question 01

Evaluate: $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$

Evaluating using cross-multiplication:

$$\Delta = (2)(-1) – (4)(-5)$$ $$\Delta = -2 – (-20) = -2 + 20 = 18$$

Result: 18

Question 02

Evaluate: (i) $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$ (ii) $\begin{vmatrix} x^2-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}$

Part (i) Solution $$\Delta = (\cos\theta)(\cos\theta) – (-\sin\theta)(\sin\theta)$$ $$\Delta = \cos^2\theta + \sin^2\theta = 1$$ Part (ii) Solution $$\Delta = (x^2-x+1)(x+1) – (x-1)(x+1)$$ $$\Delta = (x^3+1) – (x^2-1) \quad \text{[Using } a^3+b^3 \text{ and } a^2-b^2 \text{ identities]}$$ $$\Delta = x^3 – x^2 + 2$$

Results: (i) 1, (ii) $x^3 – x^2 + 2$

Question 03

If $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$, show $|2A| = 4|A|$

RHS Calculation $$|A| = (1)(2) – (2)(4) = 2 – 8 = -6$$ $$4|A| = 4(-6) = -24$$ LHS Calculation $$2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$$ $$|2A| = (2)(4) – (4)(8) = 8 – 32 = -24$$

✅ LHS = RHS. Hence Proved.

Question 04

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$, show $|3A| = 27|A|$

Evaluation of |A|

Expanding along Column 1 (most zeros): $$|A| = 1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} – 0 + 0 = 1(4 – 0) = 4$$ $$27|A| = 27 \times 4 = 108$$

Evaluation of |3A|

$$3A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}$$ Expanding along Column 1: $$|3A| = 3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} = 3(36 – 0) = 108$$

✅ LHS = RHS. Hence Proved.

Question 05

Evaluate the determinants (i) to (iv).

(i) Solution (Expand along R2)

$$\Delta = 0 + 0 – (-1) \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix} = 1(-15 – (-3)) = -12$$

(iii) Solution (Expand along R1)

$$\Delta = 0 – 1 \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2 \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix} = -1(0 – 6) + 2(-3 – 0) = 6 – 6 = 0$$

Note: A skew-symmetric determinant of odd order is always 0.

Question 06

Find |A| if $A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$

Expanding along R1: $$|A| = 1 \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} – 1 \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} + (-2) \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}$$ $$|A| = 1(-9 – (-12)) – 1(-18 – (-15)) – 2(8 – 5)$$ $$|A| = 3 – (-3) – 6 = 3 + 3 – 6 = 0$$

Result: 0

Question 07

Find $x$: (i) $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$ (ii) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$

Solution (i) $$2 – 20 = 2x^2 – 24 \implies -18 = 2x^2 – 24$$ $$2x^2 = 6 \implies x^2 = 3 \implies x = \pm\sqrt{3}$$ Solution (ii) $$10 – 12 = 5x – 6x \implies -2 = -x \implies x = 2$$

Results: (i) $\pm\sqrt{3}$, (ii) 2

Question 08 (MCQ)

If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$, then $x$ equals:

$$x^2 – 36 = 36 – 36$$ $$x^2 – 36 = 0 \implies x^2 = 36 \implies x = \pm 6$$

Correct Option: (B) ± 6