RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 4 Determinants Ex 4.1

Question 1.
\(\left|\begin{array}{rr} 2 & 4 \\ -5 & -1 \end{array}\right|\)
Answer:
\(\left|\begin{array}{rr} 2 & 4 \\ -5 & -1 \end{array}\right|\)
= 2 × (- 1) - (- 5) × 4
= - 2 + 20 = 18

Question 2.
(i) \(\left|\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right|\)
Answer:
\(\left|\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right|\)
= cos²θ - (- sin θ) sin θ
= cos²θ + sin²θ = 1

(ii) \(\left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right|\)
Answer:
\(\left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right|\)
= (x² - x + 1) (x + 1) - (x + 1) (x - 1)
= x³ + 1 - x² + 1 = x³ - x² + 2

Question 3.
If A = \(\left[\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right]\), then show that: |2A| = 4 |A|
Answer:

= 4(2 - 8) = 4 × (- 6) = - 24 ....... (2)
Form (1) and (2), we get
|2A| = 4|A|

Question 4.
If A = \(\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right]\), then show that: |3A| = 27|A|
Answer:

Expanding along R₃, we get
|3A| = 0 - 0 + 12(3 × 3 - 0 × 0) = 108
∴ |3A| = 108
∴ |3A| = 27 × 4 = 27 |A|
Thus, |3A| = 27 |A|
Hence Proved.

Question 5.
Evaluate the determinants:
(i) \(\left|\begin{array}{rrr} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right|\)
Answer:
Let |A| = \(\left|\begin{array}{rrr} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right|\)
Expanding along R₂, we get
⇒ |A| = 0 + 0 + 1{3 × (- 5) - (- 1) × 3}
Thus, |A| = 1(- 15 + 3) = - 12

(ii) \(\left|\begin{array}{rrr} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array}\right|\)
Answer:
\(\left|\begin{array}{rrr} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array}\right|\)
Expanding along R₁, we get
|A| = 3(1 × 1 - (3) × (- 2)) + 4(1 × 1 - 2 × (- 2)) + 5(1 × 3 - 2 × 1)
= 3(1 + 6) + 4(1 + 4) + 5(3 - 2)
= (3 × 7) + (4 × 5) + (5 × 1) = 46

(iii) \(\left|\begin{array}{rrr} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array}\right|\)
Answer:
Let |A| = \(\left|\begin{array}{rrr} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array}\right|\)
Expanding along R₁, we get
A = 0 - 1(- 1 × 0 - (- 2) × (- 3)) + 2(- 1 × 3 - (- 2) × 0)
= - 1(0 - 6) + 2(- 3 - 0) = 6 - 6 = 0

(iv) \(\left|\begin{array}{rrr} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right|\)
Answer:
\(\left|\begin{array}{rrr} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right|\)
Expanding along R₁, we get
|A| = 0 + 2(2 × 0 - 3 × (- 2)) + 1(2 × (- 5)- 3 × (- 1))
= 2(0 + 6) + 1(- 10 + 3)
= 2 × 6 + 1 × (- 7)
= 12 - 7 = 5

Question 6.
If A = \(\left[\begin{array}{rrr} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right]\), then find |A|.
Answer:
Let |A| = \(\left[\begin{array}{rrr} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right]\)
Expanding along R₁, we get
|A| = 1(1 × (- 9) - 4 × (- 3)) - 1(2 × (- 9) - 5 × (- 3)) - 2(2 × 4 - 5 × 1)
= |A| = 1( - 9 + 12) - 1(- 18 + 15) - 2(8 - 5)
= (1 × 3) - (1 × (- 3) - (2 × 3)
= 3 + 3 - 6 = 6 - 6 = 0
Thus, |A| = 0

Question 7.
Find the value of x, if
(i) \(\left|\begin{array}{ll} 2 & 4 \\ 5 & 1 \end{array}\right|=\left|\begin{array}{rr} 2 x & 4 \\ 6 & x \end{array}\right|\)
Answer:
Given, \(\left|\begin{array}{ll} 2 & 4 \\ 5 & 1 \end{array}\right|=\left|\begin{array}{rr} 2 x & 4 \\ 6 & x \end{array}\right|\)
⇒ (2 × 1) - (5 × 4) = (2x × x) - (6 × 4)
⇒ 2 - 20 = 2x² - 24 ⇒ - 18 = 2x² - 24
⇒ 2x² = 24 - 18
⇒ 2x² = 6
⇒ x² = 3
∴ x = ± √3

(ii) \(\left|\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}\right|=\left|\begin{array}{cc} x & 3 \\ 2 x & 5 \end{array}\right|\)
Answer:
\(\left|\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}\right|=\left|\begin{array}{cc} x & 3 \\ 2 x & 5 \end{array}\right|\)
⇒ (2 × 5) - (4 × 3) = (x × 5) - (2x × 3)
⇒ 10 - 12 = 5x - 6x
⇒ - 2 = - x ⇒ x = 2

Question 8.
If \(\left|\begin{array}{cc} x & 2 \\ 18 & x \end{array}\right|=\left|\begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array}\right|\), then x is equal to:
(A) 6
(B) ± 6
(C) - 6
(D) 0
Answer:
Given, \(\left|\begin{array}{cc} x & 2 \\ 18 & x \end{array}\right|=\left|\begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array}\right|\)
⇒ (x × x) - (18 × 2) = (6 × 6) - (18 × 2)
⇒ x² - 36 = 36 - 36 ⇒ x² - 36 = 0
⇒ x² = 36 ⇒ x = √36 ⇒ x = ±6
Thus, option (B) is correct.