RBSE Class 12 Maths Solutions Chapter 4 Determinants Ex 4.1

RBSE Class 12 Maths Solutions Chapter 4 Determinants Ex 4.1

Question 1.

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

Question 2.
(i) $\left|\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right|$
Answer:
$\left|\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right|$
= 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 = $\left[\begin{array}{cc}1& 2\\ 4& 2\end{array}\right]$, 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 = $\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right]$, 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) $\left|\begin{array}{ccc}3& -1& -2\\ 0& 0& -1\\ 3& -5& 0\end{array}\right|$
Answer:
Let |A| = $\left|\begin{array}{ccc}3& -1& -2\\ 0& 0& -1\\ 3& -5& 0\end{array}\right|$
Expanding along R2, we get
⇒ |A| = 0 + 0 + 1{3 × (- 5) - (- 1) × 3}
Thus, |A| = 1(- 15 + 3) = - 12

(ii) $\left|\begin{array}{ccc}3& -4& 5\\ 1& 1& -2\\ 2& 3& 1\end{array}\right|$
Answer:
$\left|\begin{array}{ccc}3& -4& 5\\ 1& 1& -2\\ 2& 3& 1\end{array}\right|$
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) $\left|\begin{array}{ccc}0& 1& 2\\ -1& 0& -3\\ -2& 3& 0\end{array}\right|$
Answer:
Let |A| = $\left|\begin{array}{ccc}0& 1& 2\\ -1& 0& -3\\ -2& 3& 0\end{array}\right|$
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) $\left|\begin{array}{ccc}2& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\right|$
Answer:
$\left|\begin{array}{ccc}2& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\right|$
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 = $\left[\begin{array}{ccc}1& 1& -2\\ 2& 1& -3\\ 5& 4& -9\end{array}\right]$, then find |A|.
Answer:
Let |A| = $\left[\begin{array}{ccc}1& 1& -2\\ 2& 1& -3\\ 5& 4& -9\end{array}\right]$
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) $\left|\begin{array}{cc}2& 4\\ 5& 1\end{array}\right|=\left|\begin{array}{cc}2x& 4\\ 6& x\end{array}\right|$
Answer:
Given, $\left|\begin{array}{cc}2& 4\\ 5& 1\end{array}\right|=\left|\begin{array}{cc}2x& 4\\ 6& x\end{array}\right|$
⇒ (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) $\left|\begin{array}{cc}2& 3\\ 4& 5\end{array}\right|=\left|\begin{array}{cc}x& 3\\ 2x& 5\end{array}\right|$
Answer:
$\left|\begin{array}{cc}2& 3\\ 4& 5\end{array}\right|=\left|\begin{array}{cc}x& 3\\ 2x& 5\end{array}\right|$
⇒ (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.

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