# 4-DETERMINANTS

Here we are going through the ncert solutions for class 12 maths chapter 4- determinants so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions more easily

#### Question 1:

Evaluate the determinants in Exercises 1 and 2.  = 2(−1) − 4(−5) = − 2 + 20 = 18

#### Question 2:

Evaluate the determinants in Exercises 1 and 2.

(i) (ii) (i) = (cos θ)(cos θ) − (−sin θ)(sin θ) = cos2 θ+ sin2 θ = 1

(ii) = (x2 − x + 1)(x + 1) − (x − 1)(x + 1)

x3 − x2 + x + x2 − x + 1 − (x2 − 1)

x3 + 1 − x2 + 1

x3 − x2 + 2

#### Question 3:

If , then show that The given matrix is . #### Question 4:

If , then show that The given matrix is .

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.  From equations (i) and (ii), we have: Hence, the given result is proved.

#### Question 5:

Evaluate the determinants

(i) (iii) (ii) (iv) (i) Let .

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation. (ii) Let .

By expanding along the first row, we have: (iii) Let By expanding along the first row, we have: (iv) Let By expanding along the first column, we have: #### Question 6:

If , find .

Let By expanding along the first row, we have: #### Question 7:

Find values of x, if

(i) ∣∣∣2541∣∣∣=∣∣∣2x64x∣∣∣2451=2x46x

(ii)
∣∣∣2435∣∣∣=∣∣∣x2x35∣∣∣2345=x32x5

(i)  (ii)  #### Question 8:

If , then x is equal to

(A) 6 (B) ±6 (C) −6 (D) 0  Hence, the correct answer is B.

​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ EXERCISE 4.2

Question 1:

Using the property of determinants and without expanding, prove that:  #### Question 2:

Using the property of determinants and without expanding, prove that:  Here, the two rows R1 and R3 are identical. Δ = 0.

#### Question 3:

Using the property of determinants and without expanding, prove that:  #### Question 4:

Using the property of determinants and without expanding, prove that:  By applying C→ C3 + C2, we have: Here, two columns C1 and Care proportional. Δ = 0.

#### Question 5:

Using the property of determinants and without expanding, prove that:  Applying R2 → R2 − R3, we have: Applying R1 ↔R3 and R2 ↔R3, we have:  Applying R→ R1 − R3, we have: Applying R1 ↔R2 and R2 ↔R3, we have: From (1), (2), and (3), we have: Hence, the given result is proved.

#### Question 6:

By using properties of determinants, show that: We have,  Here, the two rows R1 and Rare identical.

Δ = 0.

#### Question 7:

By using properties of determinants, show that:  Applying R→ R2 + R1 and R→ R3 + R1, we have: #### Question 8:

By using properties of determinants, show that:

(i) (ii) (i) Applying R1 → R1 − Rand R2 → R2 − R3, we have: Applying R1 → R1 + R2, we have: Expanding along C1, we have: Hence, the given result is proved.

(ii) Let .

Applying C1 → C1 − Cand C2 → C2 − C3, we have: Applying C1 → C1 + C2, we have: Expanding along C1, we have: Hence, the given result is proved.

#### Question 9:

By using properties of determinants, show that:  Applying R2 → R2 − Rand R3 → R3 − R1, we have: Applying R3 → R3 + R2, we have: Expanding along R3, we have: Hence, the given result is proved.

#### Question 10:

By using properties of determinants, show that:

(i) (ii) (i) Applying R1 → R1 + R+ R3, we have: Applying C2 → C2 − C1, C3 → C3 − C1, we have: Expanding along C3, we have: Hence, the given result is proved.

(ii) Applying R1 → R1 + R+ R3, we have: Applying C2 → C2 − Cand C3 → C3 − C1, we have: Expanding along C3, we have: Hence, the given result is proved.

#### Question 11:

By using properties of determinants, show that:

(i) (ii) (i) Applying R1 → R1 + R+ R3, we have: Applying C2 → C2 − C1, C3 → C3 − C1, we have: Expanding along C3, we have: Hence, the given result is proved.

(ii) Applying C1 → C1 + C+ C3, we have: Applying R2 → R2 − Rand R3 → R3 − R1, we have: Expanding along R3, we have: