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NCERT solution class 12 chapter 4 Determinants exercise 4.2 mathematics part 1

EXERCISE 4.2


Page No 119:

Question 1:

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

Answer:

Question 2:

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

Answer:

Here, the two rows R1 and R3 are identical.

Δ = 0.

Question 3:

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

Answer:

Question 4:

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

Answer:

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:

Answer:

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.

Page No 120:

Question 6:

By using properties of determinants, show that:

Answer:

We have,

Here, the two rows R1 and Rare identical.

∴Δ = 0.

Question 7:

By using properties of determinants, show that:

Answer:

Applying R→ R2 + R1 and R→ R3 + R1, we have:

Question 8:

By using properties of determinants, show that:

(i) 

(ii) 

Answer:

(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:

Answer:

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) 

Answer:

(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) 

Answer:

(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:

Hence, the given result is proved.

Page No 121:

Question 12:

By using properties of determinants, show that:

Answer:

Applying R1 → R1 + R+ R3, we have:

Applying C2 → C2 − Cand C3 → C3 − C1, we have:

Expanding along R1, we have:

Hence, the given result is proved.

Question 13:

By using properties of determinants, show that:

Answer:

Applying R1 → R1 + bRand R2 → R2 − aR3, we have:

Expanding along R1, we have:

Question 14:

By using properties of determinants, show that:

Answer:

Taking out common factors ab, and c from R1, R2, and Rrespectively, we have:

Applying R2 → R2 − Rand R3 → R3 − R1, we have:

Applying C1 → aC1, C→ bC2, and C3 → cC3, we have:

Expanding along R3, we have:

Hence, the given result is proved.

Question 15:

Choose the correct answer.

Let A be a square matrix of order 3 × 3, then is equal to

A.  B.  C.  D.

Answer:

Answer: C

A is a square matrix of order 3 × 3.

Hence, the correct answer is C.

Question 16:

Which of the following is correct?

A. Determinant is a square matrix.

B. Determinant is a number associated to a matrix.

C. Determinant is a number associated to a square matrix.

D. None of these

Answer:

Answer: C

We know that to every square matrix, of order n. We can associate a number called the determinant of square matrix A, where element of A.

Thus, the determinant is a number associated to a square matrix.

Hence, the correct answer is C.


 

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