SureDenNCERT Solution Exercise 2
Question 1: Using the property of determinants and without expanding, prove that:
Answer
[Here, the two columns of the determinants are identical]
Question 2: Using the property of determinants and without expanding, prove that:
Answer
Applying R1 → R1 + R2,we have:
Here, the two rows R1 and R3 are identical.
∴Δ = 0
Question 3: Using the property of determinants and without expanding, prove that:
Answer
= 0 [Two columns are identical]
Question 4: Using the property of determinants and without expanding, prove that:
Answer
By applying C3 → C3 + C2, we have:
Here, two columns C1 and C3 are proportional.
∴ Δ = 0.
Question 5: Using the property of determinants and without expanding, prove that:
Answer
= Δ1 + Δ2 (say)
Applying R2 → R2 – R3, we have:
Applying R1 → R1 – R2, we have:
Applying R1 ←→ R3 And R2 ←→ R3, we have:
Applying R1 → R1 - R3, we have:
Applying R2 → R2 – R1, 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:
Answer:
We have,
Applying R1 → cR1, we have:
Applying R1 → cR1, we have:
Here, the two rows R1 and R3 are identical.
∴ Δ = 0.
Question 7: By using properties of determinants, show that:
Answer
Applying R2 → R2 + R1 and R3 → R3 + R1, we have:
= - a2b2c2 (0 -4) = 4a2b2c2
Question 8: By using properties of determinants, show that:
Answer
Applying R1 → R1 − R3 and R2 → R2 − R3, we have:
Applying R1 → R1 + R2, we have:
Expanding along C1, we have:
Hence, the given result is proved.
Applying C1 → C1 − C3 and C2 → C2 − C3, we have:
Applying C1 → C1 + C2, we have:
Expanding along C1, we have:
= (a - b) (b - c)(c - a)(a + b + c)
Hence, the given result is proved.
Question 9: By using properties of determinants, show that:
Answer
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
Applying R3 → R3 + R2, we have:
Expanding along R3, we have:
= (x - y)(z - x)(z - y) [(-xy - yz) + (-x2 – xy + x2)]
= - (x - y)(z - x)(z - y) (xy + yz + zx)
= (x - y)(y - z)(z - x) (xy + yz + zx)
Hence, the given result is proved.
Question 10: By using properties of determinants, show that:
Answer
Applying R1 → R1 + R2 + R3, we have:
Applying C2 → C2 − C1, C3 → C3 − C1, we have:
Expanding along C3, we have:
Hence, the given result is proved.
Applying R1 → R1 + R2 + R3, we have:
Applying C2 → C2 − C1 and 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:
Answer
Applying R1 → R1 + R2 + R3, we have:
Applying C2 → C2 − C1, C3 → C3 − C1, we have:
Expanding along C3, we have:
Δ = (a + b + c)3 (-1) (-1) = (a + b + c)3
Hence, the given result is proved.
Applying C1 → C1 + C2 + C3, we have:
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
Expanding along R3, we have:
= 2(x + y + z)3 (1) (1 - 0) = 2(x + y + z)3
Hence, the given result is proved.
Question 12: By using properties of determinants, show that:
Answer
Applying R1 → R1 + R2 + R3, we have:
Applying C2 → C2 − C1 and C3 → C3 − C1, we have:
Expanding along R1, we have:
= (1 – x3) (1 – x) (1 + x + x2)
= (1 – x3) (1 – x3)
= (1 – x3)2
Hence, the given result is proved.
Question 13: By using properties of determinants, show that:
Answer
Applying R1 → R1 + bR3 and R2 → R2 − aR3, we have:
Expanding along R1, we have:
= (1 + a2 + b2)2 [1 – a2 – b2 + 2a2 – b(-2b)]
= (1 + a2 + b2)2 (1 + a2 + b2)
= (1 + a2 + b2)3
Question 14: By using properties of determinants, show that:
Answer
Taking out common factors a, b, and c from R1, R2, and R3 respectively, we have:
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
Applying C1 → aC1, C2 → 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 |kA| is equal to
A. k|A|
B. k2|A|
C. k3|A|
D. 3k|A|
Answer
A is a square matrix of order 3 x 3
= k3|A|
∴ |KA| = k3 |A|
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, A = [aij] of order n. We can associate a number called the determinant of square matrix A, where aij = (i, j)th element of A.
Thus, the determinant is a number associated to a square matrix.
Hence, the correct answer is C.