## NCERT Solutions Class 12th Maths Chapter 4: Determinants## Exercise 4.1
Expand along R = 2 × (-1) - 4 × (-5) = -2 + 20 = 18
= 3 × (3 × 12 - 6 × 0) - 0 × (0 × 12 - 6 × 0) + 3 × (0 × 0 - 6 × 0) = 3 × (36 - 0) - 0 + 3 × 0 = 3 × 36 - 0 = 108 Now, RHS = 27|A| Expand along R1 RHS = 27|A| = 27 = 27 [1 × (1 × 4 - 2 × 0) - 0 × (0 × 4 - 2 × 0) + 1 × (0 × 0 - 2 × 0) = 27 [(4 - 0) - 0 + 1 × 0] = 27 [4 - 0 + 0] = 27 × 4 = 108 LHS = RHS Hence, proved.
Expand along R = 3 × (0 × 0 - (-1) × (-5)) - (-1) × (0 × 0 - (-1) × 3) - 2 × (0 × (-5) - 0 × 3) = 3 × (0 - 5) + (0 + 3) - 2 × 0 = 3 × (-5) + 3 - 0 = -15 + 3 = -12 Expand along R = 3 × (1 × 1 - (-2) × 3) - (-4) × (1 × 1 - (-2) × 2) + 5 × (1 × 3 - 1 × 2) = 3 × (1 + 6) + 4 × (1 + 4) + 5 × (3 - 2) = 3 × 7 + 4 × 5 + 5 × 1 = 21 + 20 + 5 = 46 Expand along R = 0 × (0 × 0 - (-3) × 3) - 1 × ((-1) × 0 - (-3) × (-2)) + 2 × ((-1) × 3 - 0 × (-3)) = 0 - (0 - 6) + 2 × (-3 - 0) = 6 - 6 = 0 Expand along R = 2 × (2 × 0 - (-1) × (-5)) - (-1) × (0 × 0 - (-1) × 3) - 2 × (0 × (-5) - 2 × 3) = 2 × (0 - 5) + (0 + 3) - 2 × (0 - 6) = -10 + 3 + 12 = 5 Expand along R |A| = 1 × (1 × (-9) - (-3) × 4) - 1 × (2 × (-9) - (-3) × 5) - 2 × (2 × 4 - 1 × 5) = (-9 + 12) - (-18 + 15) - 2 × (8 - 5) = (3) - (-3) - 2 × 3 = 3 + 3 - 6 = 0
2 × 1 - 4 × 5 = 2x × x - 4 × 6 2 - 20 = 2x -18 = 2(x -9 = x x x x = ± √3 2 × 5 - 3 × 4 = x × 5 - 3 × 2x 10 - 12 = 5x - 6x -2 = -x x = 2 x × x - 2 × 18 = 6 × 6 - 2 × 18 x x x x = ±6 Hence, (B) is the correct answer. ## Exercise 4.2
We know that the area of a triangle with vertices (x
= 1/2 × [1 × (0 × 1 - 1 × 3) - 0 × (6 × 1 - 1 × 4) + 1 × (6 × 3 - 0 × 4)] = 1/2 × [(0 - 3) - 0 + (18 - 0)] = 1/2 × (18 - 3) = 1/2 × 15 = 7.5 square units
= 1/2 × [2 × (1 × 1 - 1 × 8) - 7 × (1 × 1 - 1 × 10) + 1 × (1 × 8 - 1 × 10)] = 1/2 × [2 × (1 - 8) - 7 × (1 - 10) + (8 - 10)] = 1/2 × [2 × (-7) - 7 × (-9) - 2] = 1/2 × [-14 + 63 - 2] = 1/2 × [47] = 23.5 square units
= 1/2 × [-2 × (2 × 1 - 1 × (-8)) - (-3) × (3 × 1 - 1 × (-1)) + 1 × (3 × (-8) - 2 × (-1))] = 1/2 × [-2 × (2 + 8) + 3 × (3 + 1) + (-24 + 2)] = 1/2 × [-2 × (10) + 3 × 4 - 22] = 1/2 × (-20 + 12 - 22) = 1/2 × (-30) = |-15| = 15 square units
We know that the area of a triangle formed by three collinear points is always zero. Therefore, = 1/2 × [a × {(c + a) - (a + b)} - (b + c) × (b - c) + 1 × {b(a + b) - (c + a)c}] = 1/2 × [a × {c + a - a - b} - (b = 1/2 × [a(c - b) - b = 1/2 × [ac - ab + ab - ac] = 1/2 × 0 = 0 square units Hence, the points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.
We know that the area of a triangle with vertices (x 4 = 1/2 × |k (0 × 1 - 1 × 2) - 0 (4 × 1 - 1 × 0) + 1 (4 × 2 - 0 × 0)| 8 = |k (0 - 2) - 0 + (8 - 0)| 8 = |-2k + 8| 8 - 2k = ± 8 8 - 2k = 8 ⇒ 2k = 0 ⇒
8 - 2k = -8 ⇒ 2k = 16 ⇒ Therefore, k is 0 or 8.
4 = 1/2 × |-2 (4 × 1 - 1 × k) - 0 (0 × 1 - 1 × 0) + 1 (0 × k - 4 × 0)| 8 = |-2 (4 - k) - 0 + (0 - 0)| 8 = |-8 + 2k| 2k - 8 = ± 8 2k - 8 = 8 ⇒ 2k = 16 ⇒
2k - 8 = -8 ⇒ 2k = 0 ⇒ Therefore, k is 8 or 0.
Now, points A, B, and C are collinear. We know that the area of a triangle formed by three collinear points is always zero. Therefore, Area of triangle ABC = 0 1 (6 × 1 - 1 × y) - 2 (3 × 1 - 1 × x) + 1 (3 × y - 6 × x) = 0 6 - y - 2 (3 - x) + 3y - 6x = 0 2y - 6x + 6 - 6 + 2x = 0 2y - 4x = 0 2(y - 2x) = 0 y - 2x = 0
Therefore, equation of line joining the points (1, 2) and (3, 6) is 2x = y.
Now, points A, B, and C are collinear. We know that the area of a triangle formed by three collinear points is always zero. Therefore, Area of triangle ABC = 0 3 (3 × 1 - 1 × y) - 1 (9 × 1 - 1 × x) + 1 (9 × y - 3 × x) = 0 3 (3 - y) - (9 - x) + (9y - 3x) = 0 9 - 3y - 9 + x + 9y - 3x = 0 6y - 2x = 0 2(3y - x) = 0 3y - x = 0
Therefore, equation of line joining the points (3, 1) and (9, 3) is x = 3y.
We know that the area of a triangle with vertices (x 35 = 1/2 × |2 (4 × 1 - 1 × 4) - (-6) (5 × 1 - 1 × k) + 1 (5 × 4 - 4 × k)| 70 = |2 (4 - 4) + 6 (5 - k) + (20 - 4k)| 70 = |2 (0) + 30 - 6k + 20 - 4k| 70 = |50 - 10k| 70 = 10 |5 - k| 7 = |5 - k| 5 - k = +-7 5 - k = 7 ⇒
5 - k = -7 ⇒ Hence, (D) is the correct answer. ## Exercise 4.3
The minor of an element a
Minor of a Minor of a Minor of a
Minor of a Minor of a Minor of a
Cofactor of an element a A A A A A A A A A Cofactor of an element a A A A A A A A A A Cofactors of a A A A Therefore, = 2 (7) + 0 (7) + 1 (-7) = 14 + 0 - 7 = 7 Cofactors of a A A A Therefore, = yz (z - y) + zx (x - z) + xy (y - x) = yz = x = x = x = (z - y) [x = (z - x) [x = (z - x) [x (x - z) - y (x - z)] = (z - x) (x - z) (x - y) = (x - y) (y - z) (z - x) ## Exercise 4
A A A A
A A A A
Now, A (adj A)
A A A A |A| = = 2 (3) - (-2) (4) = 6 + 8 = 14 Since, |A| ≠ 0. Therefore, A A A A A |A| = = (-1) (2) - 5 (-3) = -2 + 8 = 6 Since, |A| ≠ 0. Therefore, A
|A| = 1 {2 (5) - 4 (0)} - 2 {0 (5) - 4 (0)} + 3 {0 (0) - 2 (0)} |A| = (10 - 0) - 2 (0 - 0) + 3 (0 - 0) |A| = 10 - 0 + 0 |A| = 10 Since, |A| ≠ 0. Therefore, A
|A| = 1 {3 (-1) - 0 (2)} - 0 {3 (-1) - 0 (5)} + 0 {3 (2) - 3 (5)} |A| = -3 - 0 + 0 |A| = -3 Since, |A| ≠ 0. Therefore, A
|A| = 2 {(-1) (1) - 0 (2)} - 1 {4 (1) - 0 (-7)} + 3 {4 (2) - (-1) (-7)} |A| = 2 (-1) - (4) + 3 {8 - 7} |A| = -2 - 4 + 3(1) |A| = -3 Since, |A| ≠ 0. Therefore, A
|A| = 1 {(2) (4) - (-3) (-2)} - (-1) {0 (4) - (-3) (3)} + 2 {0 (-2) - 2 (3)} |A| = (8 - 6) + (0 + 9) + 2 (0 - 6) |A| = 2 + 9 + 2 (-6) |A| = 1 - 12 |A| = -1 Since, |A| ≠ 0. Therefore, A
|A| = 1 {(cos α) (-cos α) - (sin α) (sin α)} - 0 {0 (-cos α) - sin α (0)} + 0 {0 (sin α) - 0 (cos α)} |A| = (-cos |A| = -(cos |A| = -1 Since, |A| ≠ 0. Therefore, A A A A A B B B B Now, AB AB AB AB
Now, LHS = A Hence, proved that A A A Multiply both sides by A A A A A AI - 5I = -7A A - 5I = -7A
Now, A So, 4 + a = 0
3 + a + b = 0 3 + (-4) + b = 0 3 - 4 + b = 0 -1 + b = 0
Hence, a = -4 and b = 1.
Now, LHS = A Hence, proved that A A A Multiply both sides by A A A A A
Hence, proved A A Multiply both sides by A A A A
We know that (adj A) A = |A|I |(adj A) A| = [|A| (|A| (|A|) + 0 (0)) + 0 (0 (|A|) + 0 (0)) + 0 (0 (0) + |A| (0))] |(adj A) A| = [|A| (|A| |(adj A) A| = |A| |(adj A)| = |A| Hence, (B) is the correct answer.
We know that the matrix A is invertible. Therefore, A det (A det (A det (A det (A det (A Hence, (B) is the correct answer. ## Exercise 4.5
Given system of equations: x + 2y = 2 and 2x + 3y = 3 (1)x + 2y = 2 and 2x + 3y = 3 This system of equations can be rewritten in the form of AX = B, where = 1 (3) - 2 (2) = 3 - 4 = -1 |A| ≠ 0. So, A Hence, the given system of equations is consistent.
Given system of equations: 2x - y = 5 and x + y = 4 2x - (1)y = 5 and (1)x + (1)y = 4 This system of equations can be rewritten in the form of AX = B, where = 2 (1) - 1 (1) = 2 - 1 = 1 |A| ≠ 0. So, A Hence, the given system of equations is consistent.
Given system of equations: x + 3y = 5 and 2x + 6y = 8 (1)x + 3y = 5 and 2x + 6y = 8 This system of equations can be rewritten in the form of AX = B, where (adj A) B ≠ O No solutions exist for the given system of equations. Hence, the given pair of equations is inconsistent.
Given system of equations: x + y + z = 1, 2x + 3y + 2z = 2, and ax + ay + 2az = 4 This system of equations can be rewritten in the form of AX = B, where = 1 {3 (2a) - 2 (a)} - 1 {2 (2a) - 2 (a)} + 1 {2 (a) - 3 (a)} = {6a - 2a} - {4a - 2a} + {2a - 3a} = 4a - 2a - a = a |A| ≠ 0. So, A Hence, the given system of equations is consistent.
Given system of equations: 3x - y - 2z = 2, 2y - z = -1, and 3x - 5y = 3 3x - y - 2z = 2, 0x + 2y - z = -1, and 3x - 5y + 0z = 3 This system of equations can be rewritten in the form of AX = B, where = 3 {2 (0) - (-1) (-5)} - (-1) {0 (0) - (-1) (3)} - 2 {0 (-5) - 2 (3)} = 3 {0 - 5} + {3} - 2 {-6} = -15 + 3 + 12 = 0 |A| = 0. So, A A A A A A A A A A (adj A) B ≠ O No solutions exist for the given system of equations. Hence, the given pair of equations is inconsistent.
Given system of equations: 5x - y + 4z = 5, 2x + 3y + 5z = 2, and 5x - 2y + 6z = -1 This system of equations can be rewritten in the form of AX = B, where = 5 {3 (6) - (5) (-2)} - (-1) {2 (6) - (5) (5)} + 4 {2 (-2) - 3 (5)} = 5 {18 + 10} + {12 - 25} + 4 {-4 - 15} = 5 (28) - 13 + 4 (-19) = 140 - 13 - 76 = 51 |A| ≠ 0. So, A Hence, the given system of equations is consistent.
Given system of equations: 5x + 2y = 4 and 7x + 3y = 5 This system of equations can be rewritten in the form of AX = B, where Now, AX = B X = A x = 2 and y = -3 is the solution of given system of equations.
Given system of equations: 2x - y = -2 and 3x + 4y = 3 This system of equations can be rewritten in the form of AX = B, where Now, AX = B X = A x = -5/11 and y = 12/11 is the solution of given system of equations.
Given system of equations: 4x - 3y = 3 and 3x - 5y = 7 This system of equations can be rewritten in the form of AX = B, where Now, AX = B X = A This system of equations can be rewritten in the form of AX = B, where x = -6/11 and y = -19/11 is the solution of given system of equations.
Given system of equations: 5x + 2y = 3 and 3x + 2y = 5 This system of equations can be rewritten in the form of AX = B, where Now, AX = B X = A x = -1 and y = 4 is the solution of given system of equations.
Given system of equations: 2x + y + z = 1, x - 2y - z = 3/2 and 3y - 5z = 9 2x + y + z = 1, x - 2y - z = 3/2 and 0x + 3y - 5z = 9 This system of equations can be rewritten in the form of AX = B, where = 2 {(-2) (-5) - (-1) (3)} - 1 {1 (-5) - (-1) (0)} + 1 {1 (3) - (-2) (0)} = 2 {10 + 3} - (-5) + (3) = 2 (13) + 5 + 3 = 26 + 8 = 34 |A| ≠ 0. So, A A A A A A A A A A x = 1, y = -1/2 and z = -3/2 is the solution of given system of equations.
Given system of equations: x - y + z = 4, 2x + y - 3z = 0 and x + y + z = 2 This system of equations can be rewritten in the form of AX = B, where = 1 {1 (1) - (-3) (1)} - (-1) {2 (1) - (-3) (1)} + 1 {2 (1) - 1 (1)} = (1 + 3) + (2 + 3) + (2 - 1) = 4 + 5 + 1 = 10 |A| ≠ 0. So, A A A A A A A A A A x = 2, y = -1 and z = 1 is the solution of given system of equations.
Given system of equations: 2x + 3y + 3z = 5, x - 2y + z = -4 and 3x - y - 2z = 3 This system of equations can be rewritten in the form of AX = B, where = 2 {(-2) (-2) - 1 (-1)} - 3 {1 (-2) - 1 (3)} + 3 {1 (-1) - (-2) (3)} = 2 (4 + 1) - 3 (-2 - 3) + 3 (-1 + 6) = 2 (5) - 3 (-5) + 3 (5) = 10 + 15 + 15 = 40 |A| ≠ 0. So, A A A A A A A A A A x = 1, y = 2 and z = -1 is the solution of given system of equations.
Given system of equations: x - y + 2z = 7, 3x + 4y - 5z = -5 and 2x - y + 3z = 12 This system of equations can be rewritten in the form of AX = B, where = 1 {4 (3) - (-5) (-1)} - (-1) {3 (3) - (-5) (2)} + 2 {3 (-1) - 4 (2)} = (12 - 5) + (9 + 10) + 2 (-3 - 8) = 7 + 19 + 2 (-11) = 26 - 22 = 4 |A| ≠ 0. So, A A A A A A A A A A x = 2, y = 1 and z = 3 is the solution of given system of equations.
Given system of equations: 2x - 3y + 5z = 11 This system of equations can be rewritten in the form of AX = B, where = 2 {2 (-2) - (-4) (1)} - (-3) {3 (-2) - (-4) (1)} + 5 {3 (1) - 2 (1)} = 2 (-4 + 4) + 3 (-6 + 4) + 5 (3 - 2) = 2 (0) + 3 (-2) + 5 (1) = 0 - 6 + 5 = -1 |A| ≠ 0. So, A A A A A A A A A A x = 1, y = 2 and z = 3 is the solution of given system of equations.
Let the cost of onion per kg (in Rs) be x, the cost of wheat per kg (in Rs) be y and the cost of rice per kg (in Rs) be z. It is given that cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Therefore, 4x + 3y + 2z = 60 2x + 4y + 6z = 90 6x + 2y + 3z = 70 This system of equations can be rewritten in the form of AX = B, where = 4 {4 (3) - (6) (2)} - 3 {2 (3) - 6 (6)} + 2 {2 (2) - 4 (6)} = 4 (12 - 12) - 3 (6 - 36) + 2 (4 - 24) = 4 (0) - 3 (-30) + 2 (-20) = 0 + 90 - 40 = 50 |A| ≠ 0. So, A A A A A A A A A A x = 5, y = 8 and z = 8. Hence, the cost of onion per kg is Rs 5, cost of wheat per kg is Rs 8 and cost of rice pe kg is Rs 8. ## Miscellaneous Exercise= x {-x (x) - 1 (1)} - sin θ {-sin θ (x) - 1 (cos θ)} + cos θ {-sin θ (1) - (-x) (cos θ)} = x (-x = -x = -x = -x = -x = -x = cos α cos β {cos β (cos α) - 0 (sin α sin β)} - cos α sin β {-sin β (cos α) - 0 (sin α cos β) - sin α {-sin β (sin α sin β) - cos β (sin α cos β)} = cos α cos β {cos α cos β - 0} - cos α sin β (-sin β cos α - 0) - sin α (-sin α sin = cos = cos = cos = sin = 1 = 1 {3 (1) - 0 (-2)} - 2 {-1 (1) - 0 (0)} - 2 {-1 (-2) - 3 (0)} = (3 + 0) - 2 (-1 - 0) - 2 (2 - 0) = 3 + 2 - 4 = 1 |B| ≠ 0. So, B B B B B B B B B B = 1 {3 (5) - 1 (1)} - (-2) {-2 (5) - 1 (1)} + 1 {-2 (1) - 3 (1)} = (15 - 1) + 2 (-10 - 1) + (-2 - 3) = 14 + 2 (-11) - 5 = 9 - 22 = -13 |A| ≠ 0. So, A A A A A A A A A A = 14 {4 (-1) - (-3) (-3)} - 11 {11 (-1) - (-3) (-5)} - 5 {11 (-3) - 4 (-5)} = 14 (-4 - 9) - 11 (-11 - 15) - 5 (-33 + 20) = 14 (-13) - 11 (-26) - 5 (-13) = -182 + 286 + 65 = 169 |adj A| ≠ 0. So, (adj A) (adj A) (adj A) (adj A) (adj A) (adj A) (adj A) (adj A) (adj A) (adj A) |A = (-14/13) [-4/169 - 9/169] + (11/13) [-11/169 - 15/169] + (5/13) [-33/169 + 20/169] = (-14/13) [-13/169] + (11/13) [-26/169] + (5/13) [-13/169] = (-14/13) [-1/13] + (11/13) [-2/13] + (5/13) [-1/13] = 14/169 - 22/169 - 5/169 = -13/169 = -1/13 (A (A (A (A (A (A (A (A (A (A Hence, proved that Expand Along C = 2(x + y) [0 {y (y) - (x - y) (x)} - 0 {-x (y) - y (x)} + 1 {-x (x - y) - y (y)}] = 2(x + y) [0 - 0 - x = -2(x + y) [x = -2 (x Expand along C = 0 [y (x + y) - (-x) (x)] - 0 [-y (x + y) - 0 (x)] + 1 [-y (-x) - 0 (y)] = 0 - 0 + (xy - 0) = xy
Given system of equations: 2/x + 3/y + 10/z = 4 4/x - 6/y + 5/z = 1 6/x + 9/y - 20/z = 2 This system of equations can be rewritten in the form of AX = B, where = 2 {-6 (-20) - 5 (9)} - 3 {4 (-20) - 5 (6)} + 10 {4 (9) - (-6) (6)} = 2 (120 - 45) - 3 (-80 - 30) + 10 (36 + 36) = 2 (75) - 3 (-110) + 10 (72) = 150 + 330 + 720 = 1200 |A| ≠ 0. So, A A A A A A A A A A Now, AX = B X = A 1/x = 1/2 ⇒ x = 2 1/y = 1/3 ⇒ y = 4 1/z = 1/5 ⇒ z = 5 x = 2, y = 3 and z = 5 is the solution of given system of equations.
= x {yz - 0 (0)} - 0 {0 (z) - 0 (0)} + 0 {0 (0) - y (0)} = x (yz) = xyz |A| ≠ 0. So, A A A A A A A A A A Hence, (A) is the correct answer. = 1 [1 (1) - sin θ (-sin θ)] - sin θ [-sin θ (1) - sin θ (-1)] + 1 [-sin θ (-sin θ) - 1 (-1)] = [1 + sin = 2 + 2 sin = 2 (1 + sin It is given that 0 ≤ θ ≤ 2π. Therefore, 0 ≤ sin θ ≤ 1 = 0 ≤ sin = 1 + 0 ≤ 1 + sin = 1 ≤ 1 + sin = 2 (1) ≤ 2 (1 + sin = 2 ≤ 2 (1 + sin = 2 ≤ Det (A) ≤ 4 Hence, (D) is the correct answer. |