RD Sharma Solutions for Class 12 Maths Chapter 6 PDF
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Summary
This document provides solutions to problems in Exercise 6.1 of RD Sharma Solutions for Class 12 Maths Chapter 6, Determinants. It covers the evaluation of determinants for various types of matrices using minors and cofactors. The solutions demonstrate the methods of calculating determinants for matrices by expanding along different columns.
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RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Exercise 6.1 Page No: 6.10 1. Write the minors and cofactors of each element of the first column of the following matrices and hen...
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Exercise 6.1 Page No: 6.10 1. Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case: Solution: (i) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column.The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, Cij = (–1)i+j × Mij Given, From the given matrix we have, M11 = –1 M21 = 20 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants C11 = (–1)1+1 × M11 = 1 × –1 = –1 C21 = (–1)2+1 × M21 = 20 × –1 = –20 Now expanding along the first column we get |A| = a11 × C11 + a21× C21 = 5× (–1) + 0 × (–20) = –5 (ii) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, Cij = (–1)i+j × Mij Given From the above matrix we have M11 = 3 M21 = 4 C11 = (–1)1+1 × M11 =1×3 =3 C21 = (–1)2+1 × 4 = –1 × 4 = –4 Now expanding along the first column we get |A| = a11 × C11 + a21× C21 = –1× 3 + 2 × (–4) = –11 (iii) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Also, Cij = (–1)i+j × Mij Given, M31 = –3 × 2 – (–1) × 2 M31 = –4 C11 = (–1)1+1 × M11 = 1 × –12 = –12 C21 = (–1)2+1 × M21 = –1 × –16 = 16 C31 = (–1)3+1 × M31 = 1 × –4 = –4 Now expanding along the first column we get |A| = a11 × C11 + a21× C21+ a31× C31 = 1× (–12) + 4 × 16 + 3× (–4) = –12 + 64 –12 = 40 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (iv) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, Cij = (–1)i+j × Mij Given, M31 = a × c a – b × bc M31 = a2c – b2c C11 = (–1)1+1 × M11 = 1 × (ab2 – ac2) = ab2 – ac2 C21 = (–1)2+1 × M21 = –1 × (a2b – c2b) = c2b – a2b C31 = (–1)3+1 × M31 = 1 × (a2c – b2c) = a2c – b2c Now expanding along the first column we get |A| = a11 × C11 + a21× C21+ a31× C31 = 1× (ab2 – ac2) + 1 × (c2b – a2b) + 1× (a2c – b2c) = ab2 – ac2 + c2b – a2b + a2c – b2c RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (v) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, Cij = (–1)i+j × Mij Given, M31 = 2×0 – 5×6 M31 = –30 C11 = (–1)1+1 × M11 =1×5 =5 C21 = (–1)2+1 × M21 = –1 × –40 = 40 C31 = (–1)3+1 × M31 = 1 × –30 = –30 Now expanding along the first column we get RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants |A| = a11 × C11 + a21× C21+ a31× C31 = 0× 5 + 1 × 40 + 3× (–30) = 0 + 40 – 90 = 50 (vi) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, Cij = (–1)i+j × Mij Given, M31 = h × f – b × g M31 = hf – bg C11 = (–1)1+1 × M11 = 1 × (bc– f2) = bc– f2 C21 = (–1)2+1 × M21 = –1 × (hc – fg) = fg – hc RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants C31 = (–1)3+1 × M31 = 1 × (hf – bg) = hf – bg Now expanding along the first column we get |A| = a11 × C11 + a21× C21+ a31× C31 = a× (bc– f2) + h× (fg – hc) + g× (hf – bg) = abc– af2 + hgf – h2c +ghf – bg2 (vii) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, Cij = (–1)i+j × Mij Given, M31 = –1(1 × 0 – 5 × (–2)) – 0(0 × 0 – (–1) × (–2)) + 1(0 × 5 – (–1) × 1) M31 = –9 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants M41 = –1(1×1 – (–1) × (–2)) – 0(0 × 1 – 1 × (–2)) + 1(0 × (–1) – 1 × 1) M41 = 0 C11 = (–1)1+1 × M11 = 1 × (–9) = –9 C21 = (–1)2+1 × M21 = –1 × 9 = –9 C31 = (–1)3+1 × M31 = 1 × –9 = –9 C41 = (–1)4+1 × M41 = –1 × 0 =0 Now expanding along the first column we get |A| = a11 × C11 + a21× C21+ a31× C31 + a41× C41 = 2 × (–9) + (–3) × –9 + 1 × (–9) + 2 × 0 = – 18 + 27 –9 =0 2. Evaluate the following determinants: Solution: (i) Given RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants ⇒ |A| = x (5x + 1) – (–7) x |A| = 5x2 + 8x (ii) Given ⇒ |A| = cos θ × cos θ – (–sin θ) x sin θ |A| = cos2θ + sin2θ We know that cos2θ + sin2θ = 1 |A| = 1 (iii) Given ⇒ |A| = cos15° × cos75° + sin15° x sin75° We know that cos (A – B) = cos A cos B + Sin A sin B By substituting this we get, |A| = cos (75 – 15)° |A| = cos60° |A| = 0.5 (iv) Given ⇒ |A| = (a + ib) (a – ib) – (c + id) (–c + id) = (a + ib) (a – ib) + (c + id) (c – id) = a2 – i2 b2 + c2 – i2 d2 We know that i2 = -1 = a2 – (–1) b2 + c2 – (–1) d2 = a2 + b2 + c2 + d2 3. Evaluate: Solution: RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Since |AB|= |A||B| = 2(17 × 12 – 5 × 20) – 3(13 × 12 – 5 × 15) + 7(13 × 20 – 15 × 17) = 2 (204 – 100) – 3 (156 – 75) + 7 (260 – 255) = 2×104 – 3×81 + 7×5 = 208 – 243 +35 =0 Now |A|2 = |A|×|A| |A|2= 0 4. Show that Solution: Given Let the given determinant as A Using sin (A+B) = sin A × cos B + cos A × sin B ⇒ |A| = sin 10° × cos 80° + cos 10° x sin 80° |A| = sin (10 + 80)° |A| = sin90° |A| = 1 Hence Proved Solution: Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants = 2(1 × 1 – 4 × (–2)) – 3(7 × 1 – (–2) × (–3)) – 5(7 × 4 – 1 × (–3)) = 2(1 + 8) – 3(7 – 6) – 5(28 + 3) = 2 × 9 – 3 × 1 – 5 × 31 = 18 – 3 – 155 = –140 Now by expanding along the second column = 2(1 × 1 – 4 × (–2)) – 7(3 × 1 – 4 × (–5)) – 3(3 × (–2) – 1 × (–5)) = 2 (1 + 8) – 7 (3 + 20) – 3 (–6 + 5) = 2 × 9 – 7 × 23 – 3 × (–1) = 18 – 161 +3 = –140 Solution: Given ⇒ |A| = 0 (0 – sinβ (–sinβ)) –sinα (–sinα × 0 – sinβ cosα) – cosα ((–sinα) (–sinβ) – 0 × cosα) |A| = 0 + sinα sinβ cosα – cosα sinα sinβ |A| = 0 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Exercise 6.2 Page No: 6.57 1. Evaluate the following determinant: Solution: (i) Given RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (ii) Given = 1[(109) (12) – (119) (11)] = 1308 – 1309 =–1 So, Δ = – 1 (iii) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants = a (bc – f2) – h (hc – fg) + g (hf – bg) = abc – af2 – ch2 + fgh + fgh – bg2 = abc + 2fgh – af2 – bg2 – ch2 So, Δ = abc + 2fgh – af2 – bg2 – ch2 (iv) Given = 2[1(24 – 4)] = 40 So, Δ = 40 (v) Given RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants = 1[(– 7) (– 36) – (– 20) (– 13)] = 252 – 260 =–8 So, Δ = – 8 (vi) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (vii) Given RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (viii) Given, 2. Without expanding, show that the value of each of the following determinants is zero: RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: (i) Given, (ii) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (iii) Given, (iv) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (v) Given, (vi) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (vii) Given, (viii) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (ix) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants As, C1 = C2, hence determinant is zero (x) Given, (xi) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (xii) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (xiii) Given, (xiv) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (xv) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (xvi) Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (xvii) Given, Hence proved. RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Evaluate the following (3 – 9): Solution: Given, = (a + b + c) (b – a) (c – a) (b – c) RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants So, Δ = (a + b + c) (b – a) (c – a) (b – c) Solution: Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Given, Solution: Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants = a [a (a + x + y) + az] + 0 + 0 = a2 (a + x + y + z) So, Δ = a2 (a + x + y + z) Solution: RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Prove the following identities (11 – 45): Solution: Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Consider, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants = – (a + b + c) [(b – c) (a + b – 2c) – (c – a) (c + a – 2b)] = 3abc – a3 – b3 – c3 Therefore, L.H.S = R.H.S, Hence the proof. Solution: Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Consider, , RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Consider, L.H.S = Now by applying, R1→R1 + R2 + R3, we get, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Consider, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Consider, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Consider, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Hence, the proof. Solution: Given, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants = – xyz(x – y) (z – y) [z2 + y2 + zy – x2 – y2 – xy] = – xyz(x – y) (z – y) [(z – x) (z + x0 + y (z – x)] = – xyz(x – y) (z – y) (z – x) (x + y + z) = R.H.S Hence, the proof. Solution: Consider, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants = (a2 + b2 + c2) (b – a) (c – a) [(b + a) (– b) – (– c) (c + a)] = (a2 + b2 + c2) (a – b) (c – a) (b – c) (a + b + c) = R.H.S Hence, the proof. Solution: Consider, = [(2a + 4) (1) – (1) (2a + 6)] RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants =–2 = R.H.S Hence, the proof. Solution: Consider, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants = – (a2 + b2 + c2) (a – b) (c – a) [(– (b + a)) (– b) – (c) (c + a)] = (a – b) (b – c) (c – a) (a + b + c) (a2 + b2 + c2) = R.H.S Hence, the proof. Solution: Consider, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants = R.H.S Hence, the proof. Solution: Consider, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Consider, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Expanding the determinant along R1, we have Δ = 1[(1) (7) – (3) (2)] – 0 + 0 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants ∴Δ=7–6=1 Thus, Hence the proof. RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Exercise 6.3 Page No: 6.71 1. Find the area of the triangle with vertices at the points: (i) (3, 8), (-4, 2) and (5, -1) (ii) (2, 7), (1, 1) and (10, 8) (iii) (-1, -8), (-2, -3) and (3, 2) (iv) (0, 0), (6, 0) and (4, 3) Solution: (i) Given (3, 8), (-4, 2) and (5, -1) are the vertices of the triangle. We know that, if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by: (ii) Given (2, 7), (1, 1) and (10, 8) are the vertices of the triangle. We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants triangle is given by: (iii) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle. We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by: RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants As we know area cannot be negative. Therefore, 15 square unit is the area Thus area of triangle is 15 square units (iv) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle. We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by: 2. Using the determinants show that the following points are collinear: (i) (5, 5), (-5, 1) and (10, 7) RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants (ii) (1, -1), (2, 1) and (10, 8) (iii) (3, -2), (8, 8) and (5, 2) (iv) (2, 3), (-1, -2) and (5, 8) Solution: (i) Given (5, 5), (-5, 1) and (10, 7) We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by (ii) Given (1, -1), (2, 1) and (10, 8) We have the condition that three points to be collinear, the area of the triangle formed RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, (iii) Given (3, -2), (8, 8) and (5, 2) We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, Now, by substituting given value in above formula RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants + + Since, Area of triangle is zero Hence, points are collinear. (iv) Given (2, 3), (-1, -2) and (5, 8) We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants 3. If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab Solution: Given (a, 0), (0, b) and (1, 1) are collinear We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, ⇒ ⇒ a + b = ab Hence Proved 4. Using the determinants prove that the points (a, b), (a', b') and (a - a', b - b) are RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants collinear if a b' = a' b. Solution: Given (a, b), (a', b') and (a - a', b - b) are collinear We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, ⇒ a b' = a' b Hence, the proof. 5. Find the value of λ so that the points (1, -5), (-4, 5) and (λ, 7) are collinear. Solution: Given (1, -5), (-4, 5) and (λ, 7) are collinear We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants ⇒ - 50 – 10λ = 0 ⇒λ=–5 6. Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, -6) and (5, 4). Solution: Given (x, 4), (2, -6) and (5, 4) are the vertices of a triangle. We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants ⇒ [x (– 10) – 4(– 3) + 1(8 – 30)] = ± 70 ⇒ [– 10x + 12 + 38] = ± 70 ⇒ ±70 = – 10x + 50 Taking positive sign, we get ⇒ + 70 = – 10x + 50 ⇒ 10x = – 20 ⇒x=–2 Taking –negative sign, we get ⇒ – 70 = – 10x + 50 ⇒ 10x = 120 ⇒ x = 12 Thus x = – 2, 12 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Exercise 6.4 Page No: 6.84 Solve the following system of linear equations by Cramer’s rule: 1. x – 2y = 4 -3x + 5y = -7 Solution: Given x – 2y = 4 -3x + 5y = -7 Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st row ⇒ D = 5(1) – (– 3) (– 2) ⇒D=5–6 ⇒D=–1 Again, Solving determinant, expanding along 1st row ⇒ D1 = 5(4) – (– 7) (– 2) ⇒ D1 = 20 – 14 ⇒ D1 = 6 And Solving determinant, expanding along 1st row ⇒ D2 = 1(– 7) – (– 3) (4) ⇒ D2 = – 7 + 12 ⇒ D2 = 5 Thus by Cramer’s Rule, we have 2. 2x – y = 1 7x – 2y = -7 Solution: Given 2x – y = 1 and RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants 7x – 2y = -7 Let there be a system of n simultaneous linear equations and with n unknown given by Solving determinant, expanding along 1st row ⇒ D1 = 1(– 2) – (– 7) (– 1) ⇒ D1 = – 2 – 7 ⇒ D1 = – 9 And RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st row ⇒ D2 = 2(– 7) – (7) (1) ⇒ D2 = – 14 – 7 ⇒ D2 = – 21 Thus by Cramer’s Rule, we have 3. 2x – y = 17 3x + 5y = 6 Solution: Given 2x – y = 17 and 3x + 5y = 6 Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st row ⇒ D1 = 17(5) – (6) (– 1) ⇒ D1 = 85 + 6 ⇒ D1 = 91 Solving determinant, expanding along 1st row ⇒ D2 = 2(6) – (17) (3) ⇒ D2 = 12 – 51 ⇒ D2 = – 39 Thus by Cramer’s Rule, we have RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants 4. 3x + y = 19 3x – y = 23 Solution: Let there be a system of n simultaneous linear equations and with n unknown given by Solving determinant, expanding along 1st row ⇒ D = 3(– 1) – (3) (1) ⇒D=–3–3 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants ⇒D=–6 Again, Solving determinant, expanding along 1st row ⇒ D1 = 19(– 1) – (23) (1) ⇒ D1 = – 19 – 23 ⇒ D1 = – 42 Solving determinant, expanding along 1st row ⇒ D2 = 3(23) – (19) (3) ⇒ D2 = 69 – 57 ⇒ D2 = 12 Thus by Cramer’s Rule, we have 5. 2x – y = -2 3x + 4y = 3 Solution: Given 2x – y = -2 and 3x + 4y = 3 Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st row ⇒ D2 = 3(2) – (– 2) (3) ⇒ D2 = 6 + 6 ⇒ D2 = 12 Thus by Cramer’s Rule, we have 6. 3x + ay = 4 2x + ay = 2, a ≠ 0 Solution: Given 3x + ay = 4 and 2x + ay = 2, a ≠ 0 Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants 3x + ay = 4 2x + ay = 2, a≠0 So by comparing with the theorem, let's find D, D1 and D2 Solving determinant, expanding along 1st row ⇒ D = 3(a) – (2) (a) ⇒ D = 3a – 2a ⇒D=a Again, Solving determinant, expanding along 1st row ⇒ D1 = 4(a) – (2) (a) ⇒ D = 4a – 2a ⇒ D = 2a Solving determinant, expanding along 1st row ⇒ D2 = 3(2) – (2) (4) ⇒D=6–8 ⇒D=–2 Thus by Cramer’s Rule, we have 7. 2x + 3y = 10 x + 6y = 4 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solution: Let there be a system of n simultaneous linear equations and with n unknown given by Solving determinant, expanding along 1st row ⇒ D = 2 (6) – (3) (1) ⇒ D = 12 – 3 ⇒D=9 Again, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st row ⇒ D1 = 10 (6) – (3) (4) ⇒ D = 60 – 12 ⇒ D = 48 Solving determinant, expanding along 1st row ⇒ D2 = 2 (4) – (10) (1) ⇒ D2 = 8 – 10 ⇒ D2 = – 2 Thus by Cramer’s Rule, we have 8. 5x + 7y = -2 4x + 6y = -3 Solution: Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Now, here we have 5x + 7y = – 2 4x + 6y = – 3 So by comparing with the theorem, let's find D, D1 and D2 Solving determinant, expanding along 1st row ⇒ D = 5(6) – (7) (4) ⇒ D = 30 – 28 ⇒D=2 Again, Solving determinant, expanding along 1st row ⇒ D1 = – 2(6) – (7) (– 3) ⇒ D1 = – 12 + 21 ⇒ D1 = 9 Solving determinant, expanding along 1st row ⇒ D2 = – 3(5) – (– 2) (4) ⇒ D2 = – 15 + 8 ⇒ D2 = – 7 Thus by Cramer’s Rule, we have RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants 9. 9x + 5y = 10 3y – 2x = 8 Solution: Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st row ⇒ D = 3(9) – (5) (– 2) ⇒ D = 27 + 10 ⇒ D = 37 Again, Solving determinant, expanding along 1st row ⇒ D1 = 10(3) – (8) (5) ⇒ D1 = 30 – 40 ⇒ D1 = – 10 Solving determinant, expanding along 1st row ⇒ D2 = 9(8) – (10) (– 2) ⇒ D2 = 72 + 20 ⇒ D2 = 92 Thus by Cramer’s Rule, we have 10. x + 2y = 1 3x + y = 4 Solution: Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st row ⇒ D = 1(1) – (3) (2) ⇒D=1–6 ⇒D=–5 Again, Solving determinant, expanding along 1st row ⇒ D1 = 1(1) – (2) (4) RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants ⇒ D1 = 1 – 8 ⇒ D1 = – 7 Solving determinant, expanding along 1st row ⇒ D2 = 1(4) – (1) (3) ⇒ D2 = 4 – 3 ⇒ D2 = 1 Thus by Cramer’s Rule, we have Solve the following system of linear equations by Cramer’s rule: 11. 3x + y + z = 2 2x – 4y + 3z = -1 4x + y – 3z = -11 Solution: Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Now, here we have 3x + y + z = 2 2x – 4y + 3z = – 1 4x + y – 3z = – 11 So by comparing with the theorem, let's find D, D1, D2 and D3 Solving determinant, expanding along 1st row ⇒ D = 3[(– 4) (– 3) – (3) (1)] – 1[(2) (– 3) – 12] + 1[2 – 4(– 4)] ⇒ D = 3[12 – 3] – [– 6 – 12] + [2 + 16] ⇒ D = 27 + 18 + 18 ⇒ D = 63 Again, Solving determinant, expanding along 1st row ⇒ D1 = 2[( – 4)( – 3) – (3)(1)] – 1[( – 1)( – 3) – ( – 11)(3)] + 1[( – 1) – ( – 4)( – 11)] ⇒ D1 = 2[12 – 3] – 1[3 + 33] + 1[– 1 – 44] ⇒ D1 = 2 – 36 – 45 ⇒ D1 = 18 – 36 – 45 ⇒ D1 = – 63 Again Solving determinant, expanding along 1st row ⇒ D2 = 3[3 + 33] – 2[– 6 – 12] + 1[– 22 + 4] ⇒ D2 = 3 – 2(– 18) – 18 ⇒ D2 = 126 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants ⇒ Solving determinant, expanding along 1st row ⇒ D3 = 3[44 + 1] – 1[– 22 + 4] + 2[2 + 16] ⇒ D3 = 3 – 1(– 18) + 2(18) ⇒ D3 = 135 + 18 + 36 ⇒ D3 = 189 Thus by Cramer’s Rule, we have 12. x – 4y – z = 11 2x – 5y + 2z = 39 -3x + 2y + z = 1 Solution: Given, x – 4y – z = 11 2x – 5y + 2z = 39 -3x + 2y + z = 1 Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Now, here we have x – 4y – z = 11 2x – 5y + 2z = 39 – 3x + 2y + z = 1 So by comparing with theorem, now we have to find D, D1 and D2 Solving determinant, expanding along 1st row ⇒ D = 1[(– 5) (1) – (2) (2)] + 4[(2) (1) + 6] – 1[4 + 5(– 3)] ⇒ D = 1[– 5 – 4] + 4 – [– 11] ⇒ D = – 9 + 32 + 11 ⇒ D = 34 Again, Solving determinant, expanding along 1st row RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants ⇒ D1 = 11[(– 5) (1) – (2) (2)] + 4[(39) (1) – (2) (1)] – 1[2 (39) – (– 5) (1)] ⇒ D1 = 11[– 5 – 4] + 4[39 – 2] – 1[78 + 5] ⇒ D1 = 11[– 9] + 4(37) – 83 ⇒ D1 = – 99 – 148 – 45 ⇒ D1 = – 34 Again Solving determinant, expanding along 1st row ⇒ D2 = 1[39 – 2] – 11[2 + 6] – 1[2 + 117] ⇒ D2 = 1 – 11(8) – 119 ⇒ D2 = – 170 And, ⇒ Solving determinant, expanding along 1st row ⇒ D3 = 1[– 5 – (39) (2)] – (– 4) [2 – (39) (– 3)] + 11[4 – (– 5)(– 3)] ⇒ D3 = 1 [– 5 – 78] + 4 (2 + 117) + 11 (4 – 15) ⇒ D3 = – 83 + 4(119) + 11(– 11) ⇒ D3 = 272 Thus by Cramer’s Rule, we have = (272/34) = 8 13. 6x + y – 3z = 5 x + 3y – 2z = 5 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants 2x + y + 4z = 8 Solution: Given 6x + y – 3z = 5 x + 3y – 2z = 5 2x + y + 4z = 8 Let there be a system of n simultaneous linear equations and with n unknown given by Now, here we have 6x + y – 3z = 5 x + 3y – 2z = 5 2x + y + 4z = 8 So by comparing with theorem, now we have to find D , D1 and D2 Solving determinant, expanding along 1st Row RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants ⇒ D = 6[(4) (3) – (1) (– 2)] – 1[(4) (1) + 4] – 3[1 – 3(2)] ⇒ D = 6[12 + 2] – – 3[– 5] ⇒ D = 84 – 8 + 15 ⇒ D = 91 Again, Solve D1 formed by replacing 1st column by B matrices Here Solving determinant, expanding along 1st Row ⇒ D1 = 5[(4) (3) – (– 2) (1)] – 1[(5) (4) – (– 2) (8)] – 3[(5) – (3) (8)] ⇒ D1 = 5[12 + 2] – 1[20 + 16] – 3[5 – 24] ⇒ D1 = 5 – 36 – 3(– 19) ⇒ D1 = 70 – 36 + 57 ⇒ D1 = 91 Again, Solve D2 formed by replacing 1st column by B matrices Here Solving determinant ⇒ D2 = 6[20 + 16] – 5[4 – 2(– 2)] + (– 3)[8 – 10] ⇒ D2 = 6 – 5(8) + (– 3) (– 2) ⇒ D2 = 182 And, Solve D3 formed by replacing 1st column by B matrices Here RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st Row ⇒ D3 = 6[24 – 5] – 1[8 – 10] + 5[1 – 6] ⇒ D3 = 6 – 1(– 2) + 5(– 5) ⇒ D3 = 114 + 2 – 25 ⇒ D3 = 91 Thus by Cramer’s Rule, we have 14. x + y = 5 y+z=3 x+z=4 Solution: Given x + y = 5 y+z=3 x+z=4 Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Let Dj be the determinant obtained from D after replacing the jth column by Now, here we have x+y=5 y+z=3 x+z=4 So by comparing with theorem, now we have to find D, D1 and D2 Solving determinant, expanding along 1st Row ⇒ D = 1 – 1[– 1] + 0[– 1] ⇒D=1+1+0 ⇒D=2 Again, Solve D1 formed by replacing 1st column by B matrices Here Solving determinant, expanding along 1st Row ⇒ D1 = 5 – 1[(3) (1) – (4) (1)] + 0[0 – (4) (1)] ⇒ D1 = 5 – 1[3 – 4] + 0[– 4] ⇒ D1 = 5 – 1[– 1] + 0 ⇒ D1 = 5 + 1 + 0 ⇒ D1 = 6 Again, Solve D2 formed by replacing 1st column by B matrices Here RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant ⇒ D2 = 1[3 – 4] – 5[– 1] + 0[0 – 3] ⇒ D2 = 1[– 1] + 5 + 0 ⇒ D2 = 4 And, Solve D3 formed by replacing 1st column by B matrices Here Solving determinant, expanding along 1st Row ⇒ D3 = 1[4 – 0] – 1[0 – 3] + 5[0 – 1] ⇒ D3 = 1 – 1(– 3) + 5(– 1) ⇒ D3 = 4 + 3 – 5 ⇒ D3 = 2 Thus by Cramer’s Rule, we have RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants 15. 2y – 3z = 0 x + 3y = -4 3x + 4y = 3 Solution: Given 2y – 3z = 0 x + 3y = -4 3x + 4y = 3 Let there be a system of n simultaneous linear equations and with n unknown given by Now, here we have 2y – 3z = 0 x + 3y = – 4 3x + 4y = 3 So by comparing with theorem, now we have to find D, D1 and D2 RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st Row ⇒ D = 0 – 2[(0) (1) – 0] – 3[1 (4) – 3 (3)] ⇒ D = 0 – 0 – 3[4 – 9] ⇒ D = 0 – 0 + 15 ⇒ D = 15 Again, Solve D1 formed by replacing 1st column by B matrices Here Solving determinant, expanding along 1st Row ⇒ D1 = 0 – 2[(0) (– 4) – 0] – 3[4 (– 4) – 3(3)] ⇒ D1 = 0 – 0 – 3[– 16 – 9] ⇒ D1 = 0 – 0 – 3(– 25) ⇒ D1 = 0 – 0 + 75 ⇒ D1 = 75 Again, Solve D2 formed by replacing 2nd column by B matrices Here Solving determinant ⇒ D2 = 0 – 0[(0) (1) – 0] – 3[1 (3) – 3(– 4)] ⇒ D2 = 0 – 0 + (– 3) (3 + 12) ⇒ D2 = – 45 And, Solve D3 formed by replacing 3rd column by B matrices Here RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st Row ⇒ D3 = 0[9 – (– 4) 4] – 2[(3) (1) – (– 4) (3)] + 0[1 (4) – 3 (3)] ⇒ D3 = 0 – 2(3 + 12) + 0(4 – 9) ⇒ D3 = 0 – 30 + 0 ⇒ D3 = – 30 Thus by Cramer’s Rule, we have 16. 5x – 7y + z = 11 6x – 8y – z = 15 3x + 2y – 6z = 7 Solution: Given 5x – 7y + z = 11 6x – 8y – z = 15 3x + 2y – 6z = 7 Let there be a system of n simultaneous linear equations and with n unknown given by RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Now, here we have 5x – 7y + z = 11 6x – 8y – z = 15 3x + 2y – 6z = 7 So by comparing with theorem, now we have to find D, D1 and D2 Solving determinant, expanding along 1st Row ⇒ D = 5[(– 8) (– 6) – (– 1) (2)] – 7[(– 6) (6) – 3(– 1)] + 1[2(6) – 3(– 8)] ⇒ D = 5[48 + 2] – 7[– 36 + 3] + 1[12 + 24] ⇒ D = 250 – 231 + 36 ⇒ D = 55 Again, Solve D1 formed by replacing 1st column by B matrices Here RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Solving determinant, expanding along 1st Row ⇒ D1 = 11[(– 8) (– 6) – (2) (– 1)] – (– 7) [(15) (– 6) – (– 1) (7)] + 1[(15)2 – (7) (– 8)] ⇒ D1 = 11[48 + 2] + 7[– 90 + 7] + 1[30 + 56] ⇒ D1 = 11 + 7[– 83] + 86 ⇒ D1 = 550 – 581 + 86 ⇒ D1 = 55 Again, Solve D2 formed by replacing 2nd column by B matrices Here Solving determinant, expanding along 1st Row ⇒ D2 = 5[(15) (– 6) – (7) (– 1)] – 11 [(6) (– 6) – (– 1) (3)] + 1[(6)7 – (15) (3)] ⇒ D2 = 5[– 90 + 7] – 11[– 36 + 3] + 1[42 – 45] ⇒ D2 = 5[– 83] – 11(– 33) – 3 ⇒ D2 = – 415 + 363 – 3 ⇒ D2 = – 55 And, Solve D3 formed by replacing 3rd column by B matrices Here Solving determinant, expanding along 1st Row ⇒ D3 = 5[(– 8) (7) – (15) (2)] – (– 7) [(6) (7) – (15) (3)] + 11[(6)2 – (– 8) (3)] ⇒ D3 = 5[– 56 – 30] – (– 7) [42 – 45] + 11[12 + 24] ⇒ D3 = 5[– 86] + 7[– 3] + 11 ⇒ D3 = – 430 – 21 + 396 ⇒ D3 = – 55 Thus by Cramer’s Rule, we have RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Exercise 6.5 Page No: 6.89 Solve each of the following system of homogeneous linear equations: 1. x + y – 2z = 0 2x + y – 3z =0 5x + 4y – 9z = 0 Solution: Given x + y – 2z = 0 2x + y – 3z =0 5x + 4y – 9z = 0 Any system of equation can be written in matrix form as AX = B Now finding the Determinant of these set of equations, = 1(1 × (– 9) – 4 × (– 3)) – 1(2 × (– 9) – 5 × (– 3)) – 2(4 × 2 – 5 × 1) = 1(– 9 + 12) – 1(– 18 + 15) – 2(8 – 5) = 1 × 3 –1 × (– 3) – 2 × 3 =3+3–6 =0 Since D = 0, so the system of equation has infinite solution. Now let z = k ⇒ x + y = 2k And 2x + y = 3k Now using the Cramer’s rule RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants 2. 2x + 3y + 4z = 0 x+y+z=0 2x + 5y – 2z = 0 Solution: Given 2x + 3y + 4z = 0 x+y+z=0 2x + 5y – 2z = 0 Any system of equation can be written in matrix form as AX = B Now finding the Determinant of these set of equations, RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants = 2(1 × (– 2) – 1 × 5) – 3(1 × (– 2) – 2 × 1) + 4(1 × 5 – 2 × 1) = 2(– 2 – 5) – 3(– 2 – 2) + 4(5 – 2) = 1 × (– 7) – 3 × (– 4) + 4 × 3 = – 7 + 12 + 12 = 17 Since D ≠ 0, so the system of equation has infinite solution. Therefore the system of equation has only solution as x = y = z = 0.
