Exercise 4.5 answers: Determinants

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1. Write A = [1, 2; 2, 3] and B = [2; 3].
2. Find |A| = 1×3 − 2×2 = −1. Since |A| ≠ 0, A is non-singular, so the system has a unique solution.
3. A unique solution exists, so the system is consistent.

AnswerYes, the system is consistent (it has a unique solution).

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1. Write A = [2, −1; 1, 1] and B = [5; 4].
2. Find |A| = 2×1 − (−1)×1 = 3. Since |A| ≠ 0, A is non-singular, so the system has a unique solution.
3. A unique solution exists, so the system is consistent.

AnswerYes, the system is consistent (it has a unique solution).

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1. Write A = [1, 3; 2, 6] and B = [5; 8].
2. Find |A| = 1×6 − 3×2 = 0. So A is singular, and the direct-inverse test cannot be used.
3. Find adj A = [6, −3; −2, 1], and multiply (adj A) B = [6×5 + (−3)×8; −2×5 + 1×8] = [6; −2].
4. Since (adj A) B is not the zero column, the system has no solution.

AnswerNo, the system is inconsistent (it has no solution).

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1. Write A = [1, 1, 1; 2, 3, 2; a, a, 2a] and B = [1; 2; 4].
2. Find |A| by expanding along the first row: |A| = 1(3×2a − 2a) − 1(2×2a − 2a) + 1(2a − 3a) = a(4) − a(2) − a = a.
3. So |A| = a, which is not zero as long as a ≠ 0. Whenever a ≠ 0, A is non-singular and the system has a unique solution — the system is consistent.
4. If a = 0, the third equation becomes 0 = 4, which is impossible, and the system is inconsistent.
5. Solving for a ≠ 0 (using A⁻¹) gives x = 2 − 4/a, y = 0, z = (4 − a)/a, which satisfies all three equations.

AnswerThe system is consistent for every a ≠ 0 (unique solution x = 2 − 4/a, y = 0, z = (4 − a)/a); it is inconsistent only when a = 0.

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1. Write the system as AX = B with A = [3, −1, −2; 0, 2, −1; 3, −5, 0], X = [x; y; z] and B = [2; −1; 3].
2. Expand along the first column: |A| = 3(2×0 − (−1)×(−5)) − 0 + 3((−1)×(−1) − (−2)×2) = 3(−5) + 3(5) = 0. So A is singular, and we must check (adj A) B.
3. Cofactors: A11 = −5, A12 = −3, A13 = −6; A21 = 10, A22 = 6, A23 = 12; A31 = 5, A32 = 3, A33 = 6. Transposing gives adj A = [−5, 10, 5; −3, 6, 3; −6, 12, 6].
4. (adj A) B = [−5×2 + 10×(−1) + 5×3; −3×2 + 6×(−1) + 3×3; −6×2 + 12×(−1) + 6×3] = [−10 − 10 + 15; −6 − 6 + 9; −12 − 12 + 18] = [−5; −3; −6].
5. Since |A| = 0 and (adj A) B is not the zero column, the system has no solution, so it is inconsistent.

AnswerNo, the system is inconsistent: |A| = 0 and (adj A) B = [−5; −3; −6] ≠ O, so it has no solution.

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1. Write A = [5, −1, 4; 2, 3, 5; 5, −2, 6] and B = [5; 2; −1].
2. Find |A| by expanding along the first row: |A| = 5(3×6 − 5×(−2)) − (−1)(2×6 − 5×5) + 4(2×(−2) − 3×5) = 5(28) + 1(−13) + 4(−19) = 140 − 13 − 76 = 51.
3. Since |A| ≠ 0, A is non-singular, so the system has a unique solution.
4. A unique solution exists, so the system is consistent.

AnswerYes, the system is consistent (it has a unique solution).

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1. Write the equations as AX = B, with A = [[5,2],[7,3]], X = [[x],[y]], B = [[4],[5]].
2. Find |A| = 5×3 − 2×7 = 1. Since |A| ≠ 0, A⁻¹ exists.
3. A⁻¹ = (1/1) × [[3,−2],[−7,5]] = [[3,−2],[−7,5]].
4. X = A⁻¹B = [[3×4 − 2×5],[−7×4 + 5×5]] = [[2],[−3]].

Answerx = 2, y = −3

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1. Write as AX = B with A = [[2,−1],[3,4]], B = [[−2],[3]].
2. Find |A| = 2×4 − (−1)×3 = 11. Since |A| ≠ 0, A⁻¹ exists.
3. A⁻¹ = (1/11) × [[4,1],[−3,2]].
4. X = A⁻¹B = (1/11)[[4×(−2) + 1×3],[−3×(−2) + 2×3]] = (1/11)[[−5],[12]].

Answerx = −5/11, y = 12/11

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1. Write as AX = B with A = [[4,−3],[3,−5]], B = [[3],[7]].
2. Find |A| = 4×(−5) − (−3)×3 = −11. Since |A| ≠ 0, A⁻¹ exists.
3. A⁻¹ = (1/−11) × [[−5,3],[−3,4]].
4. X = A⁻¹B = (1/−11)[[−5×3 + 3×7],[−3×3 + 4×7]] = (1/−11)[[6],[19]].

Answerx = −6/11, y = −19/11

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1. Write as AX = B with A = [[5,2],[3,2]], B = [[3],[5]].
2. Find |A| = 5×2 − 2×3 = 4. Since |A| ≠ 0, A⁻¹ exists.
3. A⁻¹ = (1/4) × [[2,−2],[−3,5]].
4. X = A⁻¹B = (1/4)[[2×3 − 2×5],[−3×3 + 5×5]] = (1/4)[[−4],[16]] = [[−1],[4]].

Answerx = −1, y = 4

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1. Write the system as AX = B with A = [[2,1,1],[1,−2,−1],[0,3,−5]], X = [[x],[y],[z]], B = [[1],[3/2],[9]].
2. |A| = 2[(−2)(−5) − (−1)(3)] − 1[(1)(−5) − (−1)(0)] + 1[(1)(3) − (−2)(0)] = 2(13) + 5 + 3 = 34. Since |A| ≠ 0, A⁻¹ exists.
3. Cofactors of A: C11 = 13, C12 = 5, C13 = 3, C21 = 8, C22 = −10, C23 = −6, C31 = 1, C32 = 3, C33 = −5.
4. adj A is the transpose of the cofactor matrix: adj A = [[13,8,1],[5,−10,3],[3,−6,−5]].
5. A⁻¹ = (1/|A|) adj A = (1/34) × [[13,8,1],[5,−10,3],[3,−6,−5]].
6. X = A⁻¹B = (1/34) × [[13,8,1],[5,−10,3],[3,−6,−5]] × [[1],[3/2],[9]] = (1/34) × [[34],[17],[−51]] = [[1],[1/2],[−3/2]].
7. So x = 1, y = 1/2, z = −3/2. Check in the first equation: 2(1) + 1/2 − 3/2 = 1 ✓.

Answerx = 1, y = 1/2, z = −3/2

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1. Write the system as AX = B with A = [[1,−1,1],[2,1,−3],[1,1,1]], X = [[x],[y],[z]], B = [[4],[0],[2]].
2. |A| = 1[(1)(1) − (−3)(1)] − (−1)[(2)(1) − (−3)(1)] + 1[(2)(1) − (1)(1)] = 4 + 5 + 1 = 10. Since |A| ≠ 0, A⁻¹ exists.
3. Cofactors of A: C11 = 4, C12 = −5, C13 = 1, C21 = 2, C22 = 0, C23 = −2, C31 = 2, C32 = 5, C33 = 3.
4. adj A is the transpose of the cofactor matrix: adj A = [[4,2,2],[−5,0,5],[1,−2,3]].
5. A⁻¹ = (1/|A|) adj A = (1/10) × [[4,2,2],[−5,0,5],[1,−2,3]].
6. X = A⁻¹B = (1/10) × [[4,2,2],[−5,0,5],[1,−2,3]] × [[4],[0],[2]] = (1/10) × [[20],[−10],[10]] = [[2],[−1],[1]].
7. So x = 2, y = −1, z = 1. Check in the first equation: 2 − (−1) + 1 = 4 ✓.

Answerx = 2, y = −1, z = 1

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1. Write the system as AX = B with A = [[2,3,3],[1,−2,1],[3,−1,−2]], X = [[x],[y],[z]], B = [[5],[−4],[3]].
2. |A| = 2[(−2)(−2) − (1)(−1)] − 3[(1)(−2) − (1)(3)] + 3[(1)(−1) − (−2)(3)] = 2(5) − 3(−5) + 3(5) = 40. Since |A| ≠ 0, A⁻¹ exists.
3. Cofactors of A: C11 = 5, C12 = 5, C13 = 5, C21 = 3, C22 = −13, C23 = 11, C31 = 9, C32 = 1, C33 = −7.
4. adj A is the transpose of the cofactor matrix: adj A = [[5,3,9],[5,−13,1],[5,11,−7]].
5. A⁻¹ = (1/|A|) adj A = (1/40) × [[5,3,9],[5,−13,1],[5,11,−7]].
6. X = A⁻¹B = (1/40) × [[5,3,9],[5,−13,1],[5,11,−7]] × [[5],[−4],[3]] = (1/40) × [[40],[80],[−40]] = [[1],[2],[−1]].
7. So x = 1, y = 2, z = −1. Check in the first equation: 2(1) + 3(2) + 3(−1) = 5 ✓.

Answerx = 1, y = 2, z = −1

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1. Write the system as AX = B with A = [[1,−1,2],[3,4,−5],[2,−1,3]], X = [[x],[y],[z]], B = [[7],[−5],[12]].
2. |A| = 1[(4)(3) − (−5)(−1)] − (−1)[(3)(3) − (−5)(2)] + 2[(3)(−1) − (4)(2)] = 7 + 19 − 22 = 4. Since |A| ≠ 0, A⁻¹ exists.
3. Cofactors of A: C11 = 7, C12 = −19, C13 = −11, C21 = 1, C22 = −1, C23 = −1, C31 = −3, C32 = 11, C33 = 7.
4. adj A is the transpose of the cofactor matrix: adj A = [[7,1,−3],[−19,−1,11],[−11,−1,7]].
5. A⁻¹ = (1/|A|) adj A = (1/4) × [[7,1,−3],[−19,−1,11],[−11,−1,7]].
6. X = A⁻¹B = (1/4) × [[7,1,−3],[−19,−1,11],[−11,−1,7]] × [[7],[−5],[12]] = (1/4) × [[8],[4],[12]] = [[2],[1],[3]].
7. So x = 2, y = 1, z = 3. Check in the first equation: 2 − 1 + 2(3) = 7 ✓.

Answerx = 2, y = 1, z = 3

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1. |A| = 2[(2)(−2) − (−4)(1)] − (−3)[(3)(−2) − (−4)(1)] + 5[(3)(1) − (2)(1)] = 2(0) + 3(−2) + 5(1) = −1. Since |A| ≠ 0, A⁻¹ exists.
2. Cofactors of A: C11 = 0, C12 = 2, C13 = 1, C21 = −1, C22 = −9, C23 = −5, C31 = 2, C32 = 23, C33 = 13.
3. adj A is the transpose of the cofactor matrix: adj A = [[0,−1,2],[2,−9,23],[1,−5,13]].
4. A⁻¹ = (1/|A|) adj A = (1/−1) × adj A = [[0,1,−2],[−2,9,−23],[−1,5,−13]].
5. The system has exactly this coefficient matrix, so write it as AX = B with X = [[x],[y],[z]], B = [[11],[−5],[−3]].
6. X = A⁻¹B = [[0,1,−2],[−2,9,−23],[−1,5,−13]] × [[11],[−5],[−3]] = [[0 − 5 + 6],[−22 − 45 + 69],[−11 − 25 + 39]] = [[1],[2],[3]].
7. So x = 1, y = 2, z = 3. Check in the first equation: 2(1) − 3(2) + 5(3) = 11 ✓.

AnswerA⁻¹ = [[0,1,−2],[−2,9,−23],[−1,5,−13]]; x = 1, y = 2, z = 3

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1. Let the cost per kg of onion, wheat and rice be x, y and z rupees.
2. The three sentences give 4x + 3y + 2z = 60, 2x + 4y + 6z = 90, 6x + 2y + 3z = 70.
3. Write this as AX = B with A = [[4,3,2],[2,4,6],[6,2,3]], X = [[x],[y],[z]], B = [[60],[90],[70]].
4. |A| = 4[(4)(3) − (6)(2)] − 3[(2)(3) − (6)(6)] + 2[(2)(2) − (4)(6)] = 4(0) − 3(−30) + 2(−20) = 50. Since |A| ≠ 0, A⁻¹ exists.
5. Cofactors of A: C11 = 0, C12 = 30, C13 = −20, C21 = −5, C22 = 0, C23 = 10, C31 = 10, C32 = −20, C33 = 10.
6. adj A is the transpose of the cofactor matrix: adj A = [[0,−5,10],[30,0,−20],[−20,10,10]].
7. A⁻¹ = (1/|A|) adj A = (1/50) × [[0,−5,10],[30,0,−20],[−20,10,10]].
8. X = A⁻¹B = (1/50) × [[0,−5,10],[30,0,−20],[−20,10,10]] × [[60],[90],[70]] = (1/50) × [[250],[400],[400]] = [[5],[8],[8]].
9. So x = 5, y = 8, z = 8. Check: 4(5) + 3(8) + 2(8) = 60 ✓, 2(5) + 4(8) + 6(8) = 90 ✓, 6(5) + 2(8) + 3(8) = 70 ✓.

AnswerOnion costs ₹5 per kg, wheat ₹8 per kg and rice ₹8 per kg.

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