Determinants — Practice

Question 1

Using properties of determinants, prove: |a+b, a, a; a, a+b, a; a, a, a+b| = b²(3a+b).

Accepted Answer

Question 2

Find the inverse of A = [[1, 2, 1], [3, 1, −1], [2, 0, 1]] using cofactors.

Accepted Answer

|A| = 1(1×1−(−1)×0) − 2(3×1−(−1)×2) + 1(3×0−1×2) = 1(1) − 2(3+2) + 1(0−2) = 1 − 10 − 2 = −11. Since |A| = −11 ≠ 0, A⁻¹ exists. COFACTORS: C₁₁ = (+1)|1,−1;0,1| = 1. C₁₂ = (−1)|3,−1;2,1| = −(3+2) = −5. C₁₃ = (+1)|3,1;2,0| = 0−2 = −2. C₂₁ = (−1)|2,1;0,1| = −(2−0) = −2. C₂₂ = (+1)|1,1;2,1| = 1−2 = −1. C₂₃ = (−1)|1,2;2,0| = −(0−4) = 4. C₃₁ = (+1)|2,1;1,−1| = −2−1 = −3. C₃₂ = (−1)|1,1;3,−1| = −(−1−3) = 4. C₃₃ = (+1)|1,2;3,1| = 1−6 = −5. Cofactor matrix = [[1,−5,−2],[−2,−1,4],[−3,4,−5]]. adj(A) = (Cofactor matrix)ᵀ = [[1,−2,−3],[−5,−1,4],[−2,4,−5]]. A⁻¹ = adj(A)/|A| = (1/−11)[[1,−2,−3],[−5,−1,4],[−2,4,−5]].

Question 3

Using the matrix method, solve the system: x + 2y + z = 7; x + 3z = 11; 2x − 3y = 1.

Accepted Answer

Write as AX = B: A = [[1,2,1],[1,0,3],[2,−3,0]], X = [[x],[y],[z]], B = [[7],[11],[1]]. FIND |A|: Expand along R₁: |A| = 1|0,3;−3,0| − 2|1,3;2,0| + 1|1,0;2,−3| = 1(0+9) − 2(0−6) + 1(−3−0) = 9 + 12 − 3 = 18. |A| = 18 ≠ 0, so unique solution exists. COFACTORS: C₁₁ = |0,3;−3,0| = 9. C₁₂ = −|1,3;2,0| = −(0−6) = 6. C₁₃ = |1,0;2,−3| = −3. C₂₁ = −|2,1;−3,0| = −(0+3) = −3. C₂₂ = |1,1;2,0| = 0−2 = −2. C₂₃ = −|1,2;2,−3| = −(−3−4) = 7. C₃₁ = |2,1;0,3| = 6. C₃₂ = −|1,1;1,3| = −(3−1) = −2. C₃₃ = |1,2;1,0| = 0−2 = −2. adj(A) = (cofactors)ᵀ = [[9,−3,6],[6,−2,−2],[−3,7,−2]]. A⁻¹ = (1/18)[[9,−3,6],[6,−2,−2],[−3,7,−2]]. X = A⁻¹B = (1/18)[[9×7+(−3)×11+6×1],[6×7+(−2)×11+(−2)×1],[−3×7+7×11+(−2)×1]] = (1/18)[[63−33+6],[42−22−2],[−21+77−2]] = (1/18)[[36],[18],[54]] = [[2],[1],[3]]. SOLUTION: x = 2, y = 1, z = 3. VERIFY: 2 + 2 + 3 = 7 ✓; 2 + 9 = 11 ✓; 4 − 3 = 1 ✓.

Practice — Determinants

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