10 min readConcept sheet10-mark problem: find an inverse and use it to solve equations

Determinants, inverse matrices and solving systems

A square matrix has an inverse only if its determinant is not zero. Then the system AX = B is solved in one step: X equals A⁻¹B.

2 × 2 determinant|A| = ad − bc

2 × 2 inverseA⁻¹ = (1/|A|)[d −b; −c a]

CheckAA⁻¹ = I

Singular|A| = 0: no inverse

Inverse and determinant formulas

Determinant of [a b; c d]ad − bc

Inverse of [a b; c d](1/(ad − bc)) × [d −b; −c a]: swap a and d, change signs of b and c

3 × 3 determinant (row 1)a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃

CofactorCᵢⱼ = (−1)^(i+j) × minor

Sign pattern[+ − +; − + −; + − +]

AdjointTranspose of the cofactor matrix

3 × 3 inverseA⁻¹ = adj(A) ÷ |A|

Solving AX = BX = A⁻¹B

Inverse of a 2 × 2 matrix

Problem

Find the inverse of A = [4 7; 2 6] and check it.

Working

|A| = 4 × 6 − 7 × 2 = 24 − 14 = 10 (not zero, so A⁻¹ exists).
Swap 4 and 6, change the signs of 7 and 2: [6 −7; −2 4].
A⁻¹ = (1/10)[6 −7; −2 4] = [0.6 −0.7; −0.2 0.4].
Check AA⁻¹: row 1: 2.4 − 1.4 = 1, −2.8 + 2.8 = 0; row 2: 1.2 − 1.2 = 0, −1.4 + 2.4 = 1. So AA⁻¹ = I. ✓

Answer

A⁻¹ = [0.6 −0.7; −0.2 0.4]

Solve a system with the inverse

Problem

A factory makes two products, x and y (hundreds of units). Labour: 4x + 7y = 220 hours. Material: 2x + 6y = 160 kg. Solve using the inverse found above.

Working

In matrix form AX = B with A = [4 7; 2 6], X = [x; y], B = [220; 160].
X = A⁻¹B = [0.6 −0.7; −0.2 0.4][220; 160].
x = 0.6 × 220 − 0.7 × 160 = 132 − 112 = 20.
y = −0.2 × 220 + 0.4 × 160 = −44 + 64 = 20.
Check: 4(20) + 7(20) = 220 ✓; 2(20) + 6(20) = 160 ✓.

Answer

x = 20 and y = 20 (2,000 units of each).

A 3 × 3 inverse by cofactors

Problem

A factory's three departments need 1x + 2y + 3z = 14, y + 4z = 14 and 5x + 6y = 17 hours (x, y, z in hundreds of units). Find A⁻¹ for A = [1 2 3; 0 1 4; 5 6 0] and solve.

Working

|A| = 1(0 − 24) − 2(0 − 20) + 3(0 − 5) = −24 + 40 − 15 = 1.
Cofactors (with the + − + sign pattern): [−24 20 −5; 18 −15 4; 5 −4 1].
Adjoint (transpose): [−24 18 5; 20 −15 −4; −5 4 1]. As |A| = 1, A⁻¹ = adj(A).
X = A⁻¹B with B = [14; 14; 17]:
x = −24(14) + 18(14) + 5(17) = −336 + 252 + 85 = 1.
y = 20(14) − 15(14) − 4(17) = 280 − 210 − 68 = 2.
z = −5(14) + 4(14) + 1(17) = −70 + 56 + 17 = 3.
Check row 1: 1 + 4 + 9 = 14 ✓.

Answer

A⁻¹ = [−24 18 5; 20 −15 −4; −5 4 1]; x = 1, y = 2, z = 3 (100, 200 and 300 units).

Row reduction or inverse?

Row reductionWorks for any system, even when there is no unique solution. Best for a single set of equations.

Inverse matrixNeeds |A| ≠ 0. Best when the same A is used with many B, e.g. a new production target each month.

In Sri Lankan business

A Sri Lankan factory planner who solves the same resource equations every month with new targets can find A⁻¹ once and reuse it: X = A⁻¹B gives each month's production plan in one multiplication.

Exam angle

Short answer: “Find the determinant of [5 2; 3 4].” 5 × 4 − 2 × 3 = 20 − 6 = 14.

Structured question: “Find the inverse of [4 7; 2 6] and check that AA⁻¹ = I.” |A| = 10; A⁻¹ = [0.6 −0.7; −0.2 0.4]; the product gives I.

Essay: “Find the inverse of A = [1 2 3; 0 1 4; 5 6 0] by cofactors and use it to solve AX = [14; 14; 17].” |A| = 1; A⁻¹ = [−24 18 5; 20 −15 −4; −5 4 1]; X = [1; 2; 3].

Remember it

2 × 2 inverse: "swap the diagonal, change signs of the others, divide by the determinant." Always check AA⁻¹ = I.