Solving linear equations with matrices

A set of linear equations can be written as one matrix equation AX = B:

• A: the coefficients of the unknowns, row by row;
• X: the column of unknowns, [x; y; z];
• B: the column of numbers on the right side.

If |A| ≠ 0, multiply both sides by A⁻¹: X = A⁻¹B, and there is exactly one solution. The system is called consistent (it has a solution).

Example: x + y = 5 and x – y = 1. A = [1, 1; 1, –1], |A| = –2, adj A = [–1, –1; –1, 1], so A⁻¹ = [½, ½; ½, –½]. Then X = A⁻¹[5; 1] = [3; 2]: x = 3, y = 2.

If |A| = 0, there is no inverse, so find (adj A)B:

• (adj A)B ≠ O: no solution (inconsistent).
• (adj A)B = O: either no solution or infinitely many; you have to look further.

Example: 2x + 4y = 6 and x + 2y = 5. |A| = 4 – 4 = 0. adj A = [2, –4; –1, 2], and (adj A)B = [12 – 20; –6 + 10] = [–8; 4] ≠ O. No solution. (Halve the first equation: x + 2y = 3, which clashes with x + 2y = 5.)

Word problems: give each unknown a letter, turn each sentence into one equation, then use the method. Equations with 2/x, 3/y and so on become linear if you put u = 1/x and v = 1/y first.

Write the order of work clearly in the exam: |A| first (and say it is non-zero), then the cofactors, adj A, A⁻¹, and finally X = A⁻¹B. Each step carries marks.