Systems of linear equations

Maths · Matrices and determinants. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A system of linear equations can be written as AX = B, where A is the matrix of coefficients, X is the column of unknowns and B is the column of constants. If B = O the system is homogeneous; otherwise it is non-homogeneous.

When A is square and |A| ≠ 0 there is exactly one solution, found by the matrix inverse method or by Cramer's rule.

The rank of a matrix is the number of non-zero rows in its echelon form. Comparing the rank of A with the rank of the augmented matrix tells you whether a system is consistent (has a solution) or inconsistent (has none).

Rules to remember

Matrix inverse method: X = A⁻¹B (A⁻¹ on the left of B).
Cramer's rule: x = |A₁|/|A|, y = |A₂|/|A|, where A₁ is A with its first column replaced by B, A₂ is A with its second column replaced by B, and so on.
Echelon form: zero rows at the bottom, and each leading entry lies to the right of the leading entry in the row above. Rank = number of non-zero rows.
Consistent: rank of A = rank of the augmented matrix. Inconsistent: the two ranks are not equal.
Consistent with rank = number of unknowns: unique solution. Consistent with rank less than the number of unknowns: infinitely many solutions.
Homogeneous system AX = O always has the trivial solution (all unknowns zero).
Homogeneous system with square A: |A| ≠ 0 gives only the trivial solution; |A| = 0 gives non-trivial solutions.

Common mistake

In Cramer's rule students replace the wrong column. For x replace the column of x-coefficients by the constants; for y replace the column of y-coefficients. Write each equation with the constant alone on the right-hand side first.

Worked example

Use Cramer's rule to solve 3x + 2y = 12 and x − y = −1.

|A| has rows [3 2] and [1 −1]: |A| = (3 × −1) − (2 × 1) = −5.
|A₁| has rows [12 2] and [−1 −1]: (12 × −1) − (2 × −1) = −12 + 2 = −10, so x = −10 ÷ −5 = 2.
|A₂| has rows [3 12] and [1 −1]: (3 × −1) − (12 × 1) = −15, so y = −15 ÷ −5 = 3.
Check: 3(2) + 2(3) = 12 and 2 − 3 = −1.

Answer: x = 2, y = 3

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