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Systems of Equations

Simultaneous equations can be packed into a single matrix equation. Once they are in that form, one inverse matrix unlocks both unknowns in a couple of lines.

MYP 5Extended MathsVectors & MatricesCriteria A · C · D~11 min read

Written by Fayek · MYP student

Writing it as a matrix equation

Take a pair of simultaneous equations such as \(ax + by = e\) and \(cx + dy = f\). The coefficients form a 2x2 matrix, the unknowns form a column vector, and the right-hand sides form another column vector. Multiplied out, the matrix equation reproduces the two original equations exactly.

\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} e \\ f \end{pmatrix} \]

We usually call the coefficient matrix \(A\), the unknowns \(X\), and the answers \(B\), so the whole system is just \(AX = B\).

Matrix equation
A system \(AX = B\), where \(A\) holds the coefficients, \(X = \begin{pmatrix} x \\ y \end{pmatrix}\) holds the unknowns, and \(B\) holds the constants.

Solving with the inverse matrix

To get \(X\) on its own we cannot divide by a matrix, but we can multiply both sides by the inverse \(A^{-1}\). Since \(A^{-1}A = I\) and \(IX = X\), the unknowns drop out cleanly.

\[ AX = B \;\Rightarrow\; X = A^{-1}B \]

Recall the inverse of a 2x2 matrix: \(A^{-1} = \dfrac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\). So the plan is: build \(A\) and \(B\), find \(A^{-1}\), then multiply \(A^{-1}B\) to read off \(x\) and \(y\).

Keep the inverse on the left

Order matters in matrix multiplication, so write \(X = A^{-1}B\), not \(BA^{-1}\). The inverse must sit to the left of \(B\), matching the side it was applied from.

A full worked solution (Criterion D)

Worked example

Solve \(2x + 3y = 8\) and \(x + 2y = 5\) using matrices.

1
Write it as \(AX = B\): \(\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ 5 \end{pmatrix}\).
2
Determinant of \(A\): \((2)(2) - (3)(1) = 4 - 3 = 1\).
3
Inverse: \(A^{-1} = \dfrac{1}{1}\begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}\).
4
Multiply \(X = A^{-1}B\): \(\begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 8 \\ 5 \end{pmatrix} = \begin{pmatrix} (2)(8)+(-3)(5) \\ (-1)(8)+(2)(5) \end{pmatrix} = \begin{pmatrix} 16-15 \\ -8+10 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\).
5
Check in the original equations: \(2(1)+3(2) = 8\) and \(1 + 2(2) = 5\). Both hold.
\(x = 1,\; y = 2\)

When it goes wrong

The method breaks down when the coefficient matrix has a determinant of zero. That means \(A^{-1}\) does not exist, and the two equations either describe the same line (endlessly many solutions) or parallel lines that never meet (no solution).

Where this is assessed

Forming the matrix equation and computing \(A^{-1}B\) is Criterion A and C. Recognising that a zero determinant signals no unique solution, and explaining what that means for the lines, is exactly the reasoning Criterion D rewards. Always finish by substituting your \(x\) and \(y\) back into both original equations.

Check yourself

1. Write \(x + 2y = 5\) and \(3x + y = 10\) as a matrix equation. +

Place the coefficients, unknowns, and constants into \(AX = B\): \(\mathbf{\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 10 \end{pmatrix}}\).

2. What is the determinant of the coefficient matrix \(\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}\), and does a unique solution exist? +

Determinant: \((1)(1) - (2)(3) = 1 - 6 = -5\). Because this is not zero, the matrix is invertible, so a unique solution exists.

3. Solve \(3x + y = 5\) and \(x + y = 3\) with the inverse method. +

Here \(A = \begin{pmatrix} 3 & 1 \\ 1 & 1 \end{pmatrix}\) and \(B = \begin{pmatrix} 5 \\ 3 \end{pmatrix}\). Determinant: \((3)(1) - (1)(1) = 2\), so \(A^{-1} = \dfrac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 3 \end{pmatrix}\). Then \(X = A^{-1}B = \dfrac{1}{2}\begin{pmatrix} (1)(5)+(-1)(3) \\ (-1)(5)+(3)(3) \end{pmatrix} = \dfrac{1}{2}\begin{pmatrix} 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\). So \(\mathbf{x = 1,\; y = 2}\) (check: \(3(1)+2 = 5\) and \(1 + 2 = 3\)).

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