How Do You Use an Inverse Matrix to Solve a System of Linear Equations?

Write the system as a matrix equation $A\mathbf{x}=\mathbf{b}$, then multiply both sides on the left by $A^{-1}$. This gives $\mathbf{x}=A^{-1}\mathbf{b}$, provided that the coefficient matrix $A$ is invertible.

The Method

For the system

$$\begin{aligned} ax+by&=p,\\ cx+dy&=q, \end{aligned}$$

write

$$\begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}p\\q\end{pmatrix}.$$

Here, $A$ is the coefficient matrix, $\mathbf{x}$ is the column matrix of unknowns, and $\mathbf{b}$ contains the constants. The inverse exists only if the determinant is non-zero. For a $2\times2$ matrix,

$$\det(A)=ad-bc.$$

If $\det(A)\neq0$, multiply the original equation on the left by $A^{-1}$:

$$A^{-1}A\mathbf{x}=A^{-1}\mathbf{b}.$$

Because $A^{-1}A=I$ and $I\mathbf{x}=\mathbf{x}$, the solution is

$$\mathbf{x}=A^{-1}\mathbf{b}.$$

Conceptually, left multiplication by $A^{-1}$ undoes the transformation performed by $A$. The identity matrix $I$ then leaves the unknown vector unchanged. The order is essential because matrices generally represent transformations that act in sequence, and reversing that sequence changes the result.

Worked Example

Solve $2x+y=7$ and $x-y=2$. The matrix equation is

$$\begin{pmatrix}2&1\\1&-1\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}7\\2\end{pmatrix}.$$

The determinant is

$$(2)(-1)-(1)(1)=-3.$$

Since it is non-zero, the inverse exists. Using a GDC to evaluate the matrix product,

$$\mathbf{x}=A^{-1}\mathbf{b} =\begin{pmatrix}3\\1\end{pmatrix}.$$

Therefore, $x=3$ and $y=1$. Check by substitution: $2(3)+1=7$ and $3-1$, so both original equations are satisfied.

For a larger system, the same reasoning applies: enter $A$ and $\mathbf{b}$ into the GDC, calculate the inverse product, and retain full precision until the final answer.

A common misconception is that matrix multiplication order does not matter. Matrix multiplication is not commutative, so calculate $A^{-1}\mathbf{b}$, not $\mathbf{b}A^{-1}$. Also, if $\det(A)=0$, do not attempt to use an inverse because does not exist.

IB Exam Technique

This method belongs to AHL 1.14: Introduction to matrices, so it is HL-only. In a Paper 1 short-response or Paper 2 extended-response question, define $A$, $\mathbf{x}$, and $\mathbf{b}$ before using the GDC. Writing the correct matrix equation and inverse-matrix product provides essential method evidence; a bare calculator answer may not receive full credit.

If asked to verify the solution, substitute every value into every original equation. This demonstrates that the column vector satisfies the whole system, not merely one equation.