Solving Systems of Equations Using Matrices
A system of linear equations is one matrix equation in disguise — once you have the inverse, solving it is a single multiplication.
Writing a system as a matrix equation
A system of linear equations can be written $A\mathbf{x}=\mathbf{b}$, where $A$ is the matrix of coefficients, $\mathbf{x}$ is the column vector of unknowns, and $\mathbf{b}$ is the column vector of constants. If $A$ is non-singular, pre-multiplying both sides by $A^{-1}$ gives $\mathbf{x}=A^{-1}\mathbf{b}$.
You must pre-multiply both sides by $A^{-1}$ (multiply on the left), since $A^{-1}A\mathbf{x}=I\mathbf{x}=\mathbf{x}$ only works that way round — post-multiplying gives $A\mathbf{x}A^{-1}$, which isn't even always defined and doesn't isolate $\mathbf{x}$.
Write as $A\mathbf{x}=\mathbf{b}$: $\begin{pmatrix}2&1\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}5\\1\end{pmatrix}$.
$\det(A)=2(-1)-1(1)=-3$, so $A^{-1}=\dfrac{1}{-3}\begin{pmatrix}-1&-1\\-1&2\end{pmatrix}$.
So $x=2$, $y=1$.
So $x=1$, $y=1$, $z=2$. Checking: $2(1)+1=3$ ✓, $1+1+2=4$ ✓, $1+3(2)=7$ ✓.
When the coefficient matrix is singular
If $\det(A)=0$, the matrix method breaks down — the system either has no solutions (inconsistent, e.g. parallel planes that never meet) or infinitely many solutions (the equations are dependent, e.g. one plane contains the intersection line of the other two). Distinguishing between these cases requires substituting back into the original equations or using row-reduction, not the inverse-matrix method.
$A=\begin{pmatrix}1&2\\2&4\end{pmatrix}$, $\det(A)=1(4)-2(2)=0$.
Since $A$ is singular, no inverse exists. Geometrically, the two lines $x+2y=5$ and $2x+4y=9$ are parallel (same gradient) but not identical, so they never meet — the system has no solution.
So $x=3$, $y=-2$, $z=4$. Since a negative ticket price makes no sense, this signals the model itself needs revisiting — a useful reminder that solving the algebra correctly is not the same as the answer being realistic in context.
$A=\begin{pmatrix}1&-1\\2&-2\end{pmatrix}$, $\det(A)=1(-2)-(-1)(2)=-2+2=0$, so $A$ is singular and the inverse-matrix method cannot be used.
The second equation is exactly twice the first ($2x-2y=2(x-y)=2(2)=4$), so every $(x,y)$ satisfying $x-y=2$ satisfies both equations — the two lines are identical. The system is consistent with infinitely many solutions (the whole line $x-y=2$), rather than being inconsistent.
When $\det(A)=0$ for a linear system, there is no unique solution. For two variables this means the corresponding lines are either parallel and distinct (no solution) or the same line (infinitely many solutions). For three variables, the corresponding planes fail to meet at a single point — they may form a prism-like arrangement with no common point (no solution), or intersect along a shared line or coincide entirely (infinitely many solutions). Distinguishing the two singular cases always requires checking the equations directly, not just the determinant.
Exam-style questions
Question 1 6 marks
Solve the system $3x+2y=16$, $x-y=2$ using matrices.
$A=\begin{pmatrix}3&2\\1&-1\end{pmatrix}$, $\det(A)=3(-1)-2(1)=-5$ (M1, A1)
$A^{-1}=\dfrac{1}{-5}\begin{pmatrix}-1&-2\\-1&3\end{pmatrix}$ (M1)
$\begin{pmatrix}x\\y\end{pmatrix}=\dfrac{1}{-5}\begin{pmatrix}-1&-2\\-1&3\end{pmatrix}\begin{pmatrix}16\\2\end{pmatrix}=\dfrac{1}{-5}\begin{pmatrix}-20\\-10\end{pmatrix}=\begin{pmatrix}4\\2\end{pmatrix}$ (M1, A1)
$x=4$, $y=2$ (A1)
Question 2 4 marks
Using $M^{-1}=\begin{pmatrix}2&-3&1\\-3&6&-2\\1&-2&1\end{pmatrix}$, solve $M\mathbf{x}=\begin{pmatrix}5\\6\\10\end{pmatrix}$.
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2(5)-3(6)+1(10)\\-3(5)+6(6)-2(10)\\1(5)-2(6)+1(10)\end{pmatrix}=\begin{pmatrix}2\\1\\3\end{pmatrix}$ (M1, A1, A1)
$x=2$, $y=1$, $z=3$ (A1)
Question 3 5 marks
Show that the system $2x-y=4$, $-6x+3y=-8$ has no solution, giving a geometric reason.
$\det\begin{pmatrix}2&-1\\-6&3\end{pmatrix}=2(3)-(-1)(-6)=6-6=0$ (M1, A1)
Since the determinant is $0$, no inverse exists; the two lines have the same gradient (parallel) but different intercepts, so they never meet and the system has no solution. (B1, B1, B1)