Solving Linear Systems Using the Matrix Method

The Method, Step by Step

Given a system $AX=B$ with $A$ the coefficient matrix, $X$ the column of unknowns and $B$ the column of constants, provided $|A|\ne0$ (Section 7 confirms the system is then consistent with a unique solution), the solution is found in four steps:

1. Write the system in the matrix form $AX=B$.
2. Compute $|A|$ and confirm it is non-zero.
3. Compute $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A$ (Section 6).
4. Compute $X=A^{-1}B$; the entries of the resulting column $X$ are the values of the unknowns.

Multiplying both sides of $AX=B$ on the left by $A^{-1}$ justifies step 4 directly: $A^{-1}(AX)=A^{-1}B \Rightarrow (A^{-1}A)X=A^{-1}B \Rightarrow IX=A^{-1}B \Rightarrow X=A^{-1}B$, using associativity of matrix multiplication and the defining property $A^{-1}A=I$.

Worked Illustration (Two Variables)

Solve $2x+3y=7,\ x-y=1$ by the matrix method. Here $A=\begin{pmatrix}2&3\\1&-1\end{pmatrix}$, $X=\begin{pmatrix}x\\y\end{pmatrix}$, $B=\begin{pmatrix}7\\1\end{pmatrix}$.

Step 2: $|A|=2(-1)-3(1)=-2-3=-5\ne0$, so the system is consistent with a unique solution.

Step 3: $\operatorname{adj}A=\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}$ (Section 5's $2\times2$ shortcut), so $A^{-1}=-\dfrac15\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}=\begin{pmatrix}\tfrac15&\tfrac35\\[2pt]\tfrac15&-\tfrac25\end{pmatrix}$.

Step 4: $X=A^{-1}B=\begin{pmatrix}\tfrac15&\tfrac35\\[2pt]\tfrac15&-\tfrac25\end{pmatrix}\begin{pmatrix}7\\1\end{pmatrix}=\begin{pmatrix}\tfrac75+\tfrac35\\[2pt]\tfrac75-\tfrac25\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}$.

So $x=2,\ y=1$. Checking: $2(2)+3(1)=4+3=7$ ✓ and $2-1=1$ ✓.

Extending to Three Variables …