Solution of linear equations using matrices

Packing the system into one equation. Any system of 2 or 3 linear equations can be written as three separate matrices multiplied together:

where: $A$ is the $n\times n$ coefficient matrix ($n=2$ or $3$), $X$ is the column matrix of unknowns, and $B$ is the column matrix of constants on the right-hand side. Writing the system this way is purely bookkeeping — no new mathematics — but it turns "solve for $x, y, z$" into "solve for $X$," a problem matrix algebra already knows how to handle.

Solving by inverting $A$. If $A$ is invertible, multiply both sides of $AX = B$ on the left by $A^{-1}$. Since $A^{-1}A = I$ and $IX = X$, the unknowns fall out directly:

where: $A^{-1} = \dfrac{1}{|A|}\operatorname{adj}(A)$, so this step reuses the adjoint-and-determinant machinery from earlier in the chapter — nothing new is computed here that you have not already practised. The condition for this to work is simply $|A| \neq 0$, so $A$ is non-singular and $A^{-1}$ exists.

Reading the outcome from $|A|$ alone. Before grinding through an inverse, a quick glance at $|A|$ tells you which of three outcomes you are in. This is the syllabus's basic-level consistency check, and it is exactly what the figure below sorts through.

The system's fate depends entirely on two matrix quantities computed from $A$ and $B$: the determinant $|A|$, and the product $(\operatorname{adj}A)B$ when $|A|=0$. Getting from a raw system of equations to a verified numeric answer for $x$, $y$, $z$ requires converting the equations to $AX=B$ form correctly, confirming $|A|\neq0$, computing the adjoint, forming $A^{-1}$, and multiplying it into $B$ in the correct order — a chain where a slip at any link gives a wrong final answer even if every individual computation looks right in isolation.

Step 1. Write the given system as $AX = B$ by reading off the coefficients of $x$, $y$ (and $z$, if present) row by row into $A$, stacking the unknowns into $X$, and the right-hand-side constants into $B$. Any missing variable in an equation must be entered as coefficient $0$ in $A$, not left out — a row must have one entry per unknown.

Step 2. Compute $|A|$ using the standard expansion along any convenient row or column (a row or column with a zero is usually fastest). If $|A| \neq 0$, the system has a unique solution and you proceed; if $|A| = 0$, stop and instead inspect $(\operatorname{adj}A)B$ to decide between "infinitely many solutions" and "no solution," rather than attempting an inverse that does not exist.

Step 3. Compute $\operatorname{adj}(A)$: for $2\times2$, swap the diagonal entries and negate the off-diagonal pair; for $3\times3$, compute all nine cofactors $C_{ij}$, arrange them into the matrix $[C_{ij}]$, and transpose it.

Step 4. Form $A^{-1} = \dfrac{1}{|A|}\operatorname{adj}(A)$, then compute $X = A^{-1}B$ by multiplying this matrix into the column $B$ — always $A^{-1}$ times $B$, in that order, since matrix multiplication is not commutative.

Step 5. Read off $x$, $y$, $z$ from the resulting column $X$, and verify by substituting each value back into all of the original equations.

Solve $2x + y = 5$ and $3x + 2y = 8$ by the matrix inverse method, then check what the method predicts if the right-hand side were changed to a case with $|A| = 0$.

Step 1. Write $AX = B$: $A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$.

Step 2. Compute $|A| = (2)(2) - (1)(3) = 4 - 3 = 1$. Since $|A| \neq 0$, a unique solution exists and the inverse method applies directly.

Step 3. Using the $2\times2$ shortcut, $\operatorname{adj}(A) = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$, so $A^{-1} = \dfrac{1}{1}\begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$.

Step 4. Compute $X = A^{-1}B = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 2(5)-1(8) \\ -3(5)+2(8) \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$. So $x = 2$, $y = 1$, and checking, $2(2)+1=5$ and $3(2)+2(1)=8$, both correct.

Competitive stretch

Now consider the related pair $2x + y = 5$ and $4x + 2y = 8$. Here $A' = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$ has $|A'| = (2)(2)-(1)(4) = 0$, so $A'$ is singular and $X = A'^{-1}B$ cannot be formed at all — the method does not silently give a wrong answer; it simply refuses to run, which is itself the diagnostic. The first pair has the unique solution $x=2, y=1$; the second pair fails the $|A|\neq0$ test and needs the $\operatorname{adj}(A)B$ inspection instead.

• Reaching for Cramer's rule instead of $X = A^{-1}B$ — CBSE's chosen technique for this chapter is the inverse-matrix method, so practise it fluently even if you already know Cramer's rule.
• Writing down $X = A^{-1}B$ before checking $|A| \neq 0$ — the formula is meaningless if $A^{-1}$ does not exist.
• Multiplying in the wrong order, computing $B \cdot \operatorname{adj}(A)$ instead of $\operatorname{adj}(A) \cdot B$ — matrix multiplication is not commutative, and only the stated order has matching dimensions.
• Dropping a missing variable's term instead of entering it as coefficient $0$ in $A$ — this silently changes the matrix and gives a wrong $X$.

Ask the tutor: Share your own system of 2 or 3 linear equations and have the tutor walk through the matrix setup, the determinant check, and the $A^{-1}B$ computation with you step by step.