Simultaneous Linear Equations and Matrix Formulation

The key idea. A set of simultaneous linear equations is just a matrix multiplication waiting to happen. Once you write it as $A\mathbf{x} = \mathbf{b}$ you can use all the matrix machinery — inverses, determinants, row operations.

Take the system: $\begin{aligned} 2x + y - z &= 3 \\ x - 3y + 2z &= 1 \\ 4x + y + z &= 9 \end{aligned}$

Read off three things:

• The coefficient matrix $A$ — every coefficient, in order, row by row.
• The variable vector $\mathbf{x}$ — the unknowns stacked in a column.
• The constant vector $\mathbf{b}$ — the right-hand sides stacked in a column.

$A = \begin{pmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 4 & 1 & 1 \end{pmatrix},\quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix},\quad \mathbf{b} = \begin{pmatrix} 3 \\ 1 \\ 9 \end{pmatrix}$

Check it reproduces the system. Multiplying $A\mathbf{x}$ gives a column whose entries are exactly the left-hand sides $2x+y-z$, $x-3y+2z$, $4x+y+z$. Setting that column equal to $\mathbf{b}$ recovers all three equations at once.

Cambridge tip. Keep the variables in the same order ($x, y, z$) across every equation before you build $A$. If an equation is missing a variable (e.g. no $z$), put a $0$ in that slot — never leave a gap.

Once you have $A\mathbf{x} = \mathbf{b}$, the cleanest solution (for a small system whose inverse you can find) is the inverse method.

The derivation — pre-multiply by $A^{-1}$: $A\mathbf{x} = \mathbf{b} \;\Rightarrow\; A^{-1}(A\mathbf{x}) = A^{-1}\mathbf{b} \;\Rightarrow\; (A^{-1}A)\mathbf{x} = A^{-1}\mathbf{b} \;\Rightarrow\; I\mathbf{x} = A^{-1}\mathbf{b}$ $\boxed{\;\mathbf{x} = A^{-1}\mathbf{b}\;}$

Order is everything. You must pre-multiply (put $A^{-1}$ on the left). Writing $\mathbf{x} = \mathbf{b}A^{-1}$ is wrong — matrix multiplication is not commutative, and $\mathbf{b}A^{-1}$ is not even defined for these dimensions.

This only works when $A^{-1}$ exists, i.e. when $\det A \neq 0$. Such an $A$ is called non-singular. If $\det A = 0$ the inverse does not exist, and the system has either no solution or infinitely many — that is the subject of the next subtopic.

The recipe for a $3\times3$:

1. Find $\det A$. If it is $0$, stop — the inverse method does not apply.
2. Find $A^{-1}$ (cofactors → adjugate → divide by $\det A$, or use a calculator if allowed).
3. Compute $A^{-1}\mathbf{b}$. The resulting column is $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$.

Cambridge tip. Even when a calculator can invert $A$ for you, write down $\det A$ and $A^{-1}$ explicitly. Examiners award method marks for the inverse and for the product $A^{-1}\mathbf{b}$ — a bare answer can score badly if it is wrong.

The Gaussian elimination method avoids computing a full inverse. It works directly on the augmented matrix $[A\mid\mathbf{b}]$ — the coefficient matrix with the constant column tacked on the right.

Step 1 — Augment. $[A\mid\mathbf{b}] = \left(\begin{array}{ccc|c} 2 & 1 & -1 & 3 \\ 1 & -3 & 2 & 1 \\ 4 & 1 & 1 & 9 \end{array}\right)$

Step 2 — Row-reduce to upper-triangular (echelon) form using elementary row operations:

• swap two rows ($R_i \leftrightarrow R_j$),
• multiply a row by a non-zero constant ($kR_i$),
• add a multiple of one row to another ($R_i + kR_j$).

The aim is zeros below the leading diagonal: $\left(\begin{array}{ccc|c} \bullet & \ast & \ast & \ast \\ 0 & \bullet & \ast & \ast \\ 0 & 0 & \bullet & \ast \end{array}\right)$

Step 3 — Back-substitute. The bottom row reads off one variable directly (e.g. $z$); substitute upward into the middle row for $y$, then the top row for $x$.

Step 4 — Check. Substitute your $(x, y, z)$ into all three original equations. They must each balance. This catches arithmetic slips in the elimination.

Cambridge tip. Operate on whole rows at a time and write the operation you used ($R_2 \to 2R_2 - R_1$) beside each step. Examiners follow your elimination — clear annotation earns the method marks even if a number later slips.

Both methods give the same answer for a non-singular system. Choose based on the question.

When to use the inverse method ($\mathbf{x} = A^{-1}\mathbf{b}$):

• The question explicitly asks you to find $A^{-1}$ (often part (a)) and then solve (part (b)).
• You must solve $A\mathbf{x} = \mathbf{b}$ for several different $\mathbf{b}$'s — invert once, reuse $A^{-1}$.

When to use Gaussian elimination:

• You only need one solution and an inverse would be extra work.
• The system might be singular or inconsistent — row reduction reveals a row of zeros (the inverse method simply fails with no diagnosis).

Verifying — the one step students skip. Whatever method you use, substitute back: $2x + y - z, \quad x - 3y + 2z, \quad 4x + y + z$ must equal $3, 1, 9$ respectively. If even one fails, you have an arithmetic error — find it before moving on. A correct check is worth more than a rushed extra question.

Cambridge tip. In a multi-part question, the structure usually tells you which method to use: "Find $A^{-1}$. Hence solve $A\mathbf{x} = \mathbf{b}$" means use the inverse you just found — do not start a fresh elimination, or you forfeit the 'Hence' marks.