Visual walkthrough — Solving systems using matrix inversion

This is the parent recipe drawn out slowly, one picture per step. We start with two ordinary equations and end with the single boxed formula — but every symbol earns its place before we use it, and every move gets a picture.

We will keep one running example the whole way: $2x+3y=5,x-y=1.$

Two straight lines on a flat sheet. Where they cross is the answer. Watch how matrices repackage that crossing point into one clean formula.

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Step 1 — Two equations are two lines

WHAT we did: turned each equation into a line. WHY: a solution must satisfy both promises, so it must lie on both lines. Geometry replaces algebra — we now hunt for an intersection. WHAT IT LOOKS LIKE: two chalk lines crossing at one dot.

The dot in the figure is the answer we are chasing: $(x,y)=(8/5,3/5)$. Everything below is machinery to compute that dot without eyeballing the graph.

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Step 2 — Stack the numbers into a grid $A$, a stack $X$, a stack $B$

Before any symbol appears, here is what each grid is.

Reading $A$ term by term: the top row $\left[\begin{array}{cc}2& 3\end{array}\right]$ is the coefficients of the first equation ($2x+3y$); the bottom row $\left[\begin{array}{cc}1& -1\end{array}\right]$ is the second ($1x-1y$). The column $B$ holds the "$=5$" and "$=1$".

WHAT we did: peeled the numbers away from the letters. WHY: the numbers alone (the grid $A$) carry all the structure of the system. Later we can work on the grid without dragging $x,y$ around. WHAT IT LOOKS LIKE: colouring each equation and watching its numbers march into a row.

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Step 3 — What "row times column" means (so $AX$ rebuilds the system)

We have never multiplied a grid by a column yet, so here is the single rule, built from a picture.

Doing this for both rows of $A$ stacks two results into a column: $AX=\left[\begin{array}{cc}2& 3\\ 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}2x+3y\\ x-y\end{array}\right].$

Now set that column equal to $B$: $\left[\begin{array}{c}2x+3y\\ x-y\end{array}\right]=\left[\begin{array}{c}5\\ 1\end{array}\right]⟺\{\begin{array}{l}2x+3y=5\\ x-y=1\end{array}.$

WHAT we did: showed $AX=B$ is literally the two original equations. WHY: this is the whole justification for the repackaging — the compact form loses nothing. Row $i$ times $X$ reproduces equation $i$ exactly. WHAT IT LOOKS LIKE: a row sliding down the column, pairing up terms, then dropping into a slot of the answer column.

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Step 4 — $A$ is a machine that moves points

So the whole question becomes: "which arrow $X$ does the machine $A$ send to the arrow $B$?"

WHAT we did: reframed "solve the system" as "reverse-engineer a machine's input." WHY: because machines that can be reversed have a clean undo button — and that button will hand us $X$ directly. WHAT IT LOOKS LIKE: an input arrow on the left, the machine $A$, an output arrow on the right. We know the output $B$; we're missing the input.

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Step 5 — The undo machine $A^{-1}$

Why does $A^{-1}A=I$ mean "undo"? Because applying $A$ then $A^{-1}$ is the same as applying the do-nothing machine — you end up where you started.

WHAT we did: named the reverse machine and the do-nothing machine. WHY: we cannot "divide by a matrix", but we can undo a machine. $A^{-1}$ is the matrix world's version of dividing. WHAT IT LOOKS LIKE: the output arrow $B$ fed backwards through $A^{-1}$, landing on the input arrow $X$.

The concrete recipe for the undo machine (see Adjoint and Inverse of a Matrix): $A^{-1}=\frac{1}{\det A}\mathrm{adj}(A).$ Here $\det A$ is a single number measuring "how much the machine stretches or squashes area" (see Determinants), and $\mathrm{adj}(A)$ is the transpose of the cofactor matrix (see Cofactors and Minors).

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Step 6 — The derivation, one line at a time

Now the algebra, each move justified.

Start from the packed form: $AX=B.$ Here $A$ sits on the left of $X$. Remember that for matrices, order matters — $A^{-1}A$ cancels but $A{A}^{-1}$ is a different order, so we must be careful which side we act on.

Multiply both sides on the left by $A^{-1}$: $A^{-1}(AX)=A^{-1}B.$ We chose the left because $A^{-1}$ must sit right next to the $A$ it is cancelling. Sliding it in from the right would give $A^{-1}$ nothing to cancel.

Regroup (matrix multiplication is associative — brackets can shift): $(A^{-1}A)X=A^{-1}B.$

Replace $A^{-1}A$ by $I$ (Step 5's definition), and $IX$ by $X$ (do-nothing machine): $X=A^{-1}B.$

WHAT we did: peeled $A$ off $X$ by cancelling it with $A^{-1}$. WHY each step: left-multiply → to line $A^{-1}$ against $A$; regroup → to bring $A^{-1}A$ together; simplify → because $A^{-1}A=I$ and $I$ does nothing. WHAT IT LOOKS LIKE: the $A$ and $A^{-1}$ annihilating each other, leaving $X$ alone on the left.

Plugging in our numbers ($\det A=2(-1)-3(1)=-5$):

X=\frac{1}{-5}\begin{bmatrix}-1&-3\\-1&2\end{bmatrix}\begin{bmatrix}5\\1\end{bmatrix}=\frac{1}{-5}\begin{bmatrix}-8\\-3\end{bmatrix}=\begin{bmatrix}8/5\\3/5\end{bmatrix}.$$ That is exactly the crossing dot from Step 1. ✓ --- ## Step 7 — The degenerate case: when the machine flattens space Everything above needed $\det A\neq0$. Watch what breaks otherwise. > [!intuition] > $\det A$ measures how the machine changes area. If $\det A=0$, the machine **squashes the whole plane onto a single line** — every point collapses onto that line. Once space is flat, two different inputs can land on the *same* output, so there is no unique undo — $A^{-1}$ does not exist (you'd be dividing by $\det A=0$). Two sub-cases decide the fate of the system. Multiply $AX=B$ by $\operatorname{adj}(A)$ and use the identity $\operatorname{adj}(A)\,A=(\det A)I$: $$(\det A)\,X=(\operatorname{adj}A)\,B.$$ When $\det A=0$ the left side is the zero column $O$, forcing $(\operatorname{adj}A)B=O$: > [!definition] The $\det A=0$ fork > - $(\operatorname{adj}A)B\neq O$ → **impossible** → ==no solution== (lines are parallel, never meet). > - $(\operatorname{adj}A)B= O$ → ==infinitely many solutions== (the two lines are the *same* line) — or none; confirm by substitution. **WHAT** we did: covered the case the boxed formula cannot touch. **WHY**: the reader must never hit an unsolvable input and be surprised — every scenario is now shown. **WHAT IT LOOKS LIKE**: left panel, two parallel lines that never cross (no solution); right panel, two lines lying exactly on top of each other (infinitely many). See [[Consistency of Linear Systems]]. > [!mistake] Reaching for $A^{-1}$ before checking $\det A$ > **Why it tempts:** the recipe feels universal. > **The fix:** $A^{-1}$ divides by $\det A$. Always compute $\det A$ **first**. If it is zero, abandon inversion and use the fork above. --- ## The one-picture summary The whole journey on one board: system → grid $AX=B$ → machine view → cancel $A$ with $A^{-1}$ → $X=A^{-1}B$, with the $\det A=0$ trapdoor drawn to the side. > [!recall]- Feynman: tell the whole walkthrough to a friend > We had two straight lines and wanted their crossing point. First we noticed each equation is a line, and the answer is where they meet. Then we stripped the numbers into a tidy grid $A$, stacked the unknowns into a column $X$, and the answers into a column $B$ — and checked that "grid times column", done by pairing-and-adding, rebuilds the original equations exactly, so nothing was lost. Next we pictured $A$ as a machine that pushes an input arrow to an output arrow; we know the output $B$ but not the input $X$. If the machine can be reversed, its undo button is $A^{-1}$, defined so that doing $A$ then $A^{-1}$ leaves you where you started. To get $X$ alone we left-multiplied $AX=B$ by $A^{-1}$ — *left*, because the $A^{-1}$ has to touch the $A$ it cancels — and out popped $X=A^{-1}B$. Finally, the warning: if the machine flattens the plane ($\det A=0$) you can't undo it. Then either the lines are parallel (no answer) or the very same line (endless answers). > [!mnemonic] > **Grid it, machine it, undo it: $X=A^{-1}B$ — but only if $\det A$ isn't zero.** --- ## Connections - [[Solving systems using matrix inversion]] — the parent recipe this page draws out. - [[Determinants]] — the $\det A$ that decides if the machine is reversible. - [[Adjoint and Inverse of a Matrix]] — how the undo machine $A^{-1}$ is actually built. - [[Cofactors and Minors]] — the ingredients inside $\operatorname{adj}(A)$. - [[Consistency of Linear Systems]] — the $\det A=0$ fork in Step 7. - [[Cramer's Rule]] — a sibling solver gated by the same $\det A\neq0$. - [[Gaussian Elimination]] — the practical alternative for big grids.