Visual walkthrough — Cramer's rule

We will build up one idea at a time. First we agree on what the raw objects even mean.

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Step 1 — What "$A\mathbf{x}=\mathbf{b}$" is really saying

WHAT. We have a system of equations. Written compactly it is $\underset{\text{the box of columns}}{\underbrace{A}}\underset{\text{the mix amounts}}{\underbrace{\mathbf{x}}}=\underset{\text{the target}}{\underbrace{\mathbf{b}}}.$ Here $A$ is a square grid of numbers with $n$ rows and $n$ columns; $\mathbf{x}=(x_1,\dots ,x_n)$ is the list of unknowns we want; $\mathbf{b}$ is a known target vector.

WHY read it by columns. The single most important fact for everything below: multiplying $A$ by $\mathbf{x}$ is the same as mixing the columns of $A$, using each $x_i$ as "how much of column $i$": $A\mathbf{x}=x_1{\mathbf{a}}_1+x_2{\mathbf{a}}_2+\cdots +x_n{\mathbf{a}}_n,$ where ${\mathbf{a}}_k$ means "the $k$-th column of $A$." So the question "solve $A\mathbf{x}=\mathbf{b}$" is literally "how much of each column do I stir together to land on $\mathbf{b}$?"

PICTURE. Below, the two columns of a $2\times 2$ matrix are drawn as arrows. Scaling them by $x_1$ and $x_2$ and adding tip-to-tail reaches $\mathbf{b}$.

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Step 2 — Determinant = the area (volume) of the column-box

WHAT. The determinant $\det A$ is a single number attached to a square matrix. Geometrically it is the signed area (in 2D) or signed volume (in 3D) of the box (parallelogram) whose edges are the columns ${\mathbf{a}}_1,\dots ,{\mathbf{a}}_n$.

WHY we care. Cramer's rule is a ratio of two such areas. So before we can divide areas, we must picture them. The word signed matters: if you list the columns in swapped order, the area comes out with a minus sign. Keep that in mind — it is the "alternating" behaviour we lean on later.

PICTURE. The parallelogram spanned by ${\mathbf{a}}_1$ and ${\mathbf{a}}_2$; its area is $\det A$.

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Step 3 — Build the clever helper matrix $X_i$

WHAT. Start from the identity matrix $I$ (all $1$s on the diagonal, $0$s elsewhere — its columns are the standard arrows ${\mathbf{e}}_1,\dots ,{\mathbf{e}}_n$ pointing along each axis). Now overwrite only the $i$-th column with the unknown vector $\mathbf{x}$. Call the result $X_i$. For $n=3$, $i=2$: $X_2=\left(\begin{array}{ccc}1& x_1& 0\\ 0& x_2& 0\\ 0& x_3& 1\end{array}\right).$ Every column is a plain axis-arrow ${\mathbf{e}}_k$, except column $i$, which is the orange unknown $\mathbf{x}$.

WHY invent this? Two reasons that pay off in the next steps: (1) its determinant is absurdly simple, and (2) when we multiply $A{X}_i$ the axis-columns will "pick out" columns of $A$ untouched, while the orange column becomes $A\mathbf{x}=\mathbf{b}$. It is engineered to swap exactly one column for $\mathbf{b}$.

PICTURE. The box of $X_2$: two edges are still unit axis-arrows; only the middle edge is tilted to $\mathbf{x}$.

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Step 4 — Why $\det X_i=x_i$ (read it off the picture)

WHAT. The determinant of $X_i$ equals a single coordinate of the unknown: $\det X_i=x_i.$

WHY. Look at the box in Step 3. Every edge except the $i$-th is a unit axis-arrow of length $1$, mutually perpendicular. They form a unit "floor." The orange column $\mathbf{x}$ is the only slanted edge; its height above that floor is precisely its $i$-th coordinate $x_i$ (the sideways parts of $\mathbf{x}$ lie flat inside the floor and add no volume). Volume $=$ floor-area $\times$ height $=1\times x_i=x_i$.

Term by term: $\det X_i=\underset{\text{unit floor}}{\underbrace{1}}\times \underset{\begin{array}{c}\text{height of the}\\ \text{orange column}\end{array}}{\underbrace{x_i}}=x_i.$

PICTURE. The orange edge decomposed: its "flat" part slides along the floor (no volume), only its perpendicular part $x_i$ sets the height.

Recall

Why do the other components of $\mathbf{x}$ not affect $\det X_i$? ::: They lie inside the unit floor spanned by the other axis-columns, so they only shear the box — shearing never changes volume; only the perpendicular height $x_i$ counts.

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Step 5 — The heart: $A{X}_i=A_i$

WHAT. Multiply $A$ by our helper $X_i$. The rule "$A$ times a matrix = $A$ applied to each column separately" means we hit each column of $X_i$ with $A$:

• A pure axis-column ${\mathbf{e}}_k$ satisfies $A{\mathbf{e}}_k={\mathbf{a}}_k$ — it simply selects column $k$ of $A$, unchanged.
• The orange column $\mathbf{x}$ gives $A\mathbf{x}=\mathbf{b}$ (our system!).

So the product is $A$ with every column left alone except the $i$-th, which has become $\mathbf{b}$. That is exactly the matrix the parent note calls $A_i$: $A{X}_i=[{\mathbf{a}}_1\cdots \underset{i\text{-th column}}{\underbrace{\mathbf{b}}}\cdots {\mathbf{a}}_n]=A_i.$

WHY this is the trick. We never had to hand-edit a column. Matrix multiplication did the swap for us, and it swapped a column (not a row) — because $A$ acts on columns. This is precisely why the common mistake of "replace a row" is wrong.

PICTURE. Three column-slots: the unchanged ${\mathbf{a}}_1,{\mathbf{a}}_3$ pass through; the middle slot's $\mathbf{x}$ is transformed into $\mathbf{b}$.

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Step 6 — Determinants multiply, so cancel to get the answer

WHAT. Take the determinant of both sides of $A{X}_i=A_i$, using the fact that determinants multiply, $\det (MN)=\det M\cdot \det N$: $\underset{\text{original box}}{\underbrace{\det A}}\cdot \underset{=x_i}{\underbrace{\det X_i}}=\underset{\text{swapped box}}{\underbrace{\det A_i}}.$ Substitute $\det X_i=x_i$ from Step 4: $\det A\cdot x_i=\det A_i.$

WHY divide. If (and only if) $\det A\ne 0$ we may divide both sides by it, isolating the one unknown: $\boxed{x_i=\frac{\det A_i}{\det A}}$ Read the picture as areas: $x_i$ is the factor by which the original box $\det A$ must grow to become the swapped box $\det A_i$. That growth factor is the unknown.

PICTURE. Two nested boxes: the original (area $\det A$) and the swapped one (area $\det A_i$); their ratio is the height $x_i$.

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Step 7 — The degenerate case: $\det A=0$

WHAT. Everything above needed the one forbidden move — dividing by $\det A$ — to be legal. When $\det A=0$ that step is illegal, and Cramer's rule does not apply.

WHY. $\det A=0$ means the column-box is flat: the columns ${\mathbf{a}}_k$ are squashed into a lower-dimensional space (a line, or a plane). Two sub-cases:

• If $\mathbf{b}$ lies inside that flat span, infinitely many mixes reach it — infinitely many solutions.
• If $\mathbf{b}$ pokes out of the flat span, no mix can ever reach it — no solution.

Either way there is no unique $x_i$ to read off, and indeed the formula tries to compute $\frac{\text{something}}{0}$. Use Gaussian Elimination instead; see also Invertible Matrix Theorem for why $\det A\ne 0$ is exactly the "unique solution" condition.

PICTURE. A collapsed (zero-area) column-box: the two column-arrows are parallel, so $\mathbf{b}$ off the line is unreachable.

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Step 8 — Sanity check: linearity in $\mathbf{b}$

WHAT. Predict what doubling the target $\mathbf{b}$ does, then verify.

WHY it must hold. In $A_i$, only the $i$-th column is $\mathbf{b}$; the determinant is linear in that column (multilinearity, Step 2). Doubling that one column doubles $\det A_i$, while $\det A$ (which has no $\mathbf{b}$ in it) is unchanged. So $x_i^{\text{new}}=\frac{\det (2\mathbf{b}\text{ swapped in})}{\det A}=\frac{2\det A_i}{\det A}=2x_i.$ Every unknown doubles — exactly what a linear system should do. This is a free correctness check you can run on any answer.

PICTURE. Same box, but the swapped edge $\mathbf{b}\to 2\mathbf{b}$ doubles the swapped area, hence doubles the height $x_i$.

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The one-picture summary

The whole derivation on a single canvas: build $X_i$, multiply by $A$ to swap a column into $\mathbf{b}$, take determinants (which multiply), and the ratio of the two boxes is the unknown.

Recall Feynman retelling — say it to a friend

You want to know how much banana goes into a smoothie. Picture your three ingredients (banana, apple, milk) as three arrows; the "box" they make has a certain size — that's $\det A$. Now build a trick container: leave apple and milk as plain unit measuring-cups, but pour the whole finished smoothie into the banana slot. When you run the machine ($A$) on this trick container, the plain cups just hand back the real ingredients untouched, and the smoothie-slot turns into the smoothie $\mathbf{b}$ — so you get a new box whose size is $\det A_i$. The magic (from Step 4) is that the trick container's own size is exactly the banana amount, and sizes multiply when you run the machine. So (size of banana) × (size of original box) = (size of swapped box), i.e. $x_i\cdot \det A=\det A_i$. Divide, and the banana count falls out. If the original box is flat — the ingredients were secretly copies of each other — there's no single recipe, and you can't divide by zero.

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Connections

• Determinants — every box in every figure is a determinant.
• Multilinear and Alternating Maps — the "scale a column / equal columns ⇒ 0" rules of Steps 2, 4, 8.
• Cofactor Expansion — how you actually compute each $\det A_i$.
• Matrix Inverse — the same $A{X}_i=A_i$ idea packaged as $A^{-1}=\frac{1}{\det A}\mathrm{adj}A$.
• Gaussian Elimination — what to do in the Step 7 degenerate case.
• Invertible Matrix Theorem — $\det A\ne 0⟺$ the unique-solution world where Cramer lives.