Cramer's rule: Worked examples — Maths deep dive

This page is the "throw everything at it" companion to Cramer's rule. The parent note built the formula $x_{i} = \frac{det A _{i}}{det A}$ and why it works. Here we stress-test it against every kind of input you can meet: nice numbers, negative and zero coefficients, a system where $det A = 0$ (so the rule refuses to run), a limiting near-singular case, a geometry problem, a word problem, and an exam twist with a letter parameter.

The scenario matrix

Every cell here is covered by at least one worked example below. If a case can bite you, it appears.

Cell	What makes it special	Covered by
A. Clean 2×2, positive	all entries positive, $det A > 0$	Example 1
B. Mixed signs 2×2	negative coefficients, negative $det A$	Example 2
C. Zero entries / basis-like	a coefficient is $0$ (a column axis-aligned)	Example 3
D. Degenerate: $det A = 0$ (no solution)	rule cannot run — inconsistent	Example 4
D'. Degenerate: $det A = 0$ (infinitely many)	rule cannot run — $0/0$	Example 4b
E. Limiting / near-singular	$det A \to 0^{+}$ , watch $x_{i}$ blow up	Example 5
F. 3×3 full	three unknowns, three column swaps	Example 6
G. Geometry	intersection of two lines = a 2×2 system	Example 7
H. Word problem (units)	real quantities, sanity-check units	Example 8
I. Exam twist (parameter)	a letter $k$ in $A$ ; when does it break?	Example 9

Prerequisites you may want open: Determinants, Cofactor Expansion, and for the degenerate cell, Gaussian Elimination and the Invertible Matrix Theorem.

Example 1 — Cell A: clean positive 2×2

Worked example Cell A

Solve ${3 x + 2 y = 12 x + 4 y = 14$ .

Forecast: all numbers are positive and small — guess whether $x$ and $y$ come out as tidy whole numbers before you compute.

Step 1 — Write the box. $A = (3124)$ , $b = (1214)$ . Why this step? The columns of $A$ are the two "ingredients"; $b$ is the target. See the box in the figure.

Step 2 — Denominator $det A$ . $det A = (3) (4) - (2) (1) = 10$ . Why this step? This is the area of the original ingredient-box. It is $\neq = 0$ , so a unique recipe exists — the rule is allowed to run.

Step 3 — Swap column 1 for $b$ . $A_{1} = (121424)$ , $det A_{1} = (12) (4) - (2) (14) = 48 - 28 = 20$ . So $x = \frac{20}{10} = 2$ . Why this step? Covering the $x$ -column with $b$ measures "how much of column 1" builds $b$ .

Step 4 — Swap column 2 for $b$ . $A_{2} = (311214)$ , $det A_{2} = (3) (14) - (12) (1) = 42 - 12 = 30$ . So $y = \frac{30}{10} = 3$ . Why this step? Now we cover the $y$ -column (column 2) with $b$ ; that new box's area over the original box's area is exactly the amount of column 2 needed, i.e. $y$ .

Verify: $3 (2) + 2 (3) = 6 + 6 = 12$ ✓, $2 + 4 (3) = 2 + 12 = 14$ ✓.

Example 2 — Cell B: mixed signs

Worked example Cell B

Solve ${2 x - 3 y = 13 - x + 5 y = - 8$ .

Forecast: the coefficients mix positive and negative. Will the negative signs in the numerator produce clean fractions?

Step 1 — Box. $A = (2 - 1 - 3 5)$ , $b = (13 - 8)$ . Why this step? Same setup; the only difference is signs, which the determinant handles automatically.

Step 2 — $det A$ . $det A = (2) (5) - (- 3) (- 1) = 10 - 3 = 7$ . Why this step? $det A = 7 \neq = 0$ — the box has genuine area, so the rule runs. (Note: a determinant of a mixed-sign matrix can still come out positive; here it does.)

Step 3 — $x$ . $A_{1} = (13 - 8 - 3 5)$ , $det A_{1} = (13) (5) - (- 3) (- 8) = 65 - 24 = 41$ . So $x = \frac{41}{7}$ . Why this step? Cover column 1 (the $x$ -column) with $b$ ; the ratio of this box to the original box is $x$ . Nothing forces integers — a fraction is a perfectly good answer.

Step 4 — $y$ . $A_{2} = (2 - 1 13 - 8)$ , $det A_{2} = (2) (- 8) - (13) (- 1) = - 16 + 13 = - 3$ . So $y = \frac{- 3}{7}$ . Why this step? Cover column 2 (the $y$ -column) with $b$ ; its box area over the original box area gives $y$ . The negative result just means the box orientation flipped.

Verify: $2 (\frac{41}{7}) - 3 (- \frac{3}{7}) = \frac{82 + 9}{7} = \frac{91}{7} = 13$ ✓, and $- (\frac{41}{7}) + 5 (- \frac{3}{7}) = \frac{- 41 - 15}{7} = \frac{- 56}{7} = - 8$ ✓.

Example 3 — Cell C: a zero entry (axis-aligned column)

Worked example Cell C

Solve ${4 x + 0 \cdot y = 12 3 x + 2 y = 5$ .

Forecast: column 2 is $(02)$ — it points straight up the $y$ -axis. That should make the box a clean rectangle. Guess $x$ from the first equation alone.

Step 1 — Box. $A = (4302)$ , $b = (125)$ . Why this step? The zero means column 1 and column 2 are "less tangled"; the figure shows one column lying flat and one standing up.

Step 2 — $det A$ . $det A = (4) (2) - (0) (3) = 8$ . Why this step? A zero kills one product; the box area is just $4 \times 2 = 8$ (base times height of the rectangle-ish box).

Step 3 — $x$ . $A_{1} = (12502)$ , $det A_{1} = (12) (2) - (0) (5) = 24$ . So $x = \frac{24}{8} = 3$ . Why this step? Cover the $x$ -column with $b$ ; the box ratio confirms the forecast: $4 x = 12 \Rightarrow x = 3$ .

Step 4 — $y$ . $A_{2} = (43125)$ , $det A_{2} = (4) (5) - (12) (3) = 20 - 36 = - 16$ . So $y = \frac{- 16}{8} = - 2$ . Why this step? Cover the $y$ -column (column 2) with $b$ ; the ratio of this new box to the original box gives $y$ , and the negative sign records the flipped orientation.

Verify: $4 (3) = 12$ ✓, $3 (3) + 2 (- 2) = 9 - 4 = 5$ ✓.

Example 4 — Cell D: degenerate, $det A = 0$ (no solution)

Worked example Cell D (the rule refuses)

Try to solve ${x + 2 y = 3 2 x + 4 y = 10$ .

Forecast: row 2 is almost twice row 1. Is that a coincidence, or a warning sign?

Step 1 — $det A$ . $A = (1224)$ , $det A = (1) (4) - (2) (2) = 4 - 4 = 0$ . Why this step? You always compute the denominator first. Here it is $0$ — Cramer's rule is not allowed to run.

Step 2 — Read the geometry. The two columns $(12)$ and $(24)$ are parallel (one is twice the other), so the box is flat — zero area. The figure shows both column-arrows lying on the same line. Why this step? Zero volume means the columns can't reach every $b$ : either no recipe or infinitely many.

Step 3 — Diagnose which. Equation 1 doubled is $2 x + 4 y = 6$ , but equation 2 says $2 x + 4 y = 10$ . Since $6 \neq = 10$ , the two equations contradict — there is no solution. Why this step? By the Invertible Matrix Theorem, $det A = 0$ means non-unique; we use elimination (Gaussian Elimination) to see it's the "no solution" branch.

Verify: $det A_{1} = det (31024) = 12 - 20 = - 8 \neq = 0$ . A nonzero numerator over a zero denominator confirms inconsistent (blow-up, not $0/0$ ). Cramer correctly gives no finite answer.

Example 4b — Cell D': degenerate, $det A = 0$ (infinitely many)

Worked example Cell D' (the other singular branch)

Try to solve ${x + 2 y = 3 2 x + 4 y = 6$ .

Forecast: same left-hand side as Example 4, but now the right-hand side of equation 2 is exactly double equation 1. Guess whether this is "no solution" or "too many solutions".

Step 1 — $det A$ . $A = (1224)$ , $det A = (1) (4) - (2) (2) = 0$ . Why this step? Same flat box as before — the denominator is $0$ , so Cramer still cannot produce a single number.

Step 2 — Compute the numerators. $det A_{1} = det (3624) = (3) (4) - (2) (6) = 12 - 12 = 0$ , and $det A_{2} = det (1236) = (1) (6) - (3) (2) = 6 - 6 = 0$ . Why this step? When both numerators also vanish, the Cramer ratios are $0/0$ — undefined, but signalling the opposite of Example 4.

Step 3 — Diagnose. Equation 2 is exactly $2 \times$ equation 1, so it adds no new information. Every point on the line $x + 2 y = 3$ solves the system — infinitely many solutions. Why this step? $0/0$ (not $nonzero /0$ ) is the fingerprint of the infinitely-many branch; use Gaussian Elimination to read off the free variable.

Verify: pick $y = 0 \Rightarrow x = 3$ : check $2 (3) + 4 (0) = 6$ ✓. Pick $y = 1 \Rightarrow x = 1$ : check $2 (1) + 4 (1) = 6$ ✓. Two different valid solutions confirm "infinitely many".

Example 5 — Cell E: near-singular limit

Worked example Cell E (watch it blow up)

Solve ${x + y = 1 x + (1 + ε) y = 0$ for small $ε > 0$ , then let $ε \to 0^{+}$ .

Forecast: as $ε \to 0$ the two equations become nearly identical. Guess whether $x, y$ stay bounded or explode.

Step 1 — $det A$ . $A = (11 1 1 + ε)$ , $det A = (1) (1 + ε) - (1) (1) = ε$ . Why this step? This tiny denominator $ε$ is the whole story: dividing by something near zero magnifies everything.

Step 2 — $x$ . $A_{1} = (10 1 1 + ε)$ , $det A_{1} = (1) (1 + ε) - (1) (0) = 1 + ε$ . So $x = \frac{1 + ε}{ε}$ . Why this step? Cover column 1 with $b = (1, 0)^{⊤}$ ; the box ratio gives $x$ . Notice the numerator stays near $1$ while the denominator shrinks — that is what forces the blow-up.

Step 3 — $y$ . $A_{2} = (1110)$ , $det A_{2} = (1) (0) - (1) (1) = - 1$ . So $y = \frac{- 1}{ε}$ . Why this step? Cover column 2 with $b$ ; here the numerator is a fixed $- 1$ , so dividing by the tiny $ε$ sends $y$ to $- \infty$ .

Step 4 — Take the limit. As $ε \to 0^{+}$ : $x = \frac{1 + ε}{ε} \to + \infty$ and $y = \frac{- 1}{ε} \to - \infty$ . Why this step? This is the numerical-instability warning from the parent note made concrete: a small change in $ε$ swings the answer wildly.

Verify (at $ε = 0.1$ ): $x = 1.1/0.1 = 11$ , $y = - 1/0.1 = - 10$ . Check: $11 + (- 10) = 1$ ✓, $11 + 1.1 (- 10) = 11 - 11 = 0$ ✓. Huge numbers, tiny denominator — exactly the instability.

Example 6 — Cell F: full 3×3

Worked example Cell F

Solve $⎩ ⎨ ⎧ x + 2 y + z = 4 2 x + 5 y + z = 3 - x + y + 3 z = 5$ .

Forecast: three unknowns need four determinants (one denominator, three numerators). Guess a rough sign for $det A$ from the mostly-positive entries.

Step 1 — $det A$ by Cofactor Expansion along row 1. $A = 12 - 1 251113$ . $det A = 1 14 (5 \cdot 3 - 1 \cdot 1) - 2 7 (2 \cdot 3 - 1 \cdot (- 1)) + 1 7 (2 \cdot 1 - 5 \cdot (- 1)) = 14 - 14 + 7 = 7.$ Why this step? Expansion signs alternate $+, -, +$ ; that alternation is the same "alternating" property that makes swapping columns flip sign.

Step 2 — $x$ : cover column 1 with $b = (4, 3, 5)^{⊤}$ . $A_{1} = 435251113$ . $det A_{1} = 4 (15 - 1) - 2 (9 - 5) + 1 (3 - 25) = 4 (14) - 2 (4) + (- 22) = 56 - 8 - 22 = 26$ . So $x = \frac{26}{7}$ . Why this step? Replacing the $x$ -column with $b$ builds the box whose volume, divided by the original volume $det A$ , is exactly $x$ — the 3-D version of "cover the column".

Step 3 — $y$ : cover column 2. $A_{2} = 12 - 1 435113$ . $det A_{2} = 1 (9 - 5) - 4 (6 + 1) + 1 (10 + 3) = 4 - 28 + 13 = - 11$ . So $y = \frac{- 11}{7}$ . Why this step? Now the $y$ -column carries $b$ ; the signed-volume ratio gives $y$ , and the negative sign records that this swapped box has opposite orientation.

Step 4 — $z$ : cover column 3. $A_{3} = 12 - 1 251435$ . $det A_{3} = 1 (25 - 3) - 2 (10 + 3) + 4 (2 + 5) = 22 - 26 + 28 = 24$ . So $z = \frac{24}{7}$ . Why this step? Last column swap: putting $b$ in the $z$ -slot and dividing its box volume by $det A$ yields $z$ , completing the recipe.

Verify: $x + 2 y + z = \frac{26 - 22 + 24}{7} = \frac{28}{7} = 4$ ✓, $2 x + 5 y + z = \frac{52 - 55 + 24}{7} = \frac{21}{7} = 3$ ✓, $- x + y + 3 z = \frac{- 26 - 11 + 72}{7} = \frac{35}{7} = 5$ ✓.

Example 7 — Cell G: geometry (two lines meeting)

Worked example Cell G

Find where the lines $L_{1} : 2 x + y = 4$ and $L_{2} : x - y = - 1$ cross.

Forecast: two non-parallel lines meet at exactly one point. Sketch them and estimate the crossing before computing.

Step 1 — This is a 2×2 system. $A = (21 1 - 1)$ , $b = (4 - 1)$ . Why this step? "Intersection point" is the simultaneous solution — same object, different words.

Step 2 — $det A$ . $det A = (2) (- 1) - (1) (1) = - 2 - 1 = - 3$ . Why this step? $\neq = 0$ means the lines are not parallel, so exactly one crossing exists.

Step 3 — $x$ . $A_{1} = (4 - 1 1 - 1)$ , $det A_{1} = (4) (- 1) - (1) (- 1) = - 4 + 1 = - 3$ . So $x = \frac{- 3}{- 3} = 1$ . Why this step? Cover the $x$ -column with $b$ ; the ratio of this box to $det A$ gives the $x$ -coordinate of the crossing point.

Step 4 — $y$ . $A_{2} = (21 4 - 1)$ , $det A_{2} = (2) (- 1) - (4) (1) = - 2 - 4 = - 6$ . So $y = \frac{- 6}{- 3} = 2$ . Why this step? Cover the $y$ -column with $b$ ; dividing that box by $det A$ gives the $y$ -coordinate — together $(x, y)$ is the intersection.

Verify: point $(1, 2)$ : $2 (1) + 2 = 4$ ✓, $1 - 2 = - 1$ ✓ — it lies on both lines, matching the figure's red dot.

Example 8 — Cell H: word problem with units

Worked example Cell H

A café sells two combos. Combo A = 2 coffees + 1 muffin costs ₹230. Combo B = 1 coffee + 3 muffins costs ₹290. Find the price of one coffee ( $c$ ) and one muffin ( $m$ ).

Forecast: coffees usually cost more than muffins near ₹100 each — guess before solving.

Step 1 — Model. ${2 c + 1 m = 230 1 c + 3 m = 290$ , so $A = (2113)$ , $b = (230290)$ . Why this step? Each column is "how many of that item per combo"; $b$ is rupees. Units of $c, m$ come out in ₹.

Step 2 — $det A$ . $(2) (3) - (1) (1) = 6 - 1 = 5$ . Why this step? Nonzero ⇒ the two combos give genuinely independent info, so prices are pinned down uniquely.

Step 3 — $c$ . $A_{1} = (23029013)$ , $det A_{1} = (230) (3) - (1) (290) = 690 - 290 = 400$ . So $c = \frac{400}{5} = 80$ (₹). Why this step? Cover the coffee-column (column 1) with the rupee-vector $b$ ; the box ratio isolates the coffee price, since $c$ is "how much of column 1" builds $b$ .

Step 4 — $m$ . $A_{2} = (21230290)$ , $det A_{2} = (2) (290) - (230) (1) = 580 - 230 = 350$ . So $m = \frac{350}{5} = 70$ (₹). Why this step? Cover the muffin-column (column 2) with $b$ ; dividing that box by $det A$ gives the muffin price in ₹.

Verify (units + values): Combo A $= 2 (80) + 70 = 160 + 70 = 230$ ✓ ₹, Combo B $= 80 + 3 (70) = 80 + 210 = 290$ ✓ ₹. Both answers positive and sensibly priced.

Example 9 — Cell I: exam twist with a parameter

Worked example Cell I

For which value(s) of $k$ does ${k x + 2 y = 4 3 x + k y = 6$ fail to have a unique solution? Solve it for $k = 4$ .

Forecast: a unique solution exists exactly when $det A \neq = 0$ . The "fail" cases are the roots of $det A = 0$ . Guess how many such $k$ there are.

Step 1 — $det A$ as a function of $k$ . $A = (k 3 2 k)$ , $det A = k \cdot k - (2) (3) = k^{2} - 6$ . Why this step? The denominator now depends on $k$ ; wherever it vanishes, the Invertible Matrix Theorem says no unique solution.

Step 2 — Find the bad $k$ . $k^{2} - 6 = 0 \Rightarrow k = \pm 6$ . Why this step? These are exactly the values that flatten the column-box to zero volume — Cramer stalls there.

Step 3 — Solve at $k = 4$ (safe, since $16 - 6 = 10 \neq = 0$ ). $det A = 10$ . $A_{1} = (4624)$ , $det A_{1} = 16 - 12 = 4$ , so $x = \frac{4}{10} = \frac{2}{5}$ . $A_{2} = (4346)$ , $det A_{2} = 24 - 12 = 12$ , so $y = \frac{12}{10} = \frac{6}{5}$ . Why this step? With $k = 4$ the box is non-flat, so we may run the rule: cover column 1 with $b$ and divide by $det A$ for $x$ , cover column 2 with $b$ and divide for $y$ — the ratios give the actual solution.

Verify: $4 (\frac{2}{5}) + 2 (\frac{6}{5}) = \frac{8 + 12}{5} = \frac{20}{5} = 4$ ✓, $3 (\frac{2}{5}) + 4 (\frac{6}{5}) = \frac{6 + 24}{5} = \frac{30}{5} = 6$ ✓.

Recall Quick self-check

Which cell does a system of two identical equations $x + y = 1, x + y = 1$ fall into, and what does Cramer say? ::: Cell D' — $det A = 0$ and every numerator is also $0$ , so it's the $0/0$ / infinitely-many-solutions branch (contrast Example 4's nonzero-over-zero contradiction).

Recap of the scenario coverage

Connections

Cramer's rule — the parent formula and its proof.
Determinants — the volume engine behind every denominator.
Cofactor Expansion — used to compute the 3×3 in Example 6.
Gaussian Elimination — the tool to reach for in Cells D and D'.
Invertible Matrix Theorem — $det A \neq = 0 ⟺$ unique solution (Cells D, D', I).
Matrix Inverse — Cramer is the inverse via adjugate applied to $b$ .
Multilinear and Alternating Maps — why swapping a column behaves as it does.