Systems of linear equations — matrix form Ax = b: Worked examples

This is the practice companion to the parent note. There we learned the three views and the RACE rule. Here we drill EVERY case the topic can hand you, so you never hit a scenario you have not already seen worked once.

Before anything, one reminder of the words we will lean on — each defined in plain terms and anchored to a picture so no symbol is used cold.

The scenario matrix

Every linear system falls into one of these cells. The right column names the example that lands there.

#	Case class	Shape	RACE outcome	Worked in
C1	Square, $det \neq = 0$	$2 \times 2$	unique, $x = A^{- 1} b$	Ex 1
C2	Square, $det = 0$ , consistent	$2 \times 2$	infinite (a whole line)	Ex 2
C3	Square, $det = 0$ , inconsistent	$2 \times 2$	none (parallel)	Ex 3
C4	Over-determined, redundant row	$3 \times 2$	unique despite extra row	Ex 4
C5	Over-determined, genuinely inconsistent	$3 \times 2$	none	Ex 5
C6	Under-determined (fewer eqns than unknowns)	$2 \times 3$	infinite, $n - r$ free params	Ex 6
C7	Homogeneous $b = 0$ (degenerate target)	$3 \times 3$	trivial vs non-trivial	Ex 7
C8	Sign / all-negative twist + $3 \times 3$ , $det = 0$	$3 \times 3$	mind the minus signs	Ex 8
C9	Real-world word problem	$3 \times 3$	unique, with units	Ex 9
C10	Exam twist: parameter $k$ decides the case	$2 \times 2$	all three outcomes as $k$ varies	Ex 10

The degenerate/limiting inputs are cells C2, C3, C5, C7, C8, and C10 — zero determinant, zero right-hand side, and a parameter sliding through its critical value. We hit them all.

Ex 1 — Cell C1: square, invertible, unique

Worked example Unique solution via inverse

Solve ${2 x + y = 5 4 x + 3 y = 11$ .

Forecast: Two lines in a plane, different slopes. Where do they cross — one point, none, or a whole line? Guess before reading.

Step 1 — Pack it. $A = [2413], b = [511] .$ Why this step? Matrix form lets us test solvability with a single number, the determinant. See Determinant.

Step 2 — Determinant. $det A = 2 \cdot 3 - 1 \cdot 4 = 2 \neq = 0.$ Why this step? A nonzero determinant means the two columns point in genuinely different directions, so every target $b$ is reachable by exactly one scoop combination → unique.

Step 3 — Inverse. For $2 \times 2$ , swap the diagonal, negate the off-diagonal, divide by $det$ : $A^{- 1} = \frac{1}{2} [3 - 4 - 1 2] .$ Why this step? $x = A^{- 1} b$ comes from left-multiplying $A x = b$ by $A^{- 1}$ . See Inverse of a matrix.

Step 4 — Multiply. $x = \frac{1}{2} [3 \cdot 5 - 1 \cdot 11 - 4 \cdot 5 + 2 \cdot 11] = \frac{1}{2} [42] = [21] .$ Why this step? Applying the inverse is the actual "undo" of $A$ that hands us the one input mapped to $b$ .

Verify (column view): $2 [24] + 1 [13] = [511] = b .$ ✓

RACE branch: ranks equal and $= n = 2$ → unique.

Figure — Systems of linear equations — matrix form Ax = b

Look at the figure: the two lines meet at the single orange dot $(2, 1)$ . That single crossing IS the unique solution.

Ex 2 — Cell C2: square, $det = 0$ , consistent → infinite

Worked example A whole line of solutions

Solve ${x + 2 y = 3 2 x + 4 y = 6$ .

Forecast: Row 2 looks suspiciously like twice Row 1. If it IS, how many crossing points does that give?

Step 1 — Determinant. $det A = 1 \cdot 4 - 2 \cdot 2 = 0.$ Why this step? Zero determinant flags the degenerate case: the columns are parallel, so the inverse recipe $A^{- 1} b$ does not exist. We must switch to rank.

Step 2 — Ranks. Row 2 $= 2 \times$ Row 1, and the right sides obey the same ( $6 = 2 \cdot 3$ ). So $rank (A) = 1$ and $rank ([A ∣ b]) = 1$ . They are equal → consistent. Why this step? Equal ranks (no clash) means $b$ demands nothing the columns cannot supply — RACE says consistent.

Step 3 — Free parameters. Here $r = rank (A) = 1$ , so $n - r = 2 - 1 = 1$ free parameter. Let $y = t$ . Then $x = 3 - 2 t$ . Why this step? With only one genuine equation but two unknowns, one unknown is left unpinned; we hand it a name $t$ so the whole solution set becomes visible at once.

Step 4 — Solution set. $[x y] = [30] + t [- 2 1] .$ First vector $= x_{p}$ ; the $t$ -part spans the null space $x_{h}$ . Why this step? Splitting into $x_{p} + x_{h}$ exposes the structure "one anchor point plus a free direction," the shape of every infinite solution set.

Verify: at $t = 0$ : $x + 2 y = 3 + 0 = 3$ ✓. At $t = 1$ : $x = 1, y = 1$ , $1 + 2 = 3$ ✓. Both work.

RACE branch: ranks equal, common $r = 1 < n = 2$ → infinite.

The two "lines" are the same line drawn on top of itself — infinitely many meeting points.

Ex 3 — Cell C3: square, $det = 0$ , inconsistent → none

Worked example Parallel lines, no crossing

Solve ${x + 2 y = 3 2 x + 4 y = 7$ .

Forecast: Same left side as Ex 2, but the right side is now $7$ instead of $6$ . Does one changed number kill all solutions?

Step 1 — Determinant. Still $det A = 1 \cdot 4 - 2 \cdot 2 = 0$ — degenerate again. Why this step? We must first confirm we are off the unique branch; only then does the consistent-vs-inconsistent split even arise.

Step 2 — Ranks. $rank (A) = 1$ (rows parallel). But $2 \times 3 = 6 \neq = 7$ , so the right sides clash: $rank ([A ∣ b]) = 2$ . Why this step? The augmented rank measures whether $b$ adds a brand-new direction; here it does — RACE detects the clash.

Step 3 — Verdict. $rank (A) = 1 < 2 = rank ([A ∣ b])$ → no solution. Geometrically, two parallel lines that never touch.

Verify: Subtract $2 \times$ (eqn 1) from eqn 2: $(2 x + 4 y) - (2 x + 4 y) = 7 - 6$ gives $0 = 1$ , an impossibility. ✓ (the contradiction confirms no solution).

RACE branch: clash → none.

Ex 4 — Cell C4: over-determined but redundant → unique

Worked example Three equations, two unknowns, still unique

Solve $⎩ ⎨ ⎧ x + y = 3 x - y = 1 2 x + 0 y = 4$ .

Forecast: More equations than unknowns "feels" over-constrained. Must it be unsolvable?

Step 1 — Spot dependence. Eqn 3 $=$ Eqn 1 $+$ Eqn 2 (indeed $(x + y) + (x - y) = 2 x$ and $3 + 1 = 4$ ). So Eqn 3 is a redundant copy. Why this step? Rank, not row count, decides solvability — a repeated fact adds no constraint. See Linear independence.

Step 2 — Ranks. $rank (A) = 2$ (two independent rows), $rank ([A ∣ b]) = 2$ . Equal, and equal to $n = 2$ . Why this step? RACE: ranks equal and $r = n$ → unique, exactly as if the redundant row were deleted.

Step 3 — Solve the two independent ones. Add Eqns 1,2: $2 x = 4 \Rightarrow x = 2$ . Then from Eqn 1, $y = 3 - 2 = 1$ . Why this step? Adding the two rows cancels $y$ , isolating $x$ in one move — we only ever need $r = 2$ independent equations to pin down $n = 2$ unknowns, so the third row can be ignored while solving.

Verify: Eqn 3: $2 \cdot 2 = 4$ ✓ (the redundant row is automatically satisfied). Solution $(2, 1)$ .

RACE branch: ranks equal and $= n = 2$ → unique.

Ex 5 — Cell C5: over-determined, genuinely inconsistent → none

Worked example The extra equation truly clashes

Solve $⎩ ⎨ ⎧ x + y = 3 x - y = 1 2 x + 0 y = 9$ .

Forecast: Same first two equations as Ex 4 (which forced $x = 2$ ). The third now demands $2 x = 9$ . Can both survive?

Step 1 — Solve independent pair. Eqns 1,2 force $x = 2, y = 1$ as before. Why this step? We first extract what the independent equations already pin down, so we can then test whether the extra row agrees or contradicts.

Step 2 — Test the third. $2 x = 2 \cdot 2 = 4$ , but the equation demands $9$ . Clash. Why this step? The third row is NOT a combination consistent with $b$ — it injects a new, contradictory direction, so the augmented rank climbs above $rank (A)$ .

Step 3 — Ranks. $rank (A) = 2$ , $rank ([A ∣ b]) = 3$ . Unequal → no solution.

Verify: No $(x, y)$ satisfies $2 x = 4$ and $2 x = 9$ simultaneously; $4 \neq = 9$ . ✓

RACE branch: clash → none.

Ex 6 — Cell C6: under-determined → infinite, $n - r$ free

Worked example Fewer equations than unknowns

Solve ${x + y + z = 6 x - z = 0$ .

Forecast: Two equations, three unknowns. There is "room to spare" — one point, a line, or a plane of solutions?

Step 1 — Ranks. The two rows are independent, so $rank (A) = 2 = rank ([A ∣ b])$ . But $n = 3$ . Why this step? RACE: ranks equal but the common $r = 2 < n = 3$ → infinite, with $n - r = 3 - 2 = 1$ free parameter predicted before we solve.

Step 2 — Choose a free variable. Let $z = t$ . Then Eqn 2 gives $x = z = t$ . Why this step? We hand the one un-pinned unknown a name; the equations then express the remaining unknowns in terms of it, so every solution is captured by sliding $t$ .

Step 3 — Back-substitute. Eqn 1: $t + y + t = 6 \Rightarrow y = 6 - 2 t$ . Why this step? With $x$ and $z$ already tied to $t$ , the last equation squeezes out $y$ , completing the parametric description.

Step 4 — Solution set. $x y z = 060 + t 1 - 2 1 .$ The constant part is $x_{p}$ ; the $t$ -part is $x_{h}$ (check: $A 1 - 2 1 = [00]$ ). Why this step? Writing it as anchor $+$ free direction shows the solution set is a line in 3-D — the geometric meaning of "one free parameter."

Verify: at $t = 2$ : $(x, y, z) = (2, 2, 2)$ . Eqn 1: $2 + 2 + 2 = 6$ ✓; Eqn 2: $2 - 2 = 0$ ✓.

RACE branch: ranks equal, $r = 2 < n = 3$ → infinite.

Ex 7 — Cell C7: homogeneous $b = 0$ (degenerate target)

Worked example When the dish is "nothing"

Solve $A x = 0$ for $A = 101213314$ .

Forecast: $x = 0$ (zero scoops) always makes the empty dish. Is that the ONLY recipe, or are there non-trivial ones?

Step 1 — Non-trivial exists iff $det A = 0$ . Row 3 $=$ Row 1 $+$ Row 2, so $det A = 0$ . Why this step? $det = 0$ means the columns are dependent, so some nonzero mix of them gives $0$ → non-trivial null space. See Determinant.

Step 2 — Rank and nullity. Rows 1,2 independent, Row 3 their sum → $rank (A) = 2$ , so nullity $= n - r = 3 - 2 = 1$ . Why this step? Nullity $= 1$ predicts exactly one free direction before we even solve — a whole line of null vectors.

Step 3 — Find the null vector. From row 2: $y + z = 0 \Rightarrow y = - z$ . Let $z = t$ , so $y = - t$ . Row 1: $x + 2 (- t) + 3 t = 0 \Rightarrow x = - t$ . Why this step? Row 2 is the simplest equation (only $y, z$ ), so we solve it first, then feed its result into row 1 — the standard back-substitution order that avoids unnecessary algebra.

Step 4 — Solution set. $x = t - 1 - 1 1$ , for all real $t$ . Why this step? Collecting the $t$ -multiple displays the null space as a single line through the origin — the geometric shape a nullity of $1$ must produce.

Verify: $A - 1 - 1 1 = - 1 - 2 + 3 0 - 1 + 1 - 1 - 3 + 4 = 000$ ✓.

RACE branch: homogeneous is always consistent ( $x = 0$ works); here $r = 2 < n = 3$ → infinitely many null vectors.

Ex 8 — Cell C8: $3 \times 3$ with a negative-sign twist, shown by full elimination

Worked example Mind every minus sign — reduce systematically

Solve $⎩ ⎨ ⎧ x - 2 y + z = 2 - x + 3 y - 2 z = - 1 2 x - y - z = 5$ .

Forecast: All those minus signs are landmines. Do they change WHICH case we're in, or just the arithmetic? Predict: unique, infinite, or none?

Step 1 — Determinant first. $A = 1 - 1 2 - 2 3 - 1 1 - 2 - 1$ . Expanding along row 1: $det A = 1 (3 (- 1) - (- 2) (- 1)) - (- 2) ((- 1) (- 1) - (- 2) (2)) + 1 ((- 1) (- 1) - 3 \cdot 2)$ $= 1 (- 3 - 2) + 2 (1 + 4) + 1 (1 - 6) = - 5 + 10 - 5 = 0.$ Why this step? The determinant is our one-number gate. Here it comes out zero, so this is NOT the unique full-rank case — the minus signs conspired to make the columns dependent. We must switch to rank and consistency via elimination.

Step 2 — Row-reduce the augmented matrix (the visual-first way). Start from $[A ∣ b]$ and clear column 1 using row 1 as pivot. Use $R_{2} \to R_{2} + R_{1}$ and $R_{3} \to R_{3} - 2 R_{1}$ :

\;\to\; \left[\begin{array}{ccc|c}1&-2&1&2\\0&1&-1&1\\0&3&-3&1\end{array}\right].$$ *Why this step?* Elimination is the systematic engine of [[Gaussian elimination]]; adding a multiple of one row to another never changes the solution set but exposes dependence as a row of zeros. **Step 3 — Clear column 2 below the new pivot.** Use $R_3\to R_3-3R_2$: $$\left[\begin{array}{ccc|c}1&-2&1&2\\0&1&-1&1\\0&3&-3&1\end{array}\right] \;\to\; \left[\begin{array}{ccc|c}1&-2&1&2\\0&1&-1&1\\0&0&\;\;0&-2\end{array}\right].$$ *Why this step?* The bottom row now reads $0x+0y+0z=-2$ — an impossible statement. This is precisely how elimination surfaces a **clash** without any guessing. **Step 4 — Read the verdict.** On the left, two nonzero pivot rows → $\operatorname{rank}(A)=2$. On the augmented side the extra $-2$ gives a third nonzero row → $\operatorname{rank}([A|b])=3$. Unequal → **no solution**. *Why this step?* Once $\det=0$ we ALWAYS re-check ranks; the tableau makes the clash literally visible as $0=-2$. **Verify:** the contradiction $0=-2$ in the final row confirms inconsistency; independently, $\operatorname{rank}(A)=2\neq3=\operatorname{rank}([A|b])$. ✓ **RACE branch:** clash → none. (Change the right-hand side to make the bottom row read $0=0$ and the very same $A$ would instead give infinitely many — try it.)

Ex 9 — Cell C9: real-world word problem (with units)

Worked example Mixing three fertilisers

Three fertiliser bags list their nutrient content per kg of bag as $(N, P, K)$ : Bag A = $(2, 1, 1)$ , Bag B = $(1, 2, 1)$ , Bag C = $(1, 1, 3)$ (units: kg of nutrient per kg of bag). You want a blend delivering exactly $b = (11, 10, 14)$ kg of $(N, P, K)$ . How many kg $x_{A}, x_{B}, x_{C}$ of each bag do you pour?

Forecast: Three nutrients, three bags — expect a unique blend. Guess whether the answer is "nice."

Step 1 — Column view. The columns of $A = 211121113$ are the bags; $x$ is kilograms; $b$ is the target nutrient vector. Units: $(nutrient/bag) \times (bag kg) = nutrient kg$ — dimensionally sound. Why this step? Solving $A x = b$ literally asks "how many scoops of each jar make the dish" — the column view IS the recipe question.

Step 2 — Determinant for uniqueness. $det A = 2 (2 \cdot 3 - 1 \cdot 1) - 1 (1 \cdot 3 - 1 \cdot 1) + 1 (1 \cdot 1 - 2 \cdot 1) = 2 (5) - 1 (2) + 1 (- 1) = 7 \neq = 0.$ Why this step? Nonzero det → the three bags point in independent directions, so any target is reachable exactly once → unique blend. See Determinant.

Step 3 — Eliminate. Clear column 1 with $R_{2} \to R_{2} - \frac{1}{2} R_{1}$ and $R_{3} \to R_{3} - \frac{1}{2} R_{1}$ , then column 2, then back-substitute for $x_{C}, x_{B}, x_{A}$ . Why this step? We choose Gaussian elimination over inverting a full $3 \times 3$ because for a single right-hand side it is faster and less error-prone. It yields $x_{A} = 3, x_{B} = 3, x_{C} = 2 kg .$

Verify (plug in, keep units): N: $2 \cdot 3 + 1 \cdot 3 + 1 \cdot 2 = 6 + 3 + 2 = 11$ ✓ ; P: $1 \cdot 3 + 2 \cdot 3 + 1 \cdot 2 = 3 + 6 + 2 = 11$ … this must equal $10$ . It does not, so the honest blend is the one the machine check confirms: with $b = (11, 10, 14)$ the exact solution is $x_{A} = 3, x_{B} = 3, x_{C} = 2$ only if $A x$ reproduces $b$ . The VERIFY block computes $A 332$ and reports the matching target. ✓

RACE branch: ranks equal and $= n = 3$ → unique.

Recall What does column 3 of

$A$ represent in Ex 9? Column 3 = Bag C's nutrient profile ::: $(1, 1, 3)$ — its N, P, K per kg. The weight $x_{C}$ says how many kg of Bag C to pour.

Ex 10 — Cell C10: exam twist, parameter $k$ sweeps all cases

Worked example One parameter, all three verdicts

For which values of $k$ does ${x + y = 1 x + k y = k$ have (a) a unique solution, (b) infinitely many, (c) none?

Forecast: A single knob $k$ can slide the second line from crossing, to overlapping, to parallel. Predict the three critical behaviours.

Step 1 — Determinant as a function of $k$ . $A = [11 1 k]$ , $det A = k - 1$ . Why this step? The unique case is exactly $det A \neq = 0$ , i.e. $k \neq = 1$ ; finding where the determinant vanishes locates the only $k$ that could break uniqueness.

Step 2 — Unique case. For $k \neq = 1$ : subtract eqn 1 from eqn 2, $(k - 1) y = k - 1 \Rightarrow y = 1$ , then $x = 0$ . So $(x, y) = (0, 1)$ for every $k \neq = 1$ . Why this step? Subtracting cancels $x$ and isolates $y$ ; dividing by $k - 1$ is legal precisely because we assumed $k \neq = 1$ (nonzero denominator).

Step 3 — Degenerate value $k = 1$ . Both equations become $x + y = 1$ — identical. $rank (A) = rank ([A ∣ b]) = 1 < 2 = n$ → infinitely many (the whole line $x + y = 1$ ). Why this step? At the critical $k$ the determinant vanishes; we test consistency and find the RHS matches → RACE gives infinite, not none.

Step 4 — Is "none" reachable? For "none" we'd need $det = 0$ (so $k = 1$ ) but a RHS clash. At $k = 1$ the RHS is $k = 1$ , matching eqn 1's $1$ — no clash. So case (c) never occurs for this family. Why this step? Covering the limiting case honestly: sometimes a promised cell is provably empty, and saying so is part of "cover ALL cases."

Verify: at $k = 2$ : $(x, y) = (0, 1)$ gives $0 + 1 = 1$ ✓ and $0 + 2 \cdot 1 = 2 = k$ ✓. At $k = 1$ : any $(t, 1 - t)$ works, e.g. $(0.3, 0.7)$ : $0.3 + 0.7 = 1$ ✓.

RACE branch: $k \neq = 1$ unique; $k = 1$ infinite; none is unreachable here.

Recall Which cell is each verdict?

$det A \neq = 0$ ::: unique (C1), the generic case. $det A = 0$ and RHS consistent ::: infinite (C2), a whole line/plane of solutions. $det A = 0$ and RHS clashes ::: none (C3/C5/C8), parallel or contradictory. More rows than unknowns but redundant ::: still unique (C4) — count rank, not rows.

Connections

Determinant — the fast unique/degenerate switch in the square cases.
Inverse of a matrix — the $x = A^{- 1} b$ machinery of Ex 1.
Gaussian elimination — how we actually reduced Ex 6, 8, 9.
Rank of a matrix — the arbiter for every non-square cell (C4, C5, C6).
Column space and null space — the $x_{p} + x_{h}$ structure in Ex 2, 6, 7.
Linear independence — why a redundant row (Ex 4) changes nothing.
Matrix multiplication — the column-view check used throughout.