Solving systems using matrix inversion: Exercises

This page is a self-test ladder. Each problem uses only the tools built in Solving Systems Using Matrix Inversion: writing $A X = B$ , the determinant gate $det A \neq = 0$ , the inverse $A^{- 1} = \frac{1}{d e t A} adj (A)$ , and the consistency branch when $det A = 0$ . Try each one with the solution collapsed, then open it.

Recall the two workhorses we will lean on again and again:

Two pieces of notation we will use throughout, defined once here so no symbol appears unearned:

Definition The zero vector

$O$ and the cofactor sign pattern $O$ means the zero column vector — the column whose every entry is $0$ . Its height matches whatever it is compared to: in a $2 \times 2$ problem, $O = [00]$ (the $2 \times 1$ zero vector); in a $3 \times 3$ problem it is the $3 \times 1$ version. Saying " $(adj A) B = O$ " just means every entry of that product came out zero.

Cofactor of entry $(i, j)$ : take $A$ , delete row $i$ and column $j$ , compute the determinant of what's left (that determinant is the minor $M_{ij}$ ), then attach the sign $(- 1)^{i + j}$ : $C_{ij} = (- 1)^{i + j} M_{ij} .$ WHY the $(- 1)^{i + j}$ ? The sign flips like a checkerboard so that the cofactor expansion actually reproduces the determinant; without it the signed contributions of each entry would not cancel correctly. For a $3 \times 3$ the pattern is $+ - + - + - + - + .$ The cofactor matrix is the grid $[C_{ij}]$ , and the adjugate is its transpose: $adj (A) = [C_{j i}]$ . The transpose (rows become columns) is what makes the identity $A adj (A) = (det A) I$ hold.

Level 1 — Recognition

Exercise 1.1

For the system $3 x + 2 y = 7, 4 x - y = 1$ , write down the coefficient matrix $A$ , the variable matrix $X$ , and the constant matrix $B$ .

Recall Solution

WHAT we do: read the numbers straight off the equations. The coefficient matrix collects the numbers multiplying the unknowns, row by row; the constant matrix collects the right-hand sides. $A = [34 2 - 1], X = [x y], B = [71] .$ Check it reproduces the system: row 1 of $A$ times $X$ gives $3 x + 2 y$ , set equal to the first entry of $B$ , which is $7$ — exactly equation one. ✓

Exercise 1.2

For $A = [5321]$ , compute $det A$ and state whether $A^{- 1}$ exists.

Recall Solution

$det A = (5) (1) - (2) (3) = 5 - 6 = - 1$ . Since $det A = - 1 \neq = 0$ , the machine is reversible, so $A^{- 1}$ exists.

Exercise 1.3

The system $x - y = 2, 2 x - 2 y = 9$ has coefficient matrix $A = [12 - 1 - 2]$ . Without solving, is the inversion method going to work? Compute the gate.

Recall Solution

$det A = (1) (- 2) - (- 1) (2) = - 2 + 2 = 0$ . The gate is closed ( $det A = 0$ ), so $A^{- 1}$ does not exist and $X = A^{- 1} B$ cannot be used. We must fall back to the consistency branch instead.

Level 2 — Application

Exercise 2.1

Solve $2 x + y = 5, 3 x + 4 y = 10$ by matrix inversion.

Recall Solution

$A = [2314], B = [510] .$ Gate: $det A = (2) (4) - (1) (3) = 8 - 3 = 5 \neq = 0$ — safe to invert. Inverse. WHY these moves? The $2 \times 2$ formula $\frac{1}{a d - b c} [d - c - b a]$ says: swap the two diagonal entries ( $a \leftrightarrow d$ ), negate the two off-diagonal entries ( $b, c$ ), then divide by $det$ . Each move is the $2 \times 2$ special case of "transpose the cofactor matrix and divide by the determinant." $A^{- 1} = \frac{1}{5} [4 - 3 - 1 2] .$ Multiply ( $X = A^{- 1} B$ , keep $B$ on the right because $X = A^{- 1} B$ has $B$ on the right): $X = \frac{1}{5} [4 - 3 - 1 2] [510] = \frac{1}{5} [20 - 10 - 15 + 20] = \frac{1}{5} [105] = [21] .$ So $x = 2, y = 1$ . Check: $2 (2) + 1 = 5$ ✓, $3 (2) + 4 (1) = 10$ ✓.

Exercise 2.2

Solve $x + 2 y = 4, 3 x - y = 5$ by matrix inversion.

Recall Solution

$A = [13 2 - 1], B = [45] .$ Gate: $det A = (1) (- 1) - (2) (3) = - 1 - 6 = - 7 \neq = 0$ . Inverse. WHY these moves? The $2 \times 2$ inverse formula $\frac{1}{a d - b c} [d - c - b a]$ is just "transpose the cofactor matrix, divide by $det$ " in miniature: for a $2 \times 2$ , the cofactor of $a$ is $+ d$ , of $b$ is $- c$ , of $c$ is $- b$ , of $d$ is $+ a$ ; transposing that grid puts $d$ and $a$ on the diagonal (swapped) and $- b, - c$ off it (negated). So swap the diagonal $1 \leftrightarrow - 1$ , negate the off-diagonal $2, 3$ , and divide by $- 7$ : $A^{- 1} = \frac{1}{- 7} [- 1 - 3 - 2 1] .$ $X = \frac{1}{- 7} [- 1 - 3 - 2 1] [45] = \frac{1}{- 7} [- 4 - 10 - 12 + 5] = \frac{1}{- 7} [- 14 - 7] = [21] .$ So $x = 2, y = 1$ . Check: $2 + 2 (1) = 4$ ✓, $3 (2) - 1 = 5$ ✓.

Exercise 2.3

Solve the $3 \times 3$ system $x + y = 3, y + z = 5, x + z = 4$ by inversion.

Recall Solution

$A = 101110011, B = 354 .$ Gate — expand $det A$ along row 1. WHY row 1? Any row (or column) gives the same determinant, so we pick the one with the most convenient numbers; here row 1 is $(1, 1, 0)$ and that $0$ kills one whole term, saving work. Attaching the sign pattern $+, -, +$ to row 1: $det A = 1 (1 \cdot 1 - 1 \cdot 0) - 1 (0 \cdot 1 - 1 \cdot 1) + 0 = 1 (1) - 1 (- 1) + 0 = 2 \neq = 0.$ Cofactor matrix, then adjugate. Each cofactor is $(- 1)^{i + j}$ times the minor (delete its row and column, take the $2 \times 2$ determinant). Doing all nine and transposing: $adj A = 11 - 1 - 1 11 1 - 1 1 .$ Then $A^{- 1} = \frac{1}{2} adj A$ and $X = \frac{1}{2} 11 - 1 - 1 11 1 - 1 1 354 = \frac{1}{2} 3 - 5 + 4 3 + 5 - 4 - 3 + 5 + 4 = \frac{1}{2} 246 = 123 .$ So $x = 1, y = 2, z = 3$ . Check: $1 + 2 = 3$ ✓, $2 + 3 = 5$ ✓, $1 + 3 = 4$ ✓.

Level 3 — Analysis

For these, first compute $det A$ ; only if it is zero do you test $(adj A) B$ (recall $O$ = the zero column vector defined at the top). The figure below is the whole geometric story of this level: it shows two lines in the three possible arrangements, and each exercise below is exactly one panel of it.

Figure — Solving systems using matrix inversion

Read the figure left to right. In the left panel the two coloured lines cross at the single orange point — that is $det A \neq = 0$ , the unique case. In the middle panel the two lines lie exactly on top of each other (the violet dashed line sits on the thick magenta one) — that is $det A = 0$ with $(adj A) B = O$ , infinitely many meeting points. In the right panel the two lines run parallel and never touch — that is $det A = 0$ with $(adj A) B \neq = O$ , no solution. Exercises 3.1, 3.2, 3.3 walk these panels in turn: 3.1 = left (unique), 3.2 = middle (infinitely many), 3.3 = right (no solution).

Exercise 3.1

Classify $x - 2 y = 3, 2 x + y = 4$ : unique, none, or infinitely many? (This is the left panel of the figure.)

Recall Solution

$A = [12 - 2 1], B = [34]$ . $det A = (1) (1) - (- 2) (2) = 1 + 4 = 5 \neq = 0$ — gate open. A non-zero determinant means the machine is reversible, so there is exactly one solution — the unique case. The two lines are not parallel, so they cross at a single point (the orange dot in the left panel). For completeness, solve it: $adj A = [1 - 2 21]$ , so $X = \frac{1}{5} [1 - 2 21] [34] = \frac{1}{5} [3 + 8 - 6 + 4] = \frac{1}{5} [11 - 2] = [11/5 - 2/5] .$ Unique solution $(x, y) = (11/5, - 2/5)$ . Check: $11/5 - 2 (- 2/5) = 11/5 + 4/5 = 15/5 = 3$ ✓.

Exercise 3.2

Classify $x - 2 y = 3, 2 x - 4 y = 6$ : unique, none, or infinitely many? (This is the middle panel of the figure.)

Recall Solution

$A = [12 - 2 - 4], B = [36]$ . $det A = (1) (- 4) - (- 2) (2) = - 4 + 4 = 0$ — gate closed. Now test $(adj A) B$ . For a $2 \times 2$ , $adj A = [- 4 - 2 21]$ (swap the diagonal, negate the off-diagonal — the $2 \times 2$ adjugate). $(adj A) B = [- 4 - 2 21] [36] = [- 12 + 12 - 6 + 6] = [00] = O .$ Here $O$ is the $2 \times 1$ zero vector. $det A = 0$ and $(adj A) B = O$ ⇒ infinitely many solutions. Sanity check: the second equation is exactly twice the first, so they are the same line — the middle panel.

Exercise 3.3

Classify $x - 2 y = 3, 2 x - 4 y = 7$ : unique, none, or infinitely many? (This is the right panel of the figure.)

Recall Solution

Same $A$ as Ex 3.2, so $det A = 0$ . Now $B = [37]$ : $(adj A) B = [- 4 - 2 21] [37] = [- 12 + 14 - 6 + 7] = [21] \neq = O .$ $det A = 0$ and $(adj A) B \neq = O$ ⇒ no solution (inconsistent). Geometry: two parallel but distinct lines — the "3" vs "7" pushes one line off the other, the right panel.

Exercise 3.4

For which value of $k$ does $3 x + k y = 6, x + 2 y = 4$ fail to have a unique solution? For that $k$ , is it "no solution" or "infinitely many"?

Recall Solution

$A = [31 k 2]$ . Unique solution fails exactly when $det A = 0$ : $det A = (3) (2) - (k) (1) = 6 - k = 0 \Rightarrow k = 6.$ At $k = 6$ : $A = [3162]$ , $B = [64]$ , $adj A = [2 - 1 - 6 3]$ . $(adj A) B = [2 - 1 - 6 3] [64] = [12 - 24 - 6 + 12] = [- 12 6] \neq = O .$ So at $k = 6$ the system has no solution. (Equation one becomes $3 x + 6 y = 6 \Rightarrow x + 2 y = 2$ , which contradicts $x + 2 y = 4$ .)

Level 4 — Synthesis

Exercise 4.1

You are told the solution of $A X = B$ with $A = [21 3 - 1]$ is $X = [8/5 3/5]$ . Reconstruct $B$ without re-solving — and explain which direction of the machine you used.

Recall Solution

WHAT/WHY: we know the code $X$ and the rule $A$ ; the forward machine gives the output $B = A X$ . No inverse needed here — inversion is only for going backwards. $B = A X = [21 3 - 1] [8/5 3/5] = [16/5 + 9/5 8/5 - 3/5] = [25/5 5/5] = [51] .$ So $B = [51]$ . This confirms the round trip $B \to X \to A X = B$ : feeding $B = [51]$ into the reverse machine $A^{- 1}$ gives back $X = [8/5 3/5]$ , and the forward machine $A$ returns $B$ — exactly the system $2 x + 3 y = 5, x - y = 1$ .

Exercise 4.2

Solve the pair of systems $A X = B_{1}$ and $A X = B_{2}$ at once, where $A = [2314]$ , $B_{1} = [510]$ , $B_{2} = [10]$ . Use the fact that $A^{- 1}$ is computed only once.

Recall Solution

WHY do it together: the expensive step is finding $A^{- 1}$ ; once you have it, each right-hand side is just one multiplication. This is exactly the efficiency reason inversion beats re-doing elimination. From Ex 2.1, $det A = 5$ and $A^{- 1} = \frac{1}{5} [4 - 3 - 1 2]$ . $X_{1} = A^{- 1} B_{1} = \frac{1}{5} [4 - 3 - 1 2] [510] = \frac{1}{5} [105] = [21] .$ $X_{2} = A^{- 1} B_{2} = \frac{1}{5} [4 - 3 - 1 2] [10] = \frac{1}{5} [4 - 3] = [4/5 - 3/5] .$ Check $X_{2}$ : $2 (4/5) + (- 3/5) = 8/5 - 3/5 = 1$ ✓, $3 (4/5) + 4 (- 3/5) = 12/5 - 12/5 = 0$ ✓.

Exercise 4.3

Solve $2 x + y = 3, x + z = 2, y - z = 1$ by inversion, being careful to write the coefficient matrix with a zero wherever a variable is missing.

Recall Solution

Rewrite with all three variables shown: $2 x + 1 y + 0 z = 3, 1 x + 0 y + 1 z = 2, 0 x + 1 y - 1 z = 1$ . $A = 210101 01 - 1, B = 321 .$ Gate — expand along row 1 (chosen because its $0$ in column 3 drops a term). With signs $+, -, +$ : $det A = 2 (0 \cdot (- 1) - 1 \cdot 1) - 1 (1 \cdot (- 1) - 1 \cdot 0) + 0 = 2 (- 1) - 1 (- 1) = - 2 + 1 = - 1 \neq = 0.$ Adjugate — WHY transpose? WHAT we do: compute all nine cofactors $C_{ij} = (- 1)^{i + j} M_{ij}$ to form the cofactor grid $[C_{ij}]$ , then transpose it (rows become columns) to get $adj A = [C_{j i}]$ . WHY the transpose: only the transposed grid satisfies $A adj (A) = (det A) I$ , the identity that makes $\frac{1}{d e t A} adj (A)$ actually undo $A$ ; skipping it is the single most common $3 \times 3$ slip. Carrying it out here: $adj A = - 1 11 1 - 2 - 2 1 - 2 - 1 .$ Then $A^{- 1} = \frac{1}{- 1} adj A = - adj A$ , and $X = A^{- 1} B = - - 1 11 1 - 2 - 2 1 - 2 - 1 321 = - - 3 + 2 + 1 3 - 4 - 2 3 - 4 - 1 = - 0 - 3 - 2 = 032 .$ So $x = 0, y = 3, z = 2$ . Check: $2 (0) + 3 = 3$ ✓, $0 + 2 = 2$ ✓, $3 - 2 = 1$ ✓.

Level 5 — Mastery

Exercise 5.1

A shop sells three combos. Let $x, y, z$ be the prices of items A, B, C. The receipts read: $x + 2 y + 3 z = 14, 2 x + 5 y + z = 15, x + y + z = 6.$ Solve for the prices by matrix inversion, justifying the determinant gate, and confirm all prices are non-negative (so the answer is physically sensible).

Recall Solution

$A = 121251311, B = 14156 .$ Gate — expand along row 1 (all entries small, no obvious zero, so any row works; row 1 keeps arithmetic light). With signs $+, -, +$ : $det A = 1 (5 \cdot 1 - 1 \cdot 1) - 2 (2 \cdot 1 - 1 \cdot 1) + 3 (2 \cdot 1 - 5 \cdot 1)$ $= 1 (4) - 2 (1) + 3 (- 3) = 4 - 2 - 9 = - 7 \neq = 0.$ Non-zero ⇒ unique price list exists, safe to invert. Cofactor matrix, then transpose for the adjugate. Recall each cofactor carries the sign $(- 1)^{i + j}$ : $adj A = 4 - 1 - 3 1 - 2 1 - 13 51 .$ (Verify a corner: $C_{11} = (- 1)^{1 + 1} det [5111] = + 4$ ; it sits at position $(1, 1)$ of the adjugate.) Then $A^{- 1} = \frac{1}{- 7} adj A$ and $X = A^{- 1} B = \frac{1}{- 7} 4 - 1 - 3 1 - 2 1 - 13 51 14156 .$ Compute the product before dividing: $4 (14) + 1 (15) - 13 (6) - 1 (14) - 2 (15) + 5 (6) - 3 (14) + 1 (15) + 1 (6) = 56 + 15 - 78 - 14 - 30 + 30 - 42 + 15 + 6 = - 7 - 14 - 21 .$ $X = \frac{1}{- 7} - 7 - 14 - 21 = 123 .$ Prices $x = 1, y = 2, z = 3$ — all non-negative, physically sensible. Check: $1 + 2 (2) + 3 (3) = 1 + 4 + 9 = 14$ ✓, $2 (1) + 5 (2) + 3 = 15$ ✓, $1 + 2 + 3 = 6$ ✓.

Exercise 5.2

For what value(s) of $λ$ and $μ$ does the system $x + y + z = 6, x + 2 y + 3 z = 10, x + 2 y + λ z = μ$ have (a) a unique solution, (b) no solution, (c) infinitely many? Determine the exact $μ$ that goes with each degenerate case.

Recall Solution

$A = 111122 13 λ$ , $B = 610 μ$ . Gate — $det A$ expanded along row 1 (signs $+, -, +$ ): $det A = 1 (2 λ - 6) - 1 (λ - 3) + 1 (2 - 2) = 2 λ - 6 - λ + 3 + 0 = λ - 3.$ (a) Unique. $det A \neq = 0 \Rightarrow λ \neq = 3$ . For any $μ$ , a unique solution exists.

The degenerate branch: $λ = 3$ . Set $λ = 3$ ; then row 3 of $A$ becomes $(1, 2, 3)$ , which is identical to row 2. So the last two equations share the same left side $x + 2 y + 3 z$ : $x + 2 y + 3 z = 10 and x + 2 y + 3 z = μ .$ These can both hold only if $μ = 10$ . That splits the branch cleanly:

(b) No solution. If $λ = 3$ and $μ \neq = 10$ , the two equations demand $x + 2 y + 3 z$ equal two different numbers at once — a flat contradiction. No $X$ exists. (Confirm with the consistency test: at $λ = 3$ , $det A = 0$ , and one checks $(adj A) B \neq = O$ precisely when $μ \neq = 10$ .)

(c) Infinitely many. If $λ = 3$ and $μ = 10$ , the third equation is an exact copy of the second — it adds no new information. Two genuinely independent equations remain in three unknowns, leaving one free parameter, so there are infinitely many solutions. (Consistency test: $det A = 0$ and $(adj A) B = O$ .)

Summary: unique for $λ \neq = 3$ (any $μ$ ); no solution for $λ = 3, μ \neq = 10$ ; infinitely many for $λ = 3, μ = 10$ .

Connections

Determinants — every gate above was a determinant computation.
Adjoint and Inverse of a Matrix — the machinery behind each $A^{- 1}$ .
Cofactors and Minors — how the $3 \times 3$ adjugates and their $(- 1)^{i + j}$ signs were built.
Consistency of Linear Systems — the L3 and L5 branch logic.
Cramer's Rule — a parallel-track solver using the same $det A \neq = 0$ gate.
Gaussian Elimination — the go-to method once matrices get large.