Inverse of 2×2 matrix: Worked examples — Maths deep dive

Recall the two objects we will use over and over. For $A = (a c b d), det A = a d - b c, A^{- 1} = \frac{1}{a d - b c} (d - c - b a) .$

Here $det A$ (read "determinant of $A$ ") is the single number $a d - b c$ — see Determinant of a 2×2 matrix. It is the divisor, so it decides everything.

The scenario matrix

Every $2 \times 2$ inverse problem falls into one of these case classes. Each row is a distinct trap or skill; the last column says which worked example covers it.

#	Case class	What's special / what could go wrong	Covered by
C1	All entries positive, $det > 0$	The "clean" baseline case	Ex 1
C2	Negative entries, $det < 0$	Negative determinant — sign must survive to the end	Ex 2
C3	Fractional / decimal entries	Determinant is a fraction, dividing gets messy	Ex 3
C4	$det = 0$ (singular)	No inverse exists — must detect and stop	Ex 4
C5	A zero entry ( $b = 0$ or $c = 0$ , triangular)	Degenerate-looking but still invertible	Ex 5
C6	Solving a linear system $A x = b$	Application, not just inversion	Ex 6
C7	Real-world word problem	Translate words → matrix → solve	Ex 7
C8	Exam twist: find unknown $k$ making $A$ singular	Reverse the logic: set $det = 0$	Ex 8

Two figures illustrate the geometry behind the extreme rows: C4 (area collapses) and C2 (a flip).

Example 1 — C1: the clean positive case

Worked example Invert

$A = (3124)$ Forecast: all entries positive, so guess the determinant is a modest positive number and the inverse has small fractions. Guess before reading on.

Step 1 — determinant. $det A = (3) (4) - (2) (1) = 12 - 2 = 10.$ Why this step? If $det = 0$ we stop instantly — no inverse. Here $10 \neq = 0$ , so we continue.

Step 2 — adjugate. Swap the diagonal ( $3 \leftrightarrow 4$ ), negate the off-diagonal ( $2, 1$ ): $adj A = (4 - 1 - 2 3) .$ Why this step? This is the fixed pattern the derivation produced (see Adjugate and cofactors).

Step 3 — divide by det. $A^{- 1} = \frac{1}{10} (4 - 1 - 2 3) = (0.4 - 0.1 - 0.2 0.3) .$ Why this step? The adjugate alone satisfies $A \cdot adj A = (det A) I$ ; dividing by $det A = 10$ turns that $10 I$ into the exact $I$ we want, completing $A A^{- 1} = I$ .

Verify: $A A^{- 1} = \frac{1}{10} (3124) (4 - 1 - 2 3) = \frac{1}{10} (100010) = I .$ ✓

Example 2 — C2: negative entries, negative determinant

Worked example Invert

$A = (- 1 2 31)$ Forecast: there is a negative entry and the "backwards" product is large, so guess $det$ comes out negative. Does a negative determinant kill the inverse? (No — only $det = 0$ does.)

Step 1 — determinant. $det A = (- 1) (1) - (3) (2) = - 1 - 6 = - 7.$ Why this step? $- 7 \neq = 0$ , so an inverse exists. The minus sign is real and must be carried through.

Step 2 — adjugate. Swap the diagonal ( $- 1 \leftrightarrow 1$ ), negate the off-diagonal ( $3, 2$ ): $adj A = (1 - 2 - 3 - 1) .$ Why this step? This is the fixed swap/negate pattern (see Adjugate and cofactors); with a negative on the diagonal, swapping puts the $- 1$ into the bottom-right corner, so watch its sign carefully.

Step 3 — divide by $- 7$ . Dividing by a negative flips every sign of the adjugate: $A^{- 1} = \frac{1}{- 7} (1 - 2 - 3 - 1) = (- \frac{1}{7} \frac{2}{7} \frac{3}{7} \frac{1}{7}) .$ Why this step? Common slip: people forget the $- 7$ negates the adjugate entries too, so the signs above are the ones to trust.

Verify: $A A^{- 1} = \frac{1}{- 7} (- 1 2 31) (1 - 2 - 3 - 1) = \frac{1}{- 7} (- 7 0 0 - 7) = I .$ ✓

A negative determinant means the transformation flips orientation (like a mirror) as well as rescaling. Look at the figure below: the violet unit square is traced counter-clockwise, but after applying $A$ the orange parallelogram's corners run clockwise (follow the magenta arrows). That reversal — corners going the "wrong way round" — is exactly what a negative $det$ records; the area magnitude $∣ - 7∣ = 7$ is how much the square was enlarged.

Example 3 — C3: fractional and decimal entries

Worked example Invert

$A = (\frac{1}{2} \frac{1}{4} 0.4 1)$ Forecast: a mix of a fraction ( $\frac{1}{2}$ ), a decimal ( $0.4$ ), and another fraction — guess a small fractional determinant, so dividing will enlarge the numbers. Guess whether $det$ is above or below $\frac{1}{2}$ .

Step 1 — determinant. Convert the decimal first: $0.4 = \frac{2}{5}$ . Then $det A = (\frac{1}{2}) (1) - (\frac{2}{5}) (\frac{1}{4}) = \frac{1}{2} - \frac{2}{20} = \frac{10}{20} - \frac{2}{20} = \frac{8}{20} = \frac{2}{5} .$ Why this step? Turning $0.4$ into $\frac{2}{5}$ lets us use a common denominator ( $20$ ) and subtract cleanly. $\frac{2}{5} \neq = 0$ ⟹ invertible.

Step 2 — adjugate. Swap the diagonal ( $\frac{1}{2} \leftrightarrow 1$ ), negate the off-diagonal ( $\frac{2}{5}$ and $\frac{1}{4}$ ): $adj A = (1 - \frac{1}{4} - \frac{2}{5} \frac{1}{2}) .$ Why this step? Same fixed swap/negate pattern as always (see Adjugate and cofactors) — fractions and decimals don't change the recipe, only the arithmetic.

Step 3 — divide by $\frac{2}{5}$ , i.e. multiply by $\frac{5}{2}$ : $A^{- 1} = \frac{5}{2} (1 - \frac{1}{4} - \frac{2}{5} \frac{1}{2}) = (\frac{5}{2} - \frac{5}{8} - 1 \frac{5}{4}) .$ Why this step? Dividing by a fraction = multiplying by its reciprocal; that turns $- \frac{2}{5}$ into $\frac{5}{2} \cdot (- \frac{2}{5}) = - 1$ , etc.

Verify: $A A^{- 1}$ should give $I$ . Top-left entry: $\frac{1}{2} \cdot \frac{5}{2} + \frac{2}{5} \cdot (- \frac{5}{8}) = \frac{5}{4} - \frac{1}{4} = 1$ ✓ (full check in VERIFY).

Example 4 — C4: the singular case (no inverse)

Worked example Try to invert

$A = (6496)$ Forecast: look at the columns $(4 6)$ and $(6 9)$ . Is the second a multiple of the first? If yes, area collapses and there is no inverse. Decide before Step 1.

Step 1 — determinant. $det A = (6) (6) - (9) (4) = 36 - 36 = 0.$ Why this step? Always check the determinant first. $0$ is the deal-breaker.

Step 2 — conclude. $det A = 0 \Rightarrow$ == $A$ is singular, no inverse exists.== Why this step? The formula would demand dividing by $0$ (undefined). See Singular vs non-singular matrices.

Why geometrically? Column 2 $= \frac{3}{2} \times$ column 1 (since $9 = \frac{3}{2} \cdot 6$ and $6 = \frac{3}{2} \cdot 4$ ). Both columns point the same way, so $A$ squashes the whole plane onto a single line — area becomes zero and information is lost.

Verify: the columns are parallel ⟺ their cross-ratio $6 \cdot 6 - 9 \cdot 4 = 0$ , matching $det = 0$ . Consistent. ✓

The figure makes the collapse concrete. The violet unit square (area $1$ ) is fed into $A$ . Its image is not a parallelogram at all — it is the single magenta line segment: every point of the plane lands somewhere on that one line. The two column vectors, orange $(6, 4)$ and navy $(9, 6)$ , lie exactly on top of each other's direction (navy $= 1.5 \times$ orange), which is why the square has nowhere to spread into. With zero area there is no way to "un-flatten" back to the original square — hence no inverse.

Example 5 — C5: a zero entry (triangular matrix)

Worked example Invert

$A = (2053)$ Forecast: one entry is $0$ . Does a zero mean it's degenerate? Look at the columns — are they parallel? Guess whether an inverse exists.

Step 1 — determinant. $det A = (2) (3) - (5) (0) = 6 - 0 = 6.$ Why this step? The zero sits in the $c$ position, so it drops out of $a d - b c$ . $6 \neq = 0$ ⟹ invertible — a zero entry does not mean singular.

Step 2 — adjugate. Swap diagonal, negate off-diagonal: $adj A = (30 - 5 2) .$ Why this step? Negating $c = 0$ still gives $0$ — the zero simply stays put.

Step 3 — divide by $6$ . $A^{- 1} = \frac{1}{6} (30 - 5 2) = (\frac{1}{2} 0 - \frac{5}{6} \frac{1}{3}) .$ Why this step? Dividing the adjugate by $det A = 6$ rescales $A \cdot adj A = 6 I$ down to the exact $I$ , completing $A A^{- 1} = I$ — the same finishing move as every other example.

Verify: $A A^{- 1} = \frac{1}{6} (2053) (30 - 5 2) = \frac{1}{6} (6006) = I .$ ✓ The inverse of a triangular matrix is triangular in the same shape.

Example 6 — C6: solving a linear system

Example 7 — C7: real-world word problem

Worked example A café sells muffins and coffees

Two combos are on the menu. Combo A = 2 muffins + 1 coffee = £7. Combo B = 1 muffin + 3 coffees = £11. Find the price of one muffin ( $m$ ) and one coffee ( $c$ ). Forecast: guess whole-pound prices. A coffee is probably a couple of pounds.

Step 1 — build the matrix. The equations are $2 m + 1 c = 7$ and $1 m + 3 c = 11$ , i.e. $A = (2113), b = (11 7), A (c m) = b .$ Why this step? Coefficients = quantities of each item; the right side = total prices.

Step 2 — determinant. $det A = (2) (3) - (1) (1) = 6 - 1 = 5.$ Why this step? We check $det \neq = 0$ before inverting — $5 \neq = 0$ guarantees the prices are uniquely determined (no ambiguity in the menu).

Step 3 — inverse and solve. $A^{- 1} = \frac{1}{5} (3 - 1 - 1 2)$ , so $(c m) = \frac{1}{5} (3 - 1 - 1 2) (11 7) = \frac{1}{5} (- 7 + 22 21 - 11) = \frac{1}{5} (15 10) = (3 2) .$ Why this step? $x = A^{- 1} b$ recovers the unknown prices.

Verify (with units): muffin £2, coffee £3. Combo A: $2 (£2) + £3 = £7$ ✓. Combo B: $£2 + 3 (£3) = £11$ ✓. Prices are positive and sensible. ✓

Example 8 — C8: exam twist, find $k$ that makes $A$ singular

Active recall

Recall Which case classes are "safe to invert" and which are not?

Safe (invertible): C1, C2, C3, C5, C6, C7 — all have $det \neq = 0$ . Not invertible: C4 (and C8 at $k = - 6, 2$ ) — $det = 0$ .

A zero entry means the matrix is singular.

False — a zero entry (Ex 5) still gave

det = 6 \neq = 0

. Only

det = a d - b c = 0

means singular.

A negative determinant means no inverse.

False —

det = - 7

(Ex 2) is fine; it just flips orientation. Only exactly

det = 0

blocks the inverse.

To find

k

making

A

singular, you…

solve

det A = 0

for

k

(Ex 8).

To solve

A x = b

with the inverse, compute…

x = A^{- 1} b

(Ex 6).

Connections

Determinant of a 2×2 matrix — the gatekeeper checked first in every example.
Adjugate and cofactors — the swap/negate matrix used in every inversion.
Matrix multiplication — used in every "Verify" step.
Identity matrix — the target $I$ we check against.
Solving linear systems — Examples 6 and 7.
Singular vs non-singular matrices — Examples 4 and 8.
Linear transformations and area — why $det = 0$ flattens the plane (Ex 4).