Properties — row operations, multiplicativity: Worked examples

This deep dive builds on the parent properties note and leans on the cofactor/Leibniz definition, Gaussian Elimination & Row Echelon Form, Elementary Matrices, Invertibility & Singular Matrices, and Volume, Orientation & the Determinant.

The three bookkeeping symbols we will use

Every worked example below reduces a matrix to triangular form and reads off the answer with one master formula. Two little symbols do all the bookkeeping — let us define them once, in plain words, before we use them anywhere.

Definition The swap-count

$s$ and the scale-product $\prod k$

$== s ==$ is simply the number of row swaps you perform during the reduction. Count them on your fingers: each time you exchange two rows, add $1$ to $s$ . Because a single swap multiplies the determinant by $- 1$ (parent Rule 1), doing $s$ of them multiplies it by $(- 1)^{s}$ — which is $+ 1$ for an even count and $- 1$ for an odd count.
$== \prod k ==$ (read "the product of all the $k$ 's") is the product of every scaling factor $k$ you used when you did an operation of the form $R_{i} \to k R_{i}$ . If you scaled one row by $\frac{1}{3}$ and another by $2$ , then $\prod k = \frac{1}{3} \cdot 2 = \frac{2}{3}$ . If you never scaled a row, there are no factors to multiply, so by convention $\prod k = 1$ . Because scaling a row by $k$ multiplies the determinant by $k$ (parent Rule 2), scaling several rows multiplies it by their product $\prod k$ .

Careful — what does NOT count. A replacement $R_{i} \to R_{i} + k R_{j}$ (adding a multiple of another row) is not a scaling: it leaves the determinant unchanged, so its $k$ is never collected into $\prod k$ and never touches $s$ . Only the standalone move $R_{i} \to k R_{i}$ feeds $\prod k$ .

With those two symbols nailed down, every example is just careful counting.

The scenario matrix

Every determinant computation you will ever meet lands in one of these cells. The whole point of this page is to make sure not one cell is left dark.

#	Case class	What makes it tricky	Covered by
C1	Pure replacement to triangular, no swap/scale	Bookkeeping: prove nothing was lost	Ex 1
C2	A swap appears → sign bookkeeping	Off-by-one on the $(- 1)^{s}$ factor	Ex 2
C3	You scale a row to get a pivot $1$ → must divide back	Forgetting to undo the scale	Ex 3
C4	Degenerate / singular input → $det = 0$	Recognising a zero pivot appears	Ex 4
C5	Sign / orientation case → negative determinant, flipped area	Reading the minus sign as "area got mirrored"	Ex 5 (figure)
C6	Multiplicativity with mixed signs $det A < 0 < det B$	Signs multiply, they don't add	Ex 6
C7	Scalar multiple of whole matrix $det (k A) = k^{n} det A$ , incl. negative $k$	The exponent is $n$ , and $k$ can be negative	Ex 7
C8	Inverse / power corollaries, incl. a limiting "almost singular" matrix	$det A^{- 1} = 1/ det A$ blows up as $det A \to 0$	Ex 8
C9	Word problem — real-world area scaling	Translating geometry into a determinant	Ex 9 (figure)
C10	Exam twist — combine swap + scale + multiplicativity in one shot	Doing all bookkeeping simultaneously	Ex 10

Read the table top-to-bottom once. Each row is a promise; the ten examples keep them all.

Ex 1 — Cell C1: pure replacement, no swap, no scale

Worked example Statement

Compute $det A$ for $A = 123257 - 1 12$ using only replacement operations ( $R_{i} \to R_{i} + k R_{j}$ ).

Forecast: Since replacement never changes the determinant, the answer will just be the product of the pivots we reach — guess whether it is positive or negative before reading on.

Step 1. $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 3 R_{1}$ . Why this step? Replacement leaves $det$ exactly unchanged (it adds a term with two equal rows, which is $0$ ), so it is the one "free" move. We use it to zero out the first column below the pivot. $\to 100211 - 1 35$ Step 2. $R_{3} \to R_{3} - R_{2}$ . Why this step? Same free move, now clearing the entry below the second pivot to reach upper-triangular form. $\to 100210 - 1 32$ Step 3. For a triangular matrix, $det$ = product of the diagonal. We used only replacements, so the swap-count is $s = 0$ and there were no row-scalings, giving $\prod k = 1$ . The master formula then reads $det A = (- 1)^{0} \cdot \frac{1}{1} \cdot (1 \cdot 1 \cdot 2) = 2.$

Verify: Let us cofactor-expand along column 1 and deliberately show the trap first. A tempting shortcut is to write only the top entry's term, $1 \cdot (5 \cdot 2 - 1 \cdot 7) = 1 \cdot 3 = 3$ — this is wrong, because it silently drops the two other cofactor terms coming from the $2$ and the $3$ in column 1 (they are not zero in the original $A$ ; only after reduction were those entries zeroed). The honest full expansion keeps all three terms: $1 (5 \cdot 2 - 1 \cdot 7) - 2 (2 \cdot 2 - 1 \cdot 3) + (- 1) (2 \cdot 7 - 5 \cdot 3) = 1 (3) - 2 (1) - 1 (- 1) = 3 - 2 + 1 = 2.$ So the correct value is $2$ , matching the reduction. ✓

Ex 2 — Cell C2: a swap appears

Worked example Statement

Compute $det B$ for $B = 012213101$

Forecast: The top-left entry is $0$ , so we must swap to get a usable pivot. One swap means one factor of $- 1$ . Predict the sign of your final answer.

Step 1. Swap $R_{1} \leftrightarrow R_{2}$ . Why this step? You cannot pivot on a $0$ . A swap fixes it, at the known cost of multiplying $det$ by $- 1$ (from Rule 1). This is our first (and only) swap, so the swap-count is now $s = 1$ . $\to 102123011$ Step 2. $R_{3} \to R_{3} - 2 R_{1}$ (free replacement). $\to 100121011$ Step 3. $R_{3} \to R_{3} - \frac{1}{2} R_{2}$ (free replacement — note this is not a scaling of a row by itself, it is adding a multiple of $R_{2}$ to $R_{3}$ , so $det$ is untouched and it feeds neither $s$ nor $\prod k$ ). $\to 100120 01 \frac{1}{2}$ Step 4. Pivots product $= 1 \cdot 2 \cdot \frac{1}{2} = 1$ . No scalings so $\prod k = 1$ ; apply the swap factor $(- 1)^{s} = (- 1)^{1}$ : $det B = (- 1)^{1} \cdot \frac{1}{1} \cdot 1 = - 1.$

Verify: Leibniz directly: $0 (1 \cdot 1 - 0 \cdot 3) - 2 (1 \cdot 1 - 0 \cdot 2) + 1 (1 \cdot 3 - 1 \cdot 2) = 0 - 2 (1) + 1 (1) = - 1.$ ✓

Ex 3 — Cell C3: scaling a row to make a pivot $1$ , then dividing back

Worked example Statement

Compute $det C$ where $C = 310621954$ while deliberately scaling $R_{1}$ to make its pivot $1$ (so you can practise undoing it).

Forecast: If you multiply a row by $\frac{1}{3}$ you divide the whole determinant by $3$ , so at the end you must multiply back by $3$ . Guess: does the final $det$ come out as a multiple of $3$ ?

Step 1. $R_{1} \to \frac{1}{3} R_{1}$ . Why this step? A clean leading $1$ simplifies the arithmetic — but scaling a row by $\frac{1}{3}$ multiplies $det$ by $\frac{1}{3}$ (Rule 2). This is a genuine row-scaling, so it feeds the scale-product: $\prod k = \frac{1}{3}$ . By the master formula we will divide $det U$ by this $\prod k$ , i.e. multiply by $3$ at the end. $\to 110221354$ Step 2. $R_{2} \to R_{2} - R_{1}$ (free replacement). $\to 100201324$ Step 3. The second pivot is now $0$ . Swap $R_{2} \leftrightarrow R_{3}$ ; this is our first swap, so the swap-count is $s = 1$ . $\to 100210342$ Step 4. Triangular: pivots product $det U = 1 \cdot 1 \cdot 2 = 2$ . Assemble the master formula $det C = (- 1)^{s} \cdot \frac{1}{\prod k} \cdot det U$ with $s = 1$ and $\prod k = \frac{1}{3}$ : $det C = (- 1)^{1} \cdot \frac{1}{1/3} \cdot 2 = (- 1) \cdot 3 \cdot 2 = - 6.$

Verify: Direct Leibniz on $C$ : $3 (2 \cdot 4 - 5 \cdot 1) - 6 (1 \cdot 4 - 5 \cdot 0) + 9 (1 \cdot 1 - 2 \cdot 0) = 3 (3) - 6 (4) + 9 (1) = 9 - 24 + 9 = - 6.$ ✓

Ex 4 — Cell C4: degenerate / singular input, $det = 0$

Ex 5 — Cell C5: negative determinant means orientation flipped

Worked example Statement

The matrix $M = (0110)$ maps the unit square to a new shape. Find $det M$ and say what the sign means geometrically.

Forecast: $M$ sends $(1, 0) \mapsto (0, 1)$ and $(0, 1) \mapsto (1, 0)$ — it swaps the axes. Guess the area factor and its sign before reading.

Step 1. $M$ is one row swap away from $I$ : swap $R_{1} \leftrightarrow R_{2}$ gives $I$ , whose $det = 1$ . Why this step? $det I = 1$ is the normalization axiom, and a single swap ( $s = 1$ ) multiplies by $(- 1)^{1}$ . $det M = (- 1)^{1} \cdot 1 = - 1.$ Step 2. Interpret the sign. $∣ det M ∣ = 1$ means area is preserved. The minus sign means orientation flipped — the image is a mirror reflection of the original, as if the square were turned over. In the figure, watch the corner-labels $P, Q, R$ go clockwise after the map though they were counter-clockwise before.

Figure — what to observe: On the LEFT, the teal unit square has its corners $O \to P \to Q \to R$ traced counter-clockwise (the purple arrow curls CCW). On the RIGHT, after applying $M$ , the very same corner labels now trace out clockwise (the purple arrow curls CW), even though the shape has the same size. That reversal of the go-round direction is exactly what a negative determinant looks like — the plane has been flipped over like a pancake, not merely rotated.

Verify: Leibniz $2 \times 2$ : $0 \cdot 0 - 1 \cdot 1 = - 1.$ ✓

Ex 6 — Cell C6: multiplicativity with mixed signs

Ex 7 — Cell C7: $det (k A) = k^{n} det A$ , including negative $k$

Ex 8 — Cell C8: inverse, power, and the "almost singular" limit

Worked example Statement

Let $A = (21 1 ε)$ . (a) Find $det A$ . (b) Find $det (A^{- 1})$ and $det (A^{3})$ for $ε = 1$ . (c) Describe what happens to $det (A^{- 1})$ as $ε \to \frac{1}{2}$ .

Forecast: $det (A^{- 1}) = 1/ det A$ . What must happen to $1/ det A$ when $det A$ heads toward $0$ ?

Step 1. $det A = 2 ε - 1 \cdot 1 = 2 ε - 1.$ Why this step? For a $2 \times 2$ matrix $(a c b d)$ the Leibniz/cofactor definition gives $det = a d - b c$ — the two "diagonal products" minus the two "anti-diagonal products". Here $a = 2, b = 1, c = 1, d = ε$ , so $det A = 2 ε - 1$ . Step 2 (b), $ε = 1$ . $det A = 2 (1) - 1 = 1.$ Then $det (A^{- 1}) = \frac{1}{d e t A} = \frac{1}{1} = 1, det (A^{3}) = (det A)^{3} = 1^{3} = 1.$ Why: $det (A^{- 1}) = 1/ det A$ comes from $det A \cdot det A^{- 1} = det I = 1$ ; $det (A^{k}) = (det A)^{k}$ by iterating multiplicativity. Step 3 (c), the limit — sign matters. Write $det A = 2 ε - 1$ , so $det (A^{- 1}) = \frac{1}{2 ε - 1}$ . As $ε \to \frac{1}{2}$ the denominator $\to 0$ , and its sign depends on which side you approach from:

If $ε ↓ \frac{1}{2}$ (from above, $ε > \frac{1}{2}$ ), then $2 ε - 1 > 0$ and tiny, so $det (A^{- 1}) \to + \infty$ .
If $ε ↑ \frac{1}{2}$ (from below, $ε < \frac{1}{2}$ ), then $2 ε - 1 < 0$ and tiny, so $det (A^{- 1}) \to - \infty$ .

So the blow-up is $+ \infty$ from the right and $- \infty$ from the left — the two-sided limit does not exist. Geometrically the parallelogram is being crushed toward a line from opposite orientations, so the inverse map's stretch factor grows without bound, positive on one side and negative (orientation-flipped) on the other.

Verify: at $ε = 1$ : $det A = 1$ , $det A^{- 1} = 1$ , $det A^{3} = 1$ ; and $det A = 2 ε - 1 = 0$ exactly at $ε = \frac{1}{2}$ . ✓

Ex 9 — Cell C9: word problem, real-world area scaling

Worked example Statement

A graphic designer applies the transform $T = (3012)$ to a company logo whose original area is $12 cm^{2}$ . What is the new area, and did the logo get mirrored?

Forecast: The determinant is the area-scaling factor. Predict the new area before computing.

Step 1. $det T = 3 \cdot 2 - 1 \cdot 0 = 6.$ Why this step? The determinant is exactly the signed area-scaling factor of the linear map, so any shape's area is multiplied by $∣ det T ∣$ . Step 2. New area $= ∣ det T ∣ \times 12 = 6 \times 12 = 72 cm^{2} .$ Step 3. Sign of $det T$ is positive, so orientation is preserved — the logo is stretched and sheared, not mirrored. The figure shows the unit square (area 1) becoming a parallelogram of area 6.

Figure — what to observe: The small teal square is the original unit cell (area $1$ ). Apply $T$ and it becomes the orange parallelogram: the two basis arrows land at $T e_{1} = (3, 0)$ (purple, along the axis) and $T e_{2} = (1, 2)$ (ink, tilted). Notice the arrows still turn the same way as the original axes — no mirror flip — which is why $det T = + 6$ is positive. The orange region visibly holds six copies of the teal square, confirming the area grew by the factor $det T = 6$ . Multiply your real logo's $12 cm^{2}$ by that same $6$ .

Verify: $det T = 6 > 0$ and $6 \times 12 = 72$ . Units: (dimensionless factor) $\times cm^{2} = cm^{2}$ . ✓

Ex 10 — Cell C10: exam twist, everything at once

Worked example Statement

Given that $det P = 4$ for a $3 \times 3$ matrix $P$ , and $E_{1}$ (swap $R_{1} \leftrightarrow R_{3}$ ), $E_{2}$ (scale $R_{2}$ by $5$ ) are elementary matrices, compute $det (2 E_{1} E_{2} P) .$

Forecast: Four effects stack: the scalar $2$ on a $3 \times 3$ (that's $2^{3}$ ), a swap ( $- 1$ ), a scale-by- $5$ ( $5$ ), and $det P = 4$ . Multiply them — don't add.

Step 1. Peel the scalar: $det (2 X) = 2^{3} det X = 8 det X$ where $X = E_{1} E_{2} P$ . Why this step? $2 X$ scales all $n = 3$ rows (Rule 2 / Cell C7), giving the factor $2^{n} = 2^{3} = 8$ . Step 2. Use multiplicativity: $det (E_{1} E_{2} P) = det E_{1} \cdot det E_{2} \cdot det P$ . Why this step? $det (A B) = det A det B$ chained across all three factors. Step 3. Fill in the elementary determinants (parent's table): a swap matrix has $det E_{1} = - 1$ , a scale-by- $5$ matrix has $det E_{2} = 5$ , and we are told $det P = 4$ . Substituting, $det (E_{1} E_{2} P) = (- 1) \cdot 5 \cdot 4 = - 20.$ Step 4. Combine with the scalar factor from Step 1: $det (2 E_{1} E_{2} P) = 8 \cdot (- 20) = - 160.$

Verify: $8 \times (- 1) \times 5 \times 4 = - 160$ . ✓

Active recall

Recall Which cell, which factor? (cover the answers)

You swapped rows 3 times computing a $det$ — what factor? ::: $(- 1)^{3} = - 1$ .
$det (3 A)$ for a $4 \times 4$ with $det A = 2$ ? ::: $3^{4} \cdot 2 = 162$ .
$A$ has two proportional rows — $det$ ? ::: $0$ (singular).
$det A = - 3$ , $det B = 2$ , then $det (A B)$ ? ::: $- 6$ .
As $det A \to 0^{+}$ , what does $det (A^{- 1})$ do? ::: $\to + \infty$ (and $\to - \infty$ if $det A \to 0^{-}$ ).
$det T = 6$ positive — mirrored or not? ::: Not mirrored (orientation preserved).