Intuition What this page is forProperties of determinants gave you the eight rules. This page throws every kind of determinant a problem can be at you and shows which rule kills it fastest. Before each solution you'll forecast the answer — guessing first is how the rules stick.
Every determinant problem you meet is one of these "cells". We will hit all of them.
Cell
What triggers it
The rule that wins
Example
A — Make zeros then read diagonaldense numbers, no obvious pattern
P6 + P8
Ex 1
B — Spot an instant zerotwo equal / proportional rows or cols
P3 (+P4)
Ex 2
C — Zero row / degeneratea full row or column of zeros
P7
Ex 3
D — Pull out common factorsa row/col is a clean multiple
P4
Ex 4
E — Sign-tracking under swapsyou rearrange rows to simplify
P2
Ex 5
F — Whole-matrix scaling (the trap)det ( k A ) det ( A + B ) P4 corollary, P5 caveat
Ex 6
G — Product & inversedet ( A B ) det ( A − 1 ) P9
Ex 7
H — Real-world word problemarea / volume scaling
geometry of det
Ex 8
I — Exam twist (symbolic/letters)entries are a , b , c x
P6 factoring
Ex 9
Each cell is worth understanding as "which single fact makes the arithmetic disappear?"
Δ = 2 4 6 1 3 5 3 8 16
Forecast: guess before reading. Is this a big number, a small number, or zero? The rows look like each next row is roughly double... watch out.
Step 1. R 2 → R 2 − 2 R 1 R 3 → R 3 − 3 R 1 Why this step? P6 (add a multiple of one row to another) does not change the value , and it manufactures zeros in column 1 — turning a hard 3 × 3 = 2 0 0 1 1 2 3 2 7
Step 2. R 3 → R 3 − 2 R 2 Why this step? Same rule, now clearing the last entry below the diagonal.
= 2 0 0 1 1 0 3 2 3
Step 3. Upper-triangular ⟹ multiply the diagonal (P8).
Δ = 2 ⋅ 1 ⋅ 3 = 6.
Verify: direct expansion along column 1 of the original: 2 ( 3 ⋅ 16 − 8 ⋅ 5 ) − 1 ( 4 ⋅ 16 − 8 ⋅ 6 ) + 3 ( 4 ⋅ 5 − 3 ⋅ 6 ) = 2 ( 48 − 40 ) − 1 ( 64 − 48 ) + 3 ( 20 − 18 ) = 16 − 16 + 6 = 6.
3 1 7 6 2 14 5 9 4
Forecast: stare at column 2 versus column 1. Notice anything doubled?
Step 1. Column 2 = ( 6 , 2 , 14 ) = 2 ⋅ ( 3 , 1 , 7 ) = 2 ⋅ Why this step? One column is a scalar multiple of another. Pull the 2 = 2 3 1 7 3 1 7 5 9 4
Step 2. Now columns 1 and 2 are equal .
Why this step? Equal columns force the determinant to 0 = 2 ⋅ 0 = 0.
Verify: proportional columns always give 0 0
5 0 1 − 2 0 8 7 0 3
Forecast: a whole row is zeros. What must the answer be, and why geometrically ?
Step 1. Row 2 is all zeros.
Why this step? P7: a zero row gives det = 0 k = 0 0 0 = 0.
Verify: expand along row 2 — every cofactor is multiplied by a 0 0 0
3 1 2 6 0 2 9 2 0
Forecast: row 1 is 3 × ( 1 , 2 , 3 ) 3 3
Step 1. Row 1 = 3 ( 1 , 2 , 3 ) 3 Why this step? Smaller numbers = fewer arithmetic slips. P4 lets a common row factor come out front.
= 3 1 1 2 2 0 2 3 2 0
Step 2. R 2 → R 2 − R 1 R 3 → R 3 − 2 R 1 Why this step? Clear column 1 below the top to prepare an easy expansion.
= 3 1 0 0 2 − 2 − 2 3 − 1 − 6
Step 3. Expand along column 1: only the top-left survives.
Why this step? Two zeros in column 1 mean one term.
= 3 ⋅ 1 ⋅ [ ( − 2 ) ( − 6 ) − ( − 1 ) ( − 2 ) ] = 3 ( 12 − 2 ) = 3 ⋅ 10 = 30.
Verify: direct expansion of the original along row 1: 3 ( 0 ⋅ 0 − 2 ⋅ 2 ) − 6 ( 1 ⋅ 0 − 2 ⋅ 2 ) + 9 ( 1 ⋅ 2 − 0 ⋅ 2 ) = 3 ( − 4 ) − 6 ( − 4 ) + 9 ( 2 ) = − 12 + 24 + 18 = 30.
0 0 2 0 5 7 4 1 3
Forecast: it's almost triangular but pointing the wrong way (zeros are top-left instead of bottom-left). If we flip it into triangular form, what happens to the sign?
Step 1. Swap R 1 ↔ R 3 Why this step? This makes the zeros sit in the lower-left, giving upper-triangular form — but every swap multiplies the answer by − 1 (P2). Track it: running factor = − 1 = − 2 0 0 7 5 0 3 1 4
Step 2. Now upper-triangular ⟹ product of diagonal (P8).
Why this step? Diagonal read-off, no expansion.
= − ( 2 ⋅ 5 ⋅ 4 ) = − 40.
Verify: the original is lower -anti form; expand along column 1 (only the 2 ( − 1 ) 3 + 1 = + 2 ⋅ 0 5 4 1 = 2 ( 0 − 20 ) = − 40. + 40 − 40
Let A = ( 1 3 2 4 ) det A = 1 ⋅ 4 − 2 ⋅ 3 = − 2 det ( 3 A ) det ( A + A ) = det A + det A
Forecast: many students say det ( 3 A ) = 3 det A = − 6
Step 1 (a). 3 A both rows by 3 Why this step? P4 gives one factor of k n = 2 det ( k A ) = k n det A det ( 3 A ) = 3 2 det A = 9 ⋅ ( − 2 ) = − 18.
Step 2 (b). A + A = 2 A det ( 2 A ) = 2 2 ( − 2 ) = − 8 Why this step? Compute the honest left side.
Step 3 (b). Right side: det A + det A = ( − 2 ) + ( − 2 ) = − 4 Why this step? Compare. Since − 8 = − 4 false . P5 is linearity in one row at a time , not the whole matrix.
Verify: 3 A = ( 3 9 6 12 ) det = 3 ⋅ 12 − 6 ⋅ 9 = 36 − 54 = − 18. 2 A = ( 2 6 4 8 ) det = 16 − 24 = − 8 = − 4.
A = ( 2 1 0 3 ) B = ( 1 0 4 2 ) det ( A B ) det ( A − 1 )
Forecast: det A det B det ( A B )
Step 1. det A = 2 ⋅ 3 = 6 det B = 1 ⋅ 2 = 2 Why this step? Read diagonals; no work.
Step 2. Product rule P9: det ( A B ) = det A ⋅ det B = 6 ⋅ 2 = 12. Why this step? P9 says the scaling factors multiply — stretching by 6 2 12
Step 3. Inverse: det ( A − 1 ) = det A 1 = 6 1 . Why this step? From P9 with A − 1 A = I det I = 1 det ( A − 1 ) det A = 1
Verify: A B = ( 2 1 0 3 ) ( 1 0 4 2 ) = ( 2 1 8 10 ) det = 2 ⋅ 10 − 8 ⋅ 1 = 20 − 8 = 12. 6 1 ⋅ 6 = 1.
A designer draws a triangle on graph paper with corners P ( 1 , 1 ) Q ( 5 , 2 ) R ( 2 , 6 ) M = ( 2 0 0 3 ) × 2 × 3 stretched triangle?
Forecast: the machine stretches width by 2 3
Step 1. Original area via the determinant formula (see Area of a Triangle using Determinants ):
Area = 2 1 Q x − P x R x − P x Q y − P y R y − P y = 2 1 4 1 1 5 . Why this step? Two edge vectors from P
Step 2. Evaluate: 4 1 1 5 = 20 − 1 = 19 = 2 19 = 9.5 Why this step? Plain 2 × 2
Step 3. Apply the machine. By Linear Transformations & Scaling of Area , area scales by ∣ det M ∣ det M = 2 ⋅ 3 = 6. Why this step? P8 (triangular) gives det M ∣ det M ∣ the area multiplier — the whole point of a determinant.
Step 4. Stretched area = 6 × 9.5 = 57. Why this step? Multiply original area by the scaling factor.
Verify (units & sanity): areas are in square units. The width-2 3 2 × 3 = 6 det M 9.5 × 6 = 57 6 ×
Prove without brute expansion:
1 1 1 a b c a 2 b 2 c 2 = ( b − a ) ( c − a ) ( c − b ) . Vandermonde determinant.)
Forecast: if a = b must contain a factor ( b − a )
Step 1. R 2 → R 2 − R 1 R 3 → R 3 − R 1 Why this step? Subtracting the first row creates a zero in column 1 and reveals common factors.
= 1 0 0 a b − a c − a a 2 b 2 − a 2 c 2 − a 2 .
Step 2. Factor ( b − a ) ( c − a ) x 2 − a 2 = ( x − a ) ( x + a ) Why this step? P4 pulls a common factor out of a whole row.
= ( b − a ) ( c − a ) 1 0 0 a 1 1 a 2 b + a c + a .
Step 3. Expand along column 1 (only the top-left 1 2 × 2 1 1 b + a c + a = ( c + a ) − ( b + a ) = c − b . Why this step? Two zeros in column 1 collapse the expansion to one 2 × 2
Step 4. Combine.
= ( b − a ) ( c − a ) ( c − b ) .
Verify: set a = 1 , b = 2 , c = 4 ( 2 − 1 ) ( 4 − 1 ) ( 4 − 2 ) = 1 ⋅ 3 ⋅ 2 = 6 1 1 1 1 2 4 1 4 16 = 1 ( 2 ⋅ 16 − 4 ⋅ 4 ) − 1 ( 1 ⋅ 16 − 4 ⋅ 1 ) + 1 ( 1 ⋅ 4 − 2 ⋅ 1 ) = 1 ( 16 ) − 1 ( 12 ) + 1 ( 2 ) = 6.
Common mistake The three traps this page inoculates you against
F-trap: det ( k A ) = k n det A k det A every row scales by k each time .E-trap: forgetting the sign flip on a swap (Ex 5). One swap = × ( − 1 ) F-trap 2: det ( A + B ) = det A + det B
Recall Which rule kills which cell?Dense numbers, no pattern ::: P6 to triangularise, then P8 (Ex 1, Cell A)
Two equal or proportional rows/columns ::: P4 then P3 → 0 0 det ( k A ) n × n k n det A det ( A B ) det ( A − 1 ) det A det B 1/ det A M ∣ det M ∣ R 1
Which cell is my problem?
Cell B: P4 then P3 gives zero
Cell F: k to the n, no additivity
Cell H: multiply area by det
Cell I: subtract rows and factor