LINEAR-ALGEBRA · Unit 17 · Video 2 · Interactive Practice

Three Axioms Are All You Need

IKey Formulas

Statement	Name	Role
$\det[\ldots;\, t\,\mathbf{r};\, \ldots] = t\,\det[\ldots;\, \mathbf{r};\, \ldots]$	Axiom 1 — Linearity in each row	Starting assumption
Swapping two rows flips the sign of $\det$	Axiom 2 — Sign Flip	Starting assumption
$\det(I) = 1$	Axiom 3 — Normalization	Starting assumption
Two equal rows $\Rightarrow \det = 0$	Property 4	Derived from Axioms 1 + 2
Elimination preserves $\det$	Property 5	Derived from Axiom 1 + Property 4
Zero row $\Rightarrow \det = 0$	Property 6	Derived from Axiom 1

IIVisualization 1 — Equal Rows Collapse the Area

For a $2\times 2$ matrix, $\det$ is the signed area of the parallelogram its two rows span.

💡 Swapping two equal rows leaves the matrix unchanged yet must flip the sign (Axiom 2), so $\det = -\det$ — forcing $\det = 0$ (Property 4).

IIIVisualization 2 — Elimination Shears, Area Holds

Replacing Row 2 with $\text{Row 2} - \ell\,\text{Row 1}$ shears the parallelogram — but does its signed area change?

Multiplier ℓ 0

💡 The tip of Row 2 slides parallel to Row 1, so its height above that base — and therefore the area — never moves. This is Property 5.

IVVisualization 3 — Elimination vs. Scaling

One of these row operations leaves $\det$ fixed; the other multiplies it by $k$ — which is which?

Factor k 1

💡 Gaussian elimination uses only the operation that preserves $\det$ , so a matrix keeps its determinant all the way down to triangular form.

VQuiz Questions

Question 1

Consider the matrix $A = \begin{bmatrix} 4 & 1 & 3 \\ 4 & 1 & 3 \\ 2 & 5 & 7 \end{bmatrix}$ . What is $\det(A)$ ?

✅ Correct! Identical rows force the determinant to zero (Property 4).

❌ Not quite. Look carefully at rows 1 and 2 — they're identical. What does Property 4 tell us?

Show solution

Solution:

Rows 1 and 2 are both $[4,\, 1,\, 3]$ — they are identical.

By Property 4: if a matrix has two equal rows, $\det = 0$ .

Proof recap: Swapping the two identical rows gives the same matrix, so $\det(A)$ is unchanged. But Axiom 2 says any swap flips the sign: $\det(A) = -\det(A)$ . Therefore $2\det(A) = 0$ , giving $\det(A) = \boxed{0}$ .

No computation needed — the property gives the answer instantly.

Question 2

You know that $\det(A) = 5$ . You perform the elimination step: Row 3 $\leftarrow$ Row 3 $- 4 \cdot$ Row 1. What is the determinant of the resulting matrix?

✅ Correct! Elimination preserves the determinant regardless of the multiplier.

❌ Not quite. You may be confusing elimination with row scaling. Elimination subtracts a multiple of one row from another — the determinant doesn't change.

❌ Not quite. Property 5 says elimination preserves the determinant exactly. The multiplier value doesn't matter.

Show solution

Solution:

The operation "Row 3 $\leftarrow$ Row 3 $- 4\cdot$ Row 1" is an elimination step — subtracting a multiple of one row from another.

By Property 5, elimination preserves the determinant for any value of $\ell$ :

$\det(A') = \det(A) = \boxed{5}$

Common mistake: Confusing this with row scaling ( $4 \cdot$ Row), which would multiply det by 4 to give 20. But elimination subtracts a multiple — it doesn't scale.

Question 3

True or False: A matrix with a zero row has determinant zero because we can write the zero row as $0 \cdot [1, 1, \ldots, 1]$ and factor out the $0$ using Axiom 1 (linearity).

✅ Correct! Property 6 follows directly from the scalar part of Axiom 1 (linearity).

❌ Not quite. Think about what the scalar part of linearity does: det[...; t r; ...] = t det[...; r; ...]. What happens when t = 0?

Show solution

Solution: True

This is exactly the proof of Property 6. Rewrite the zero row as $0 \cdot [1, 1, 1]$ :

$\det\begin{bmatrix} \text{row}_1 \\ \mathbf{0} \\ \text{row}_3 \end{bmatrix} = \det\begin{bmatrix} \text{row}_1 \\ 0 \cdot [1,1,1] \\ \text{row}_3 \end{bmatrix} = 0 \cdot \det\begin{bmatrix} \text{row}_1 \\ [1,1,1] \\ \text{row}_3 \end{bmatrix} = 0$

The scalar part of Axiom 1 pulls the $0$ out front, and $0$ times anything is $0$ .

Question 4

Starting with $A = \begin{bmatrix} 3 & 1 \\ 5 & 4 \end{bmatrix}$ where $\det(A) = 7$ , you perform two operations in sequence:

Elimination: Row 2 $\leftarrow$ Row 2 $- 2 \cdot$ Row 1
Scaling: Row 2 $\leftarrow 3 \cdot$ Row 2

What is the determinant of the final matrix?

✅ Correct! Elimination keeps det = 7, then scaling by 3 gives 3 × 7 = 21.

❌ Not quite. Elimination preserves det, but the scaling step changes it. What does Axiom 1 say about multiplying a row by a scalar?

❌ Not quite. Only one of the two operations changes the determinant. Which one?

❌ Not quite. Apply each step separately: elimination preserves det, scaling multiplies det by the scale factor.

Show solution

Solution:

Apply each operation step by step:

Step 1 — Elimination (Row 2 $\leftarrow$ Row 2 $- 2\cdot$ Row 1):

By Property 5, elimination preserves the determinant: $\det = 7$ .

New Row 2: $[5 - 2(3),\; 4 - 2(1)] = [-1,\; 2]$ .

Step 2 — Scaling (Row 2 $\leftarrow 3\cdot$ Row 2):

By Axiom 1 (scalar part), multiplying a row by $k$ multiplies $\det$ by $k$ :

$\det = 3 \times 7 = 21$

Verification: Final matrix $= \begin{bmatrix} 3 & 1 \\ -3 & 6 \end{bmatrix}$ , so $\det = 3(6) - 1(-3) = 18 + 3 = \boxed{21}$ ✓

Solved: 0 / 4