LINEAR-ALGEBRA · Unit 17 · Video 1 · Interactive Practice

Three Rules Build Every Determinant

IKey Formulas

Property	Statement	Meaning
1. Normalization	$\det(I) = 1$	The identity matrix has determinant 1
2. Row Swap	Swap two rows $\Rightarrow \det$ flips sign	Each swap multiplies det by $-1$
3A. Scalar Factor	Scale one row by $t$ $\Rightarrow \det$ scales by $t$	Determinant is proportional to each row
3B. Row Additivity	Split one row as sum $\Rightarrow \det$ splits as sum	Determinant adds when one row is a sum

IIVisualization 1 — Scaling a Row Scales the Area

Multiplying one row by a scalar $t$ multiplies the parallelogram's signed area — the determinant — by that same $t$ .

Scalar t 1

Property 3A: $\det(t\,\mathbf{r}_1,\ \mathbf{r}_2) = t\,\det(\mathbf{r}_1,\ \mathbf{r}_2)$ . At $t = 0$ the two rows become parallel and the area collapses to zero.

IIIVisualization 2 — Swapping Rows Flips the Sign

The determinant is the signed area of the parallelogram the two rows span; exchanging the rows reverses its orientation.

Property 2: each row swap multiplies $\det$ by $-1$ . A corollary — if two rows are equal, swapping changes nothing yet must negate $\det$ , forcing $\det = 0$ .

IVVisualization 3 — Determinant Is Not Additive

With $A = I$ and $B = kI$ , does $\det(A+B)$ equal $\det(A) + \det(B)$ ?

Scalar k 1

The two agree only at $k = 0$ . Their difference is exactly $(1+k)^2 - (1+k^2) = 2k$ — the cross term that row-by-row linearity leaves out.

VQuiz Questions

Question 1

The $3 \times 3$ identity matrix has $\det(I) = 1$ . You perform two row swaps: first swap rows 1 and 3, then swap rows 2 and 3.

What is the determinant of the resulting permutation matrix?

✅ Correct! Two swaps flip the sign twice: $(+1) \to (-1) \to (+1)$ . Even permutations always give $\det = +1$ .

❌ Not quite. Each swap flips the sign once. Two sign flips bring it back to the original sign.

Show solution

Solution:

Start with $\det(I) = 1$ (Property 1).

Swap 1 (rows 1 $\leftrightarrow$ 3): det flips to $-1$ (Property 2)
Swap 2 (rows 2 $\leftrightarrow$ 3): det flips to $+1$ (Property 2)

Two swaps = two sign flips: $1 \times (-1) \times (-1) = +1$ .

This is an even permutation (even number of swaps), so $\det = +1$ .

Question 2

True or False: For any $n \times n$ matrices $A$ and $B$ , $\det(A + B) = \det(A) + \det(B)$ .

✅ Correct! The determinant is linear in each row separately, not in the entire matrix.

❌ Not quite. Try $A = B = I$ : does $\det(I + I) = \det(I) + \det(I)$ ?

Show solution

Solution: False.

Counterexample: Let $A = B = I_{2 \times 2}$ .

$\det(A) = 1$ , $\;\det(B) = 1$ , $\;\Rightarrow\; \det(A) + \det(B) = 2$
$A + B = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ , $\;\Rightarrow\; \det(A + B) = 4$
$4 \neq 2$ ✗

Property 3 allows splitting one row at a time with all other rows fixed. Adding entire matrices changes all rows simultaneously — that is not what the axiom permits.

Question 3

Given $M = \begin{pmatrix} 6 & 4 \\ 1 & 3 \end{pmatrix}$ , the first row can be written as $2 \cdot [3, 2]$ .

Using Property 3A (scalar factor), what is $\det(M)$ ?

✅ Correct! Factor out 2 from row 1, compute the simpler determinant (= 7), then multiply: $2 \times 7 = 14$ .

❌ Not quite. First factor out the scalar 2 from row 1 (Property 3A), then compute the determinant of the simplified matrix, and multiply.

Show solution

Solution:

Row 1 $= [6, 4] = 2 \cdot [3, 2]$ . By Property 3A:

$\det\begin{pmatrix} 6 & 4 \\ 1 & 3 \end{pmatrix} = 2 \cdot \det\begin{pmatrix} 3 & 2 \\ 1 & 3 \end{pmatrix}$

Compute: $\det\begin{pmatrix} 3 & 2 \\ 1 & 3 \end{pmatrix} = 3 \times 3 - 2 \times 1 = 9 - 2 = 7$ .

Therefore: $\det(M) = 2 \times 7 = \mathbf{14}$ .

Check: $6 \times 3 - 4 \times 1 = 18 - 4 = 14$ ✓

Question 4

A square matrix has two identical rows. Which property directly implies its determinant must be zero?

✅ Correct! Swapping identical rows flips the sign but leaves the matrix unchanged, so $\det = -\det$ , which forces $\det = 0$ .

❌ Not quite. Think about which property involves row swaps. What happens when the two rows you swap are identical?

Show solution

Solution: Property 2 (Row swap flips sign).

Here is the elegant argument:

Let $M$ have two identical rows (rows $i$ and $j$ ).
Swap rows $i$ and $j$ : by Property 2, $\det(M_{\text{swapped}}) = -\det(M)$ .
But the rows are identical, so swapping changes nothing: $M_{\text{swapped}} = M$ .
Therefore $\det(M) = -\det(M)$ .
The only number equal to its own negative is zero.

$\det(M) = -\det(M) \;\Longrightarrow\; 2\det(M) = 0 \;\Longrightarrow\; \det(M) = 0$

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