LINEAR-ALGEBRA · Unit 17 · Video 3 · Interactive Practice

Multiply the Diagonal: The Only Determinant Trick You Actually Need

IKey Formulas

Formula	Name	Key Fact
$\det(U) = u_{11} \cdot u_{22} \cdots u_{nn}$	Triangular determinant	Product of diagonal entries
$\det(A) = (-1)^r \, p_1 \, p_2 \cdots p_n$	Elimination formula	$r$ = row swaps, $p_i$ = pivots
$R_i \leftarrow R_i + c\,R_j$	Row replacement (P5)	Does not change det
$R_i \leftrightarrow R_j$	Row swap (P2)	Flips sign of det

IIVisualization 1 — Triangular Determinant = Diagonal Product

In a triangular matrix only the diagonal entries reach the determinant — the entries above never do.

💡 Row-replacement clears every above-diagonal entry without changing the determinant, so only the diagonal product survives.

IIIVisualization 2 — Elimination Tracks the Determinant

Elimination drives $A$ to upper-triangular form; each move either preserves the determinant or flips its sign.

Matrix Step 0

IVVisualization 3 — Why Libraries Use Elimination

Cofactor expansion costs $O(n!)$ ; elimination costs $O(n^3)$ — the gap explodes as $n$ grows.

Matrix size n 5

VQuiz Questions

Question 1

Consider the upper triangular matrix:

$U = \begin{bmatrix} 2 & 3 & 1 \\ 0 & 5 & 4 \\ 0 & 0 & 3 \end{bmatrix}$

What is $\det(U)$ ?

✅ Correct! The determinant of a triangular matrix is the product of its diagonal entries: 2 × 5 × 3 = 30.

❌ Not quite. For a triangular matrix, only the diagonal entries matter — multiply them together.

Show solution

Solution:

For an upper triangular matrix, the determinant equals the product of the diagonal entries:

$\det(U) = 2 \times 5 \times 3 = 30$

The off-diagonal entries $(3, 1, 4)$ do not affect the determinant. You can verify this using the argument from the video: use row replacement to zero out entries above the diagonal (Property 5 — det unchanged), then factor out each diagonal entry (Property 3A), leaving the identity matrix (Property 1 — det = 1).

Question 2

During Gaussian elimination of a matrix $A$ , you perform one row swap and arrive at the upper triangular form:

$U = \begin{bmatrix} 3 & 1 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & 5/2 \end{bmatrix}$

What is $\det(A)$ ?

✅ Correct! The pivot product is 15, but the single row swap flips the sign to −15.

❌ Almost! You got the pivot product right, but forgot the sign flip from the row swap. Each swap multiplies the determinant by −1.

❌ Not quite. Apply the formula: det(A) = (−1)^r × (product of pivots), where r = number of row swaps.

Show solution

Solution:

Use the elimination formula: $\det(A) = (-1)^r \times p_1 \times p_2 \times p_3$

Step 1 — Product of pivots: $p_1 \cdot p_2 \cdot p_3 = 3 \times 2 \times \tfrac{5}{2} = 15$

Step 2 — Sign correction: $r = 1$ swap, so $(-1)^1 = -1$

Step 3 — Final answer: $\det(A) = (-1) \times 15 = -15$

Each row swap flips the sign of the determinant (Property 2). Forgetting this sign correction is the most common mistake!

Question 3

True or False: If Gaussian elimination produces a zero pivot and no row swap can fix it (no nonzero entry below in that column), then $\det(A) = 0$ .

✅ Correct! A zero pivot with no possible swap means a zero in the diagonal product, giving det = 0.

❌ Not quite. If a pivot is zero, the entire product of pivots becomes zero — and so does the determinant.

Show solution

Answer: True

Here is the reasoning:

If no row swap can produce a nonzero pivot, the elimination results in a row of all zeros in the upper triangular form $U$ .
That zero row means at least one diagonal entry (pivot) is $0$ .
The product of the pivots includes a factor of $0$ , so the product is $0$ .
Therefore $\det(A) = (-1)^r \times 0 = 0$ .

A zero determinant means the matrix is singular — it has no inverse. This is the "binary test" from the video: $\det(A) = 0 \iff A$ is singular.

Question 4

You perform Gaussian elimination on a $4 \times 4$ matrix $A$ , making 3 row swaps. The pivots of the resulting upper triangular matrix are $2, -1, 4, 3$ .

What is $\det(A)$ ?

✅ Correct! The negative from the odd number of swaps and the negative pivot cancel, giving det(A) = 24.

❌ Close! You may have either forgotten the sign correction or missed one of the negatives. Carefully track: (−1)³ × (2)(−1)(4)(3) = (−1)(−24) = 24.

❌ Not quite. First multiply all four pivots together, then apply the (−1)^r sign correction for the row swaps.

Show solution

Solution:

Apply $\det(A) = (-1)^r \times p_1 \cdot p_2 \cdot p_3 \cdot p_4$ .

Step 1 — Product of pivots: $2 \times (-1) \times 4 \times 3 = -24$

Step 2 — Sign correction: $r = 3$ swaps, so $(-1)^3 = -1$

Step 3 — Combine: $\det(A) = (-1) \times (-24) = 24$

The two negatives — one from the odd number of swaps and one from the negative pivot — cancel each other out, giving a positive determinant.

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