Computing the determinant from the pivots
Slide 4 showed that elimination steps never change det, and that a row swap only flips its sign. That gives the fast way to compute it. Think of a leaning stack of books: push it upright and it is the same pile, but now you can measure it with a ruler. Elimination pushes A upright into the triangular matrix U from lesson 2, with the same det. For a triangular matrix, det is just the numbers on the diagonal multiplied together, and those numbers are the pivots. So det A = ± (product of the pivots), with one minus sign for each row swap. This is how every computer finds a determinant.
Lesson 6 · slide 5Det from pivots- Elimination steps (subtract a multiple of a row) keep det the same.
- Each row exchange flips the sign.
- For a triangular matrix, multiply down the diagonal.
- So det A = ± the product of the pivots. A zero pivot that no swap can fix means det A = 0.
- Two more facts: det AB = det A · det B, and det Aᵀ = det A.
A triangular matrix: multiply the diagonal
Start with a 2 by 2 triangular matrix, U = 2103. The formula gives ad − bc = 2·3 − 1·0 = 6, the two diagonal numbers multiplied. The picture says the same: the base lies along the x-axis with length 2, and the height is 3.
Why this works for any size: use rule 3 to pull each diagonal number out of its row. That leaves 1s on the diagonal. Elimination upward then clears the numbers above the diagonal (subtract multiples of lower rows from higher ones; slide 4 showed, using rule 3, that such steps keep det), and leaves I, with det I = 1. So det U is the product of the diagonal. (A 0 on the diagonal means a missing pivot. Elimination then makes a zero row, so det U = 0, still the product.)
Eliminate, then multiply
A 2 by 2 first. A = 2145:
Now a 3 by 3, the matrix from slide 4:
K = 2114−60−272 from lesson 2 had pivots 2, −8 and 1. So det K = 2 · (−8) · 1 = −16.
A bigger example: one row swap
A = 021110243 has 0 in the first pivot spot, so we must swap rows (lesson 2, slide 4). Each swap costs one minus sign.
A computer does exactly this for a 1000 by 1000 matrix: eliminate, count the swaps, multiply 1000 pivots.
Products: det AB = det A · det B
Doing B and then A (AB acts on x as A(Bx): B first) stretches areas by det B, then by det A. So det AB = det A · det B.
From A⁻¹A = I we get det A⁻¹ · det A = 1, so det A⁻¹ = 1 / det A. For the A above, A⁻¹ = 1−½0½ and det A⁻¹ = 1 · ½ = ½ ✓.
Keep reading with the full course
Finish lesson 6 and you will be able to: see a matrix as a move of the plane, and build one from where two arrows land.
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Examples
Example 1 · A row exchange
A = 0231: the first pivot position holds 0.
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