Properties of Determinants

Main idea

Each row operation does something simple to the determinant: a swap flips its sign, scaling a row scales it, and adding a multiple of one row to another leaves it alone. And since doing $B$ then $A$ scales area twice, $\det(AB) = \det A\det B$ .

\det(AB) = \det A\,\det B, \qquad \det(A^T) = \det A, \qquad \det(A^{-1}) = \frac{1}{\det A}

\det(cA) = {\htmlData{sym=0}{c^n}}\det A \quad (A \text{ is } n\times n)

What the symbols mean

$c^n$
Scaling every row
Each of the n rows contributes a factor of c.

Explanation

The properties of determinants are simply the rules for how the determinant reacts when you change a matrix: swap rows, scale a row, multiply two matrices, and so on. They're good for one big thing: computing determinants fast, without the long cofactor expansion. The formulas look like a list to memorize, but each one is the area picture from last lesson: $\det(AB) = \det A\,\det B$ (stretching twice multiplies the stretches), $\det(A^T) = \det A$ (rows and columns play the same role), $\det(A^{-1}) = \frac{1}{\det A}$ (undoing a stretch shrinks by the same factor) and $\det(cA) = c^n\det A$ (each of the $n$ directions gets scaled by $c$ ). Once you see the picture, there's very little to memorize.

Row operations.

Replacement (add a multiple of a row): determinant unchanged.
Interchange (swap two rows): determinant changes sign.
Scaling a row by $c$ : determinant multiplied by $c$ .

Think about why replacement changes nothing. Adding a multiple of one edge of a parallelogram to the other slides it along itself: same base, same height, same area. There's an animation further down that shows exactly that. Scaling one edge by $c$ scales the area by $c$ , and swapping two edges flips the orientation, so the sign changes.

Efficient computing. Row reduce to triangular form, multiply the diagonal, and adjust for any swaps and scalings. That's how computers do it, because it takes about $n^3$ steps while cofactor expansion takes about $n!$ .

Multiplicativity has a clean meaning: doing $B$ scales volume by $\det B$ , then $A$ scales it by $\det A$ , so together they scale it by the product.

A common mistake. It's tempting to write $\det(2A) = 2\det A$ , because pulling a 2 out usually works that way. It isn't: $2A$ doubles every row, and each row contributes its own factor of 2. For a $3\times3$ matrix that's $2^3 = 8$ , which makes sense, since doubling every side of a box makes its volume 8 times bigger.

See it

A parallelogram with edges a₁ and a₂ + c·a₁, as c slides from −1.5 to 1.5. As you can see, the shape leans back and forth along a₁, but the area meter stays fixed at 2.75 the whole time: same base, same height. That's why row replacement leaves det A alone.

Worked examples

Example 1

$A$ is $3\times3$ with $\det A = 5$ . Find $\det(2A)$ and $\det(A^{-1})$ .

1
Doubling $A$ doubles each of its 3 rows, and each row contributes a factor of 2: $\det(2A) = 2^3\cdot5$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

$\det A = 5$ . What is the determinant after swapping two rows?

Problem 2+5 XP

$A$ is $3\times3$ with $\det A = 5$ . Find $\det(2A)$ .

Problem 3+5 XP

$\det A = 3$ and $\det B = -2$ . Find $\det(AB)$ .

Problem 4+5 XP

$\det A = 4$ . Find $\det(A^{-1})$ .

Problem 5+5 XP

$\det A = 7$ . Find $\det(A^T)$ .

Problem 6+5 XP

A 3D model's transformation matrix has determinant $4$ . The model's volume is $25$ cm³. What is its volume after the transformation is applied twice?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

$A$ is $3\times3$ with $\det A = 2$ . Find $\det\left(3A^{-1}A^TA\right)$ .

Where it's used

Determinant properties prove that a product of invertible matrices is invertible, and they explain why numerical software computes determinants by row reduction rather than cofactors.

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