Determinants: Cofactor Expansion, Row Operations and Area

One number that decides invertibility

The $2 \times 2$ inverse formula divides by $ad - bc$, and the inverse exists exactly when that number is not zero. The number is the determinant:

$$\det\begin{bmatrix} a & b\\ c & d \end{bmatrix} = ad - bc$$

Every square matrix has a determinant, and the same rule holds in any size: $A$ is invertible exactly when $\det A \ne 0$.

For a $3 \times 3$ matrix, cofactor expansion along the first row reduces the problem to $2 \times 2$ determinants. Each entry $a_{1j}$ multiplies the determinant of the matrix left after deleting row $1$ and column $j$, with signs alternating $+,\ -,\ +$:

$$\det A = a_{11}\det A_{11} - a_{12}\det A_{12} + a_{13}\det A_{13}$$

The expansion works along any row or column, with signs following the checkerboard pattern that starts with $+$ in the top-left corner. A row or column full of zeros is the best one to choose.

Why row reduction is the fast way

Cofactor expansion of an $n \times n$ matrix calls for $n$ smaller determinants, each calling for $n - 1$ more, which grows faster than any computer can handle once $n$ reaches the twenties. Row operations change the determinant in three predictable ways:

Row operation	Effect on $\det$
swap two rows	multiplies it by $-1$
multiply a row by $c$	multiplies it by $c$
add a multiple of one row to another	no change

And for a triangular matrix, expanding down the first column keeps a single term at every step, so the determinant is the product of the diagonal. Reduce to triangular form using mostly the third operation, which costs nothing, keep track of any swaps and scalings, and multiply the diagonal. That takes about $n^3$ steps instead of $n!$.

The determinant as area

The columns of $A$ are where $A$ sends $\mathbf{e}_1$ and $\mathbf{e}_2$, so $A$ turns the unit square into the parallelogram on its columns. That parallelogram has area $|\det A|$. In three dimensions the unit cube becomes a box of volume $|\det A|$. Since a linear map stretches every small square by the same factor, $|\det A|$ is the factor by which $A$ multiplies every area. A negative determinant also flips orientation, as a mirror does.

This picture explains the product rule. Doing $B$ scales areas by $|\det B|$, and then $A$ scales them by $|\det A|$, so

$$\det(AB) = \det A \cdot \det B \qquad \det\left(A^{-1}\right) = \frac{1}{\det A}$$

and a matrix that flattens the plane onto a line, with zero area, cannot be undone.

Worked examples

Common mistakes

Practice problems