Computing Determinants with Row Operations

Definition Let $A$ be an $n \times n$ matrix and let $B$ be obtained from $A$ by one row operation.

1. If $B$ swaps two rows of $A$, then $\det B = -\det A$.
2. If $B$ multiplies one row of $A$ by a scalar $k$, then $\det B = k \det A$.
3. If $B$ adds a multiple of one row of $A$ to a different row, then $\det B = \det A$.

These rules hold for every $n$ and follow from cofactor expansion and linearity, described below. A determinant ledger records the factor of each operation, $-1$ for a swap, $k$ for a scaling and $1$ for a replacement. If operations with factors $f_1, \dots, f_m$ turn $A$ into $B$, then $\det B = f_1 f_2 \cdots f_m \det A$. When every factor is nonzero, as in elimination, this can be solved for $\det A$. Multiplying a row by $0$ cannot be undone and is never an elimination step.

Definition A square matrix is upper triangular if every entry below the main diagonal is $0$, and lower triangular if every entry above it is $0$. The determinant of an $n \times n$ triangular matrix $A = [a_{ij}]$, upper or lower, is the product of its diagonal entries,

In particular it is $0$ exactly when some diagonal entry is $0$. A diagonal matrix is both upper and lower triangular.

Definition Let $A$ be an $n \times n$ matrix with $n \ge 2$. Deleting row $i$ and column $j$ of $A$ leaves an $(n - 1) \times (n - 1)$ submatrix $A_{ij}$. Its determinant $M_{ij} = \det A_{ij}$ is the $(i, j)$ minor of $A$, and the $(i, j)$ cofactor of $A$ is

Some books call the submatrix $A_{ij}$ itself the minor matrix. The sign $(-1)^{i + j}$ is $+$ when $i + j$ is even and $-$ when $i + j$ is odd, which gives a checkerboard pattern that starts with $+$ in the top left corner,

Definition The determinant of a $1 \times 1$ matrix $[a]$ is $a$. For $n \ge 2$ the determinant of an $n \times n$ matrix $A$ is its cofactor expansion along the first row,

where each cofactor is, up to sign, a determinant of size $n - 1$. Expansion along any row $i$ or any column $j$ gives the same number,

For $n = 2$ this is $ad - bc$, and for $n = 2$ and $n = 3$ it equals the oriented area or volume of the columns. It follows that $\det A^{\top} = \det A$, so every rule for rows also holds for columns. Expanding all the way down writes $\det A$ as a sum of $n!$ signed products, each taking exactly one entry from every row and every column.

Definition Fix every column of an $n \times n$ matrix except column $j$, and let $D(\mathbf{x})$ be the determinant of the matrix whose column $j$ is $\mathbf{x}$. Then for all vectors $\mathbf{u}, \mathbf{v}$ in $\mathbb{R}^n$ and every scalar $c$,

This follows from expansion along column $j$, whose cofactors do not involve column $j$. The same holds for each row. Together with the swap rule, linearity has two consequences. A matrix with two equal columns has determinant $0$, since swapping them changes the sign but not the matrix. A matrix whose columns are linearly dependent has determinant $0$, since expanding by linearity in a column that is a combination of the others gives only determinants with a repeated column. The row version explains the replacement rule. Adding $c$ times row $k$ to a different row $i$ splits the determinant into $\det A$ plus $c$ times a determinant with two equal rows, which is $0$. The rule concerns one column at a time. For $n \ge 2$ the determinant $\det(A + B)$ need not equal $\det A + \det B$, and $\det(cA) = c^n \det A$ for every $n \times n$ matrix $A$.

Definition For $n \times n$ matrices $A$ and $B$,

To prove it, recall that multiplying $B$ on the left by an elementary matrix $E$ performs the matching row operation on $B$, so $\det(EB) = \det E \, \det B$, where $\det E$ is $-1$ for a swap, $k$ for scaling a row by $k \ne 0$, and $1$ for a replacement. If $A$ is invertible, it is a product of elementary matrices $A = E_1 E_2 \cdots E_m$. Applying the rule $m$ times gives $\det(AB) = \det E_1 \cdots \det E_m \det B$, and the case $B = I$ shows that $\det E_1 \cdots \det E_m = \det A$. If $A$ is singular, its columns are dependent, so $\det A = 0$. The columns of $AB$ lie in $\operatorname{Col} A$, whose dimension is less than $n$, so they are dependent too and $\det(AB) = 0$.