Chapter outline · Determinants

For a square matrix $A=[a_{ij}]$ of size $n\ge2$, and every fixed row index $i$,

$$\det(A)=\sum_{j=1}^{n}(-1)^{i+j}a_{ij}M_{ij}.$$

The corresponding column statement is: for every fixed column index $j$,

$$\det(A)=\sum_{i=1}^{n}(-1)^{i+j}a_{ij}M_{ij}.$$

Thus the first-row formula is the recursive definition, while an arbitrary-row or arbitrary-column formula is a theorem saying that the same determinant is independent of the chosen expansion line. The consequences below use this theorem as stated. The complete proof is in MATH1550: arbitrary cofactor expansion from the recursive determinant. Read it after the recursive definition here: grouping the Leibniz sum by $\sigma(1)$ gives the same base case and first-row recurrence, so induction identifies the two definitions; deleting $\sigma(i)=j$ changes the sign by $(-1)^{i+j}$; grouping again gives any row or column. It does not assume the row-swap or product laws that follow here. Return to the zero-row and triangular consequences below once this bridge is clear.

If every entry in one row or one column of a square matrix is zero, then its determinant is zero.

For a square matrix of size $n\ge2$, suppose all entries in row $k$ are zero except possibly $c_{k\ell}$. Then

$$\det(C)=c_{k\ell}(-1)^{k+\ell}\det\bigl(C(k\mid\ell)\bigr).$$

The same formula holds when all entries in column $\ell$ are zero except possibly $c_{k\ell}$. The formula includes the case $c_{k\ell}=0$.

If $T$ is upper triangular or lower triangular, then

$$\det(T)=t_{11}t_{22}\cdots t_{nn}.$$

Fix every row except row $p$. If the new row is $\alpha u+\beta v$, then

$$\det(\text{same rows except }\alpha u+\beta v) =\alpha\det(\text{same rows except }u) +\beta\det(\text{same rows except }v).$$

If a square matrix has two identical rows, then its determinant is $0$.

For an $n\times n$ matrix $A$, with $1\le p,q\le n$ and $p\ne q$ whenever two row indices occur:

If $B$ results from $A$ by $R_p\leftrightarrow R_q$, then $\det(B)=-\det(A)$.

If $B$ results from $A$ by $R_p\leftarrow\beta R_p$, then $\det(B)=\beta\det(A)$, including the identity at $\beta=0$.

If $B$ results from $A$ by $R_q\leftarrow R_q+\alpha R_p$ with $p\ne q$, then $\det(B)=\det(A)$.

Suppose a sequence of elementary operations changes $A$ into $U$. If the sequence contains $s$ swaps and scales rows by nonzero factors $\beta_1,\ldots,\beta_q$, then

$$\det(U)=(-1)^s\left(\prod_{j=1}^{q}\beta_j\right)\det(A).$$

Consequently,

$$\det(A)=\frac{\det(U)}{(-1)^s\prod_{j=1}^{q}\beta_j}.$$

Row replacements do not appear in the product because their multiplier is $1$.

If $A$ and $B$ are square matrices of the same size, then

$$\det(AB)=\det(A)\det(B).$$

For a square matrix $A$,

$$A\text{ is invertible}\quad\Longleftrightarrow\quad\det(A)\ne0.$$

If $A$ is invertible, then

$$\det(A^{-1})=\frac1{\det(A)}.$$

For every positive integer $m$ and every square $A$,

$$\det(A^m)=(\det A)^m.$$

If $A$ is invertible, the same identity holds for every integer $m$, including negative $m$.

For every square matrix $A$,

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

For any fixed column $j$ of a square matrix $A=[a_{ij}]$, with $n\ge2$, where $M_{ij}$ is the minor determinant and $C_{ij}=(-1)^{i+j}M_{ij}$ is the cofactor,

$$\det(A)=a_{1j}C_{1j}+a_{2j}C_{2j}+\cdots+a_{nj}C_{nj},$$

Equivalently,

$$\det(A)=\sum_{i=1}^n (-1)^{i+j}a_{ij}M_{ij}.$$

Let $B$ be obtained from a square matrix $A$ by one elementary column operation.

Swapping two distinct columns $p$ and $q$ multiplies the determinant by $-1$.

Multiplying column $p$ by $\beta$ multiplies the determinant by $\beta$. For a reversible elementary operation, $\beta\neq0$.

Replacing column $q$ by $C_q+cC_p$, with $p\neq q$, leaves the determinant unchanged.

For every square matrix $A$, including singular matrices,

$$A\,\operatorname{adj}(A)=\det(A)I.$$

If $\det(A)\neq0$, then $A$ is invertible and

$$A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A).$$

Let $A\in M_n(\mathbb{R})$ satisfy $\det(A)\ne0$, and let $b\in\mathbb{R}^n$. For each $j$, let $M_j$ be the matrix obtained by replacing the $j$th column of $A$ with $b$.

If $x$ is the unique solution of

$$Ax=b,$$

then

$$x_j=\frac{\det(M_j)}{\det(A)}.$$