Properties of Determinants

Cofactor expansion defines the determinant but costs $\sim n!$ operations. The practical route is row reduction, and it works because the three elementary row operations change the determinant in three known ways. Those rules also prove that $\det A\ne 0$ characterizes invertibility, that $A$ and its transpose share a determinant, and that the determinant of a product is the product of the determinants. All of it follows from a single theorem about row operations.

Row operations and the determinant

The three operations that reduce a matrix to echelon form have three distinct effects. A replacement leaves the determinant untouched. An interchange flips its sign. A scaling by $k$ multiplies it by $k$. The proof, deferred to the end of the lesson, runs by induction on the size of $A$ using cofactor expansion along a row the operation does not disturb.¹

A useful reading of the scaling rule runs backward: a common factor in one row can be pulled out in front of the determinant. If a row is $[5k,2k,3k]$, then $k$ factors out and the row becomes $[5,2,3]$ with the determinant carrying a leading $k$.

Determinant by row reduction

Reduce a square matrix $A$ to an echelon form $U$ using only replacements and interchanges. Replacements do nothing to the determinant; if $r$ interchanges are used, the sign flips $r$ times. Since $U$ is triangular, its determinant is the product of its diagonal entries — the pivots. Combining these facts:

If $A$ is invertible the diagonal entries $u_{11},\dots ,u_{nn}$ are all pivots and their product is nonzero; if $A$ is not invertible, at least $u_{nn}=0$ and the product is zero. The echelon form is not unique and the individual pivots are not unique, but their product is unique up to the sign $(-1{)}^r$.

A cofactor expansion on the original matrix would have cost far more arithmetic. When a common factor appears, pulling it out with the scaling rule further simplifies the numbers before reduction continues.

The invertibility criterion

The pivot formula for $\det A$ settles invertibility.

This adds the statement $\det A\ne 0$ to the list of equivalent conditions in the Invertible Matrix Theorem. Two corollaries follow at once. When the columns of $A$ are linearly dependent, $A$ is not invertible, so $\det A=0$. Because the rows of $A$ are the columns of $A^{\mathsf{T}}$ and $A^{\mathsf{T}}$ is singular exactly when $A$ is, linearly dependent rows also force $\det A=0$. In practice, dependence is often visible: two equal rows, two equal columns, or a zero row or column each give a zero determinant.

The single replacement does not change the determinant, and it exposes a repeated row: rows 2 and 3 of the new matrix are identical, and any repeated row makes the determinant zero.¹ Whenever a replacement exposes two identical rows, the matrix is singular and $\det A=0$.

Combining row reduction with cofactor expansion

The two techniques cooperate. Use a pivot to clear a column, then expand along that column to shrink the determinant, then repeat. For

adding $R_2$ to $R_4$ clears the $(4,1)$ entry, after which a cofactor expansion down column 1 keeps only the $2$ in row 2 (with sign $(-1{)}^{2+1}$):

A replacement $R_2\leftarrow R_2-3R_1$ inside the $3\times 3$ produces a column with two zeros, and a final expansion down that column finishes:

Row operations create the zeros; cofactor expansion exploits them.

Column operations and the transpose

Because cofactor expansion works down columns just as well as across rows, every statement about rows has a mirror image for columns. The transpose makes the connection precise.

The proof is by induction: the cofactor of $a_{1j}$ in $A$ equals the cofactor of $a_{j1}$ in $A^{\mathsf{T}}$, since both involve the same smaller determinant, so the first-row expansion of $\det A$ equals the first-column expansion of $\det A^{\mathsf{T}}$.¹ A row operation on $A^{\mathsf{T}}$ is a column operation on $A$, so the row-operation rules hold verbatim with row replaced by column everywhere. For numerical work only row operations are used, but column operations are convenient in proofs.

The multiplicative property

The determinant turns matrix products into ordinary products of numbers.

The proof splits on invertibility. If $A$ is singular, so is $AB$, and both sides are zero by the determinant test for invertibility. If $A$ is invertible it is a product of elementary matrices $A=E_p\cdots E_1$, and repeatedly applying the rule $\det (EA)=(\det E)(\det A)$ — the elementary-matrix form of the row-operation rules — peels the factors off one at a time.¹

The multiplicative law has immediate consequences worth recording. Taking $B=A^{-1}$ gives $\det (A^{-1})=1/\det A$ for invertible $A$. For similar matrices, $\det (P^{-1}AP)=\det A$, since the $P^{-1}$ and $P$ factors contribute reciprocal determinants — a fact the characteristic polynomial depends on. And $\det (A^2)=(\det A{)}^2$, so if $A^2=I$ then $\det A=\pm 1$.

Linearity in a single column

Holding all but one column of $A$ fixed makes the determinant a linear function of the remaining column. Writing $A=[{\mathbf{a}}_1\cdots \mathbf{x}\cdots {\mathbf{a}}_n]$ with the $j$th column variable, define $T(\mathbf{x})=\det A$. Then

for all scalars $c$ and vectors $\mathbf{u},\mathbf{v}$. The first identity is the scaling rule applied to columns; the second follows from a cofactor expansion down column $j$, whose entries appear linearly. The determinant is therefore multilinear in its columns: linear in each one separately.

Testing independence with a determinant

The determinant test for invertibility turns any square set of vectors into a one-number independence test. The vectors ${\mathbf{v}}_1,\dots ,{\mathbf{v}}_n$ in ${\mathbb{R}}^n$ are the columns of a square matrix $V=[{\mathbf{v}}_1\cdots {\mathbf{v}}_n]$; they are linearly independent exactly when $V$ is invertible, that is, exactly when $\det V\ne 0$. A single determinant decides it.

No echelon form or spanning argument was needed beyond the determinant itself. The same test drives the characteristic equation, where $\det (A-\lambda I)=0$ is the condition that $A-\lambda I$ have dependent columns.

Cost of the row-reduction method

Evaluating an $n\times n$ determinant by row reduction takes about $\frac{2}{3}{n}^3$ arithmetic operations, against the $\sim n!$ of cofactor expansion.¹ A $25\times 25$ determinant needs roughly $10,000$ operations by reduction — a fraction of a second — versus the $500,000$ years cofactor expansion would demand. Most software computes $\det A$ through this factorization route; the determinant is essentially a byproduct of the same LU factorization used to solve $A\mathbf{x}=\mathbf{b}$.

Summary

• Row operations: replacement leaves $\det$ unchanged, interchange negates it, scaling a row by $k$ multiplies it by $k$.
• Reducing $A$ to echelon form $U$ with $r$ interchanges gives $\det A=(-1{)}^r\prod u_{ii}$, the product of the pivots up to sign — an $\sim n^3$ algorithm.
• Determinant test for invertibility: $A$ is invertible if and only if $\det A\ne 0$; dependent rows or columns, or a repeated/zero line, force $\det A=0$.
• Transpose: $\det A^{\mathsf{T}}=\det A$, so every row rule has a column counterpart.
• Multiplicative property: $\det (AB)=(\det A)(\det B)$, yielding $\det (A^{-1})=1/\det A$, $\det (P^{-1}AP)=\det A$, and $\det (rA)=r^n\det A$; but $\det (A+B)\ne \det A+\det B$ in general.
• The determinant is multilinear in the columns of $A$.