---
title: Properties of Determinants
module: Determinants
moduleNumber: 3
lessonNumber: 2
order: 302
summary: >
  Row operations act on the determinant in three predictable ways, and this
  turns row reduction into a fast algorithm: the determinant is the product of
  the pivots times a sign for the interchanges. The same properties yield the
  invertibility test det A is nonzero, the transpose identity, and the
  multiplicative law det(AB) equals det A times det B.
topics: [Determinants]
sources:
  - book: Lay
    ref: "§3.2 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 $\det A \neq 0$ 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

> **Theorem (Row operations).** Let $A$ be a square matrix.
> - If a multiple of one row of $A$ is added to another row to produce $B$, then
>   $\det B = \det A$.
> - If two rows of $A$ are interchanged to produce $B$, then $\det B = -\det A$.
> - If one row of $A$ is multiplied by $k$ to produce $B$, then
>   $\det B = k\det A$.

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.[^lay-32]

$$
% caption: The three elementary row operations and their exact effect on the
% determinant: replacement is free, interchange flips the sign, scaling by $k$
% scales the determinant by $k$.
\begin{tikzpicture}[scale=1.0, font=\small,
  card/.style={draw, minimum width=34mm, minimum height=22mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[card] (r) at (0,0) {replacement\\[6pt] $\det B = \det A$};
\node[card] (s) at (4.2,0) {interchange\\[6pt] $\det B$ negated};
\node[card, draw=acc, thick] (c) at (8.4,0) {scaling by $k$\\[6pt] $\det B = k\det A$};
\end{tikzpicture}
$$

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:

$$
\det A = (-1)^{r}\,\det U = (-1)^{r} \cdot (\text{product of the pivots in } U).
$$

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 $(-1)^r$.

$$
% caption: Row reduction to echelon form: replacements leave $\det$ unchanged,
% each interchange contributes a factor $-1$, and the triangular result is the
% product of its pivots.
\begin{tikzpicture}[scale=1.0, font=\small,
  m/.style={draw, minimum width=20mm, minimum height=14mm, align=center, font=\footnotesize}]
\definecolor{acc}{HTML}{4A6FA5}
\node[m] (a) at (0,0) {$A$};
\node[m] (b) at (3.4,0) {replace};
\node[m] (c) at (6.8,0) {swap};
\node[m, draw=acc, thick] (u) at (10.2,0) {$U$ echelon};
\draw[->, acc, thick] (a) -- (b) node[midway, above, font=\scriptsize] {det same};
\draw[->, acc, thick] (b) -- (c) node[midway, above, font=\scriptsize] {negated};
\draw[->, acc, thick] (c) -- (u) node[midway, above, font=\scriptsize] {det same};
\node[font=\footnotesize, anchor=north, align=center] at (10.2,-1.1)
  {det from pivot\\ product, up to sign};
\end{tikzpicture}
$$

> **Worked example.** Compute the determinant of
> $$
> A = \begin{bmatrix} 1 & -4 & 2 \\ -2 & 8 & -9 \\ -1 & 7 & 0 \end{bmatrix}
> $$
> by reduction to echelon form. Two replacements in column 1 — add $2R_1$ to
> $R_2$, add $R_1$ to $R_3$ — leave the determinant unchanged:
> $$
> \det A = \begin{vmatrix} 1 & -4 & 2 \\ 0 & 0 & -5 \\ 0 & 3 & 2 \end{vmatrix}.
> $$
> Interchanging rows 2 and 3 reverses the sign, producing a triangular matrix:
> $$
> \det A = -\begin{vmatrix} 1 & -4 & 2 \\ 0 & 3 & 2 \\ 0 & 0 & -5 \end{vmatrix}
>        = -(1)(3)(-5) = 15.
> $$
> One interchange means $r = 1$, hence the leading $-1$; the triangular
> determinant is the pivot product $(1)(3)(-5)$.

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.

> **Theorem (Determinant test for invertibility).** A square matrix $A$ is
> invertible if and only if $\det A \neq 0$.

This adds the statement "$\det A \neq 0$" to the list of equivalent conditions in
the [Invertible Matrix Theorem](/linear-algebra/matrix-algebra/matrix-inverse-and-invertibility).
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.

$$
A = \begin{bmatrix} 3 & -1 & 2 & -5 \\ 0 & 5 & -3 & -6 \\ -6 & 7 & -7 & 4 \\ -5 & -8 & 0 & 9 \end{bmatrix}
\;\xrightarrow{\,R_3 \,+\, 2R_1\,}\;
\begin{bmatrix} 3 & -1 & 2 & -5 \\ 0 & 5 & -3 & -6 \\ 0 & 5 & -3 & -6 \\ -5 & -8 & 0 & 9 \end{bmatrix}.
$$

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.[^lay-32] 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

$$
A = \begin{bmatrix} 0 & 1 & 2 & -1 \\ 2 & 5 & -7 & 3 \\ 0 & 3 & 6 & 2 \\ -2 & -5 & 4 & -2 \end{bmatrix},
$$

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 $(-1)^{2+1}$):

$$
\det A = -2\begin{vmatrix} 1 & 2 & -1 \\ 3 & 6 & 2 \\ 0 & -3 & 1 \end{vmatrix}.
$$

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:

$$
\det A = -2\begin{vmatrix} 1 & 2 & -1 \\ 0 & 0 & 5 \\ 0 & -3 & 1 \end{vmatrix}
       = (-2)(1)\begin{vmatrix} 0 & 5 \\ -3 & 1 \end{vmatrix}
       = -2\,(0 + 15) = -30.
$$

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.

> **Theorem (Transpose).** If $A$ is an $n \times n$ matrix, then
> $\det A^{\mathsf T} = \det A$.

$$
% caption: The transpose swaps the entry in position $(i,j)$ with the one in
% $(j,i)$; the main diagonal is fixed, so the first-row expansion of $\det A$
% is the first-column expansion of $\det A^{\mathsf T}$.
\begin{tikzpicture}[scale=0.85, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\begin{scope}
  \foreach \d in {0,1,2}
    \fill[acc!14] (\d-0.45,-\d+0.45) rectangle (\d+0.45,-\d-0.45);
  \foreach \r/\rr in {0/1,1/2,2/3}
    \foreach \c/\cc in {0/1,1/2,2/3}
      \node at (\c,-\r) {$a_{\rr\cc}$};
  \draw[black, thick] (-0.5,0.5) rectangle (2.5,-2.5);
  \node[anchor=north] at (1,-2.9) {$A$};
\end{scope}
\draw[->, acc, very thick] (3.1,-1) -- (4.5,-1) node[midway, above] {transpose};
\begin{scope}[xshift=5.4cm]
  \foreach \d in {0,1,2}
    \fill[acc!14] (\d-0.45,-\d+0.45) rectangle (\d+0.45,-\d-0.45);
  \foreach \r/\rr in {0/1,1/2,2/3}
    \foreach \c/\cc in {0/1,1/2,2/3}
      \node at (\c,-\r) {$a_{\cc\rr}$};
  \draw[black, thick] (-0.5,0.5) rectangle (2.5,-2.5);
  \node[anchor=north] at (1,-2.9) {$A$ transpose};
\end{scope}
\end{tikzpicture}
$$

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
$\det A^{\mathsf T}$.[^lay-32] 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.

> **Theorem (Multiplicative property).** If $A$ and $B$ are $n \times n$
> matrices, then $\det(AB) = (\det A)(\det B)$.

$$
% caption: The determinant is multiplicative: the determinant of a product equals
% the product of the determinants, so det carries matrix multiplication to scalar
% multiplication.
\begin{tikzpicture}[scale=1.0, font=\small,
  box/.style={draw, minimum width=16mm, minimum height=11mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box] (a) at (0,1.3) {$A$};
\node[box] (b) at (2.6,1.3) {$B$};
\node[box, draw=acc, thick] (ab) at (1.3,-0.6) {$AB$};
\node[box] (da) at (6.2,1.3) {$\det A$};
\node[box] (db) at (8.8,1.3) {$\det B$};
\node[box, draw=acc, thick] (dab) at (7.5,-0.6) {$\det(AB)$};
\draw[->, black] (a) -- (ab);
\draw[->, black] (b) -- (ab);
\draw[->, black] (da) -- (dab);
\draw[->, black] (db) -- (dab);
\draw[->, acc, thick] (2.4,0.4) -- (5.9,0.4) node[midway, above, font=\footnotesize] {$\det$};
\node[font=\footnotesize, anchor=north] at (7.5,-1.5) {$= (\det A)(\det B)$};
\end{tikzpicture}
$$

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.[^lay-32]

> **Worked example.** Verify $\det(AB) = (\det A)(\det B)$ for
> $A = \begin{bmatrix} 6 & 1 \\ 3 & 2 \end{bmatrix}$ and
> $B = \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}$. The product is
> $$
> AB = \begin{bmatrix} 25 & 20 \\ 14 & 13 \end{bmatrix}, \qquad
> \det(AB) = 25\cdot 13 - 20\cdot 14 = 45.
> $$
> Separately $\det A = 12 - 3 = 9$ and $\det B = 8 - 3 = 5$, and
> $9 \cdot 5 = 45$, matching.

> **Warning.** There is no analogous rule for sums. In general
> $\det(A + B) \neq \det A + \det B$. Determinants respect products, not addition.

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](/linear-algebra/eigenvalues/the-characteristic-equation)
depends on. And $\det(A^2) = (\det A)^2$, so if $A^2 = I$ then $\det A = \pm 1$.

| Identity | Reason |
| --- | --- |
| $\det(A^{-1}) = 1/\det A$ | $\det(A)\det(A^{-1}) = \det I = 1$ |
| $\det(P^{-1}AP) = \det A$ | product rule cancels $\det P^{-1}\cdot\det P$ |
| $\det(A^{\mathsf T}) = \det A$ | Transpose invariance |
| $\det(rA) = r^{n}\det A$ | scaling all $n$ rows by $r$ |
| $\det(AB) = \det(BA)$ | both equal $(\det A)(\det B)$ |

## 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

$$
T(c\mathbf x) = c\,T(\mathbf x), \qquad
T(\mathbf u + \mathbf v) = T(\mathbf u) + T(\mathbf v)
$$

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](/linear-algebra/linear-systems/linear-independence) exactly
when $V$ is invertible, that is, exactly when $\det V \neq 0$. A single determinant
decides it.

> **Worked example.** Decide whether
> $$
> \mathbf v_1 = \begin{bmatrix} 5 \\ -7 \\ 9 \end{bmatrix},\quad
> \mathbf v_2 = \begin{bmatrix} -3 \\ 3 \\ -5 \end{bmatrix},\quad
> \mathbf v_3 = \begin{bmatrix} 2 \\ -7 \\ 5 \end{bmatrix}
> $$
> are linearly independent. Adding row 1 to row 2 creates a zero in the second
> column, and a cofactor expansion down column 2 of the reduced matrix evaluates
> to
> $$
> \det[\,\mathbf v_1\ \mathbf v_2\ \mathbf v_3\,]
> = 3(35) + 5(-21) = 105 - 105 = 0.
> $$
> The determinant is zero, so the matrix is singular and the three vectors are
> linearly dependent.

No echelon form or spanning argument was needed beyond the
determinant itself. The same test drives the
[characteristic equation](/linear-algebra/eigenvalues/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 $\tfrac{2}{3}n^3$
arithmetic operations, against the $\sim n!$ of cofactor expansion.[^lay-32] 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](/linear-algebra/matrix-algebra/partitioned-matrices-and-lu)
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 $\det A \neq 0$; 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
  $\det(A+B) \neq \det A + \det B$ in general.
- The determinant is multilinear in the columns of $A$.

[^lay-32]: Lay, Lay & McDonald, _Linear Algebra and Its Applications_, 5th ed., §3.2 — Properties of Determinants: Theorem 3 (row operations) and its elementary-matrix proof, the pivot-product formula $\det A = (-1)^r\prod u_{ii}$, Theorem 4 (invertibility iff $\det A\neq0$), Theorem 5 ($\det A^{\mathsf T}=\det A$), Theorem 6 (multiplicative property) with the sum warning, the column-linearity property, and the $\tfrac{2}{3}n^3$ operation count.