3.2 Properties of determinants

Big idea

Cofactor expansion defines the determinant but is too slow to use past a small matrix. Row reduction is fast, and it turns out each of the three row operations changes the determinant in a completely predictable way. Put those two facts together and you get the working method: reduce to triangular form, keep a running record of what you did, multiply the diagonal, and undo the recorded changes.

The rules are short. Adding a multiple of one row to another leaves the determinant alone. Swapping two rows flips its sign. Multiplying a row by a number multiplies the determinant by that same number. Since the first operation is the one row reduction uses most, most of the work costs nothing at all.

The second half of the section collects the rules that let you compute with determinants instead of computing them. A transpose does not change the determinant. A product of matrices has the product of the determinants. From those two you get the determinant of a power, of an inverse, and of a scalar multiple, without touching the entries.

Decoder

A row replacement leaves the determinant unchanged, a row interchange reverses its sign, and scaling a row by $k$ scales the determinant by $k$.

“Row replacement” is the operation that adds a multiple of one row to a different row. It is the workhorse of elimination, and it is free. “Interchange” is swapping two rows, and each swap costs one sign flip, so two swaps cancel. “Scaling” multiplies every entry of one row by the same nonzero number, and the determinant picks up exactly that factor once, not once per entry.

The catch to watch is the last one. If you divide a row by $3$ to make a pivot into a $1$, the determinant of the new matrix is one third of the old one, so the original determinant is three times what you end up with. Row reduction that only uses replacements and swaps needs no bookkeeping beyond counting swaps.

Definitions and results

Effect of the three row operations. Let $B$ come from $A$ by one row operation. If the operation adds a multiple of one row to another, $\det B = \det A$. If it interchanges two rows, $\det B = -\det A$. If it multiplies one row by $k$, $\det B = k \det A$.

Determinant by row reduction. Reduce $A$ to an echelon form $U$ using only replacements and $r$ interchanges. Then $\det A = (-1)^r \det U$, and since $U$ is triangular, $\det U$ is the product of its diagonal entries, the pivots. If you also scaled rows, divide out each scaling factor at the end.

Invertibility test. $A$ is invertible exactly when $\det A \ne 0$. The reduction says why: $A$ is invertible exactly when its echelon form has $n$ pivots, and a triangular matrix with a zero on the diagonal has determinant zero.

Free zero determinants. If two rows of $A$ are equal, or one row is a multiple of another, or a row is all zeros, then $\det A = 0$. Each of these makes the reduction produce a zero row. The same holds with “column” in place of “row”.

Transpose. $\det A^T = \det A$. This is the reason every row rule has a matching column rule: run the row rule on $A^T$. So column replacements are free, column swaps flip the sign, and scaling a column scales the determinant.

Products. For square matrices of the same size, $\det(AB) = (\det A)(\det B)$. There is no comparable rule for sums; $\det(A + B)$ is not $\det A + \det B$ in general.

Consequences of the product rule. $\det(A^k) = (\det A)^k$. If $A$ is invertible then $\det(A^{-1}) = 1/\det A$, since $A A^{-1} = I$ and $\det I = 1$. For an $n \times n$ matrix and a scalar $c$, $\det(cA) = c^n \det A$, because scaling the whole matrix scales each of the $n$ rows.

Linearity in one column. Hold every column of a matrix fixed except one, and the determinant is a linear function of that remaining column: it respects sums and scalar multiples there. This is the property that drives the volume and Cramer results in the next section.

Worked examples

Reducing to triangular form

Take

$$ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 3 & 7 & 11 \end{bmatrix} $$

Subtract twice row one from row two, then three times row one from row three. Both are replacements, so the determinant does not move:

$$ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 1 & 2 \end{bmatrix} $$

Subtract row two from row three, another free operation:

$$ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} $$

No swaps, no scaling, so $\det A$ is the product of the diagonal: $1 \cdot 1 \cdot 1 = 1$.

Check by cofactor expansion along the first row: $1(5 \cdot 11 - 7 \cdot 7) - 2(2 \cdot 11 - 7 \cdot 3) + 3(2 \cdot 7 - 5 \cdot 3) = 1(6) - 2(1) + 3(-1) = 1$. The two methods agree, and the reduction needed three subtractions against the expansion’s three $2 \times 2$ determinants.

Counting a row swap

Take

$$ B = \begin{bmatrix} 0 & 2 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & -1 \end{bmatrix} $$

The top-left entry is zero, so swap rows one and two. That is one interchange, so record a factor of $-1$:

$$ \begin{bmatrix} 1 & 1 & 3 \\ 0 & 2 & 1 \\ 2 & 0 & -1 \end{bmatrix} $$

Subtract twice row one from row three: the last row becomes $0, -2, -7$. Add row two to row three: the last row becomes $0, 0, -6$. Both replacements are free, so the triangular form is

$$ \begin{bmatrix} 1 & 1 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & -6 \end{bmatrix} $$

Its determinant is $1 \cdot 2 \cdot (-6) = -12$. One swap happened, so $\det B = (-1)^1(-12) = 12$.

Check by expanding $B$ along its first row: $-2 \cdot \det\begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} + 1 \cdot \det\begin{bmatrix} 1 & 1 \\ 2 & 0 \end{bmatrix} = -2(-1 - 6) + (0 - 2) = 14 - 2 = 12$. Matches.

Computing with determinants instead of entries

Take

$$ C = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}, \qquad \det C = 6 - 1 = 5 $$

Now read off four more determinants with no new arithmetic on entries. The square: $\det(C^2) = 5^2 = 25$. The scalar multiple: $C$ is $2 \times 2$, so $\det(3C) = 3^2 \cdot 5 = 45$. The inverse: $\det(C^{-1}) = 1/5$. The transpose: $\det(C^T) = 5$.

Check the first two directly. $C^2 = \begin{bmatrix} 5 & 5 \\ 5 & 10 \end{bmatrix}$, whose determinant is $50 - 25 = 25$. And $3C = \begin{bmatrix} 6 & 3 \\ 3 & 9 \end{bmatrix}$, whose determinant is $54 - 9 = 45$. Both agree.

Note what the scalar rule does not say. Tripling a $2 \times 2$ matrix multiplies its determinant by $9$, not by $3$, because both rows got scaled. For a $3 \times 3$ matrix the factor would be $27$.

Practice

First the rules on their own: predict how a stated row operation, a transpose, a product, or a scalar multiple changes a known determinant.

Then the method: reduce to triangular form, track swaps and scalings, and multiply the diagonal.

Videos

Watch the part where a negative determinant is explained as an orientation flip. That is the geometric content of the sign change from a row interchange.

Quiz

Five items on row operations, reduction to triangular form, and the product and transpose rules.