Math Core

Cofactor expansion defines the determinant, but it is slow: a $10 \times 10$ matrix with no zeros would need millions of multiplications. In this lesson you'll see how row operations change a determinant, which gives a fast way to compute it and leads to the central theorem of the unit: a square matrix is invertible exactly when its determinant is nonzero.

How row operations change the determinant

You know from the last lesson that a triangular matrix has an easy determinant: the product of its diagonal. Row reduction turns any square matrix into a triangular one (an echelon form is upper triangular). So the only question is how each of the three elementary row operations affects the determinant.

Row operations and determinants

Let $A$ be a square matrix.

Replacement: if $B$ comes from $A$ by adding a multiple of one row to another row, then $\det B = \det A$ .
Interchange: if $B$ comes from $A$ by swapping two rows, then $\det B = -\det A$ .
Scaling: if $B$ comes from $A$ by multiplying one row by $k$ , then $\det B = k \det A$ .

You can check each rule on a $2 \times 2$ matrix. Swapping the rows of $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ gives $cb - da = -(ad - bc)$ . Scaling the first row by $k$ gives $(ka)d - (kb)c = k(ad - bc)$ . Adding $m$ times row 1 to row 2 gives $a(d + mb) - b(c + ma) = ad - bc$ , since the $mab$ terms cancel. The general proof uses cofactor expansion and induction on $n$ .

In terms of geometry, these rules make sense too. A replacement is a shear, and shears slide a parallelogram without changing its area. Scaling one edge by $k$ scales the area by $k$ . A swap reverses orientation, which flips the sign.

The scaling rule is usually used in reverse: you may factor a common number out of a single row, putting it in front of the determinant.

Worked example: Row reducing a 3 × 3 matrix

Compute $\det A$ for $A = \begin{bmatrix} 1 & 2 & -1 \\ 2 & 1 & 3 \\ -3 & 0 & 4 \end{bmatrix}$ .

Use only replacements, which don't change the determinant:

\begin{aligned} \det A &= \begin{vmatrix} 1 & 2 & -1 \\ 2 & 1 & 3 \\ -3 & 0 & 4 \end{vmatrix} = \begin{vmatrix} 1 & 2 & -1 \\ 0 & -3 & 5 \\ 0 & 6 & 1 \end{vmatrix} \quad (R_2 - 2R_1,\ R_3 + 3R_1) \\ &= \begin{vmatrix} 1 & 2 & -1 \\ 0 & -3 & 5 \\ 0 & 0 & 11 \end{vmatrix} \quad (R_3 + 2R_2) \\ &= 1(-3)(11) = -33. \end{aligned}

Worked example: A 4 × 4 matrix that needs a swap

Compute $\det \begin{bmatrix} 1 & -2 & 3 & 1 \\ 2 & -4 & 7 & 5 \\ -1 & 3 & -2 & 0 \\ 3 & -5 & 9 & 6 \end{bmatrix}$ .

Clear column 1 with $R_2 - 2R_1$ , $R_3 + R_1$ and $R_4 - 3R_1$ (no change):

\begin{vmatrix} 1 & -2 & 3 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 3 \end{vmatrix}.

The $(2, 2)$ entry is $0$ , so swap rows 2 and 3. That flips the sign:

-\begin{vmatrix} 1 & -2 & 3 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & 1 & 0 & 3 \end{vmatrix}.

Now $R_4 - R_2$ gives row 4 $= (0, 0, -1, 2)$ , and then $R_4 + R_3$ gives $(0, 0, 0, 5)$ :

-\begin{vmatrix} 1 & -2 & 3 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 5 \end{vmatrix} = -(1 \cdot 1 \cdot 1 \cdot 5) = -5.

Common mistake

When you scale a row during row reduction, the new determinant is $k$ times the old one, so to keep an equality you must write $\det A = \dfrac{1}{k}\det B$ . It is easy to multiply when you should divide. The safest habit is to avoid scaling entirely: use only replacements and swaps, then multiply the diagonal and attach $(-1)^{\text{number of swaps}}$ .

Determinants and invertibility

Suppose you reduce $A$ to an echelon form $U$ using only replacements and $r$ row swaps. Then

\det A = (-1)^r \det U = (-1)^r \cdot (\text{product of the diagonal entries of } U).

If $A$ is invertible, every column has a pivot, so every diagonal entry of $U$ is a nonzero pivot and $\det A \neq 0$ . If $A$ is not invertible, $U$ has at least one zero on its diagonal (in the square case, a missing pivot forces a zero row at the bottom), so $\det A = 0$ .

The invertibility test

A square matrix $A$ is invertible if and only if $\det A \neq 0$ .

This adds a new statement to the Invertible Matrix Theorem. In particular, if the columns (or rows) of $A$ are linearly dependent, then $\det A = 0$ . So a matrix with two equal rows, or with one row a multiple of another, has determinant $0$ without any computation.

Worked example: When is a matrix singular?

For which values of $k$ is $A = \begin{bmatrix} k & 1 & 0 \\ 1 & k & 1 \\ 0 & 1 & k \end{bmatrix}$ not invertible?

Expand across row 1:

\det A = k\begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} - 1\begin{vmatrix} 1 & 1 \\ 0 & k \end{vmatrix} = k(k^2 - 1) - k = k^3 - 2k = k(k^2 - 2).

This is zero when $k = 0$ or $k = \pm\sqrt{2}$ . For those three values $A$ is singular; for every other $k$ it is invertible.

Transposes and column operations

Transpose

For any square matrix $A$ , $\det A^T = \det A$ .

The reason: cofactor expansion across row 1 of $A^T$ is the same computation as expansion down column 1 of $A$ . The consequence is that every row rule has a column twin. Swapping two columns flips the sign, scaling a column scales the determinant, and adding a multiple of one column to another changes nothing. You may mix row and column operations freely when computing a determinant (never when solving a system, though).

Products, inverses and scalar multiples

The multiplicative property

If $A$ and $B$ are $n \times n$ matrices, then

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

Geometrically this is natural: if $B$ scales area by a factor of $\det B$ and $A$ scales area by $\det A$ , then doing $B$ and then $A$ scales area by the product. Two useful consequences follow.

Inverses. Since $AA^{-1} = I$ and $\det I = 1$ , you get $(\det A)(\det A^{-1}) = 1$ , so $\det(A^{-1}) = \dfrac{1}{\det A}$ .
Scalar multiples. The matrix $cA$ multiplies every row of $A$ by $c$ . That's $n$ scalings, so $\det(cA) = c^n \det A$ for an $n \times n$ matrix.

Worked example: Combining the properties

Let $A$ and $B$ be $3 \times 3$ matrices with $\det A = 3$ and $\det B = -2$ . Find $\det(2AB^T)$ and $\det(A^{-1}B^3)$ .

\det(2AB^T) = 2^3 \det A \det B^T = 8(3)(-2) = -48.

\det(A^{-1}B^3) = \frac{1}{\det A}(\det B)^3 = \frac{1}{3}(-8) = -\frac{8}{3}.

Common mistake

The determinant does not respect addition: in general $\det(A + B) \neq \det A + \det B$ . Try $A = B = I_2$ : $\det(2I) = 4$ , but $\det I + \det I = 2$ . The same example shows why $\det(cA)$ is $c^n\det A$ , not $c\det A$ .

Tip

Before computing any determinant, scan for shortcuts: a zero row or column, two proportional rows or columns (determinant $0$ ), or a triangular shape (multiply the diagonal).

Practice

Practice 1

Let $A$ be a $3 \times 3$ matrix with $\det A = 4$ . The matrix $B$ is obtained from $A$ by these steps, in order: add $-5$ times row 1 to row 2; swap rows 1 and 2; multiply row 3 by $3$ . Find $\det B$ .