Determinant: cofactor expansion and computing with row operations

The determinant assigns a single number to a square matrix, and that number says a lot about the matrix: if it is zero the matrix is singular, the system does not have a unique solution, the columns are dependent. This week we learn to compute the number; next week its properties and uses (Cramer, adjoint). The problem book’s favorite method is reducing to triangular form with row operations, so the emphasis is there.

Golden rules:

1. When expanding by cofactors, choose the row or column with the most zeros.
2. Draw the sign table in the corner of the page: $+$ at the top left, then a checkerboard.
3. In row operations, prefer $R_i \leftarrow R_i + cR_j$. It does not change the determinant.
4. If you swapped rows, track $(-1)$; if you multiplied a row by $c$, track the factor $c$.

2×2 and 3×3 determinants

$\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.$

$\det A$ and $|A|$ are two ways of writing the same thing. ($|A|$ here is not an absolute value; a determinant can be negative.)

Minor and cofactor

For an $n \times n$ matrix $A$:

• The minor $M_{ij}$: the determinant of the $(n-1) \times (n-1)$ matrix that remains when row $i$ and column $j$ are deleted from $A$.
• The cofactor $C_{ij} = (-1)^{i+j}M_{ij}$.

The sign $(-1)^{i+j}$ gives a checkerboard pattern:

$\begin{bmatrix} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{bmatrix}$

Cofactor expansion

The determinant can be expanded along any row or column and always gives the same result:

$\det A = a_{i1}C_{i1} + a_{i2}C_{i2} + \dots + a_{in}C_{in} \quad (\text{row } i), \qquad \det A = a_{1j}C_{1j} + \dots + a_{nj}C_{nj} \quad (\text{column } j).$

Since the term of a zero entry is zero anyway, choosing the row with the most zeros shortens the work.

Example 1. $A = \begin{bmatrix} 2 & -1 & 3 \\ 1 & 0 & 4 \\ -2 & 5 & 1 \end{bmatrix}$. There is a zero in column 2, so let us expand there. The signs for column 2 are $-, +, -$.

C12 = (-1)^(1+2) · det[1 4; -2 1] = -(1 + 8)  = -9
C22 = its factor a22 = 0, no need to compute
C32 = (-1)^(3+2) · det[2 3; 1 4]  = -(8 - 3)  = -5
det A = (-1)(-9) + 0 + (5)(-5) = 9 - 25 = -16

Sarrus’ rule (only 3×3). Rewrite the first two columns to the right. Add the products of the three downward diagonals and subtract the products of the three upward ones. For the same matrix: down $0 + 8 + 15 = 23$, up $0 + 40 - 1 = 39$, $\det = 23 - 39 = -16$. Sarrus’ rule does not work for $4 \times 4$.

Example 2 (4×4, use the zeros).

$\det\begin{bmatrix} 1 & 0 & 2 & 0 \\ 3 & 1 & 0 & 2 \\ 0 & 0 & 4 & 0 \\ 2 & 0 & 1 & 3 \end{bmatrix} = 4 \cdot (+1) \cdot \det\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 2 \\ 2 & 0 & 3 \end{bmatrix} = 4 \cdot 1 \cdot (3 - 0) = 12.$

Row 3 has only one entry ($a_{33} = 4$, sign $+$). We expanded the remaining $3 \times 3$ along row 1: only $1 \cdot (1 \cdot 3 - 2 \cdot 0) = 3$.

Triangular matrices

The determinant of an upper triangular, lower triangular or diagonal matrix is the product of the diagonal entries:

$\det\begin{bmatrix} 2 & 5 & -1 \\ 0 & -3 & 4 \\ 0 & 0 & 6 \end{bmatrix} = 2 \cdot (-3) \cdot 6 = -36.$

Why? If you expand along the first column, only $a_{11}$ remains, and the remaining matrix is triangular too; repeat. This rule is the basis of the next method.

Determinant with row operations

Cofactor expansion needs about $n!$ multiplications for $n \times n$; it is laborious for $4 \times 4$ and a nightmare for $5 \times 5$. A better way: make the matrix triangular with row operations, multiply the diagonal, and account for the effect of the operations you did.

In practice it is more natural to think of taking out a common factor from a row: $\begin{vmatrix} 2 & 4 \\ 1 & 3 \end{vmatrix} = 2\begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix}$.

Example 3. $\det A$, $A = \begin{bmatrix} 2 & 4 & -2 & 6 \\ 1 & 3 & 0 & 4 \\ 0 & 1 & 2 & 1 \\ 3 & 6 & 1 & 10 \end{bmatrix}$.

Take 2 out of row 1: $\det A = 2 \det A'$, where row 1 of $A'$ is $(1, 2, -1, 3)$. Then only addition operations:

$\begin{bmatrix} 1 & 2 & -1 & 3 \\ 1 & 3 & 0 & 4 \\ 0 & 1 & 2 & 1 \\ 3 & 6 & 1 & 10 \end{bmatrix} \xrightarrow[R_4 \leftarrow R_4 - 3R_1]{R_2 \leftarrow R_2 - R_1} \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 4 & 1 \end{bmatrix} \xrightarrow{R_3 \leftarrow R_3 - R_2} \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 4 & 1 \end{bmatrix} \xrightarrow{R_4 \leftarrow R_4 - 4R_3} \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

The product of the diagonal is 1. With the 2 we took out at the start: $\det A = 2 \cdot 1 = 2$.

PITFALL · “I reduced it to I, so det = 1”. Reducing $A$ to $I$ does not make $\det A = 1$. Along the way, every swap leaves a $(-1)$ and every scaling $\frac1c R_i$ leaves a factor $c$. Collect them all, then decide.

PITFALL · mixed operations. Written in one step, $R_2 \leftarrow 3R_2 - R_1$ looks innocent, but it hides $R_2 \leftarrow 3R_2$: the determinant is multiplied by 3. Harmless in systems, an error in determinants.

Column operations

Since $\det A^T = \det A$ (next week), the three rules above apply to columns as well. If one column is a multiple of another, the determinant is 0; swapping two columns flips the sign.

Quick signs that the determinant is zero

Situations that let you say $\det A = 0$ without computing:

Example: in $\begin{bmatrix} 1 & -2 & 3 \\ 4 & 0 & 5 \\ -2 & 4 & -6 \end{bmatrix}$, row 3 $= -2 \times$ row 1, so the determinant is 0 (Q9).

Parametric determinants

Questions like “find the values of $t$ for which $\det = 0$” serve two purposes: finding singularity (week 5) and finding eigenvalues (week 13, the same calculation $\det(A - \lambda I) = 0$). Method: catch a common factor with row operations and keep the factored form.

$\begin{vmatrix} t & 1 & 1 \\ 1 & t & 1 \\ 1 & 1 & t \end{vmatrix} \xrightarrow{R_1 \leftarrow R_1 + R_2 + R_3} \begin{vmatrix} t+2 & t+2 & t+2 \\ 1 & t & 1 \\ 1 & 1 & t \end{vmatrix} = (t+2)\begin{vmatrix} 1 & 1 & 1 \\ 1 & t & 1 \\ 1 & 1 & t \end{vmatrix}$

Then $R_2 \leftarrow R_2 - R_1$, $R_3 \leftarrow R_3 - R_1$: the diagonal is $1, t - 1, t - 1$. The result is $(t + 2)(t - 1)^2$. Much shorter than expanding everything and solving a cubic polynomial.

Common mistakes