Week 4: Determinants

Defining Properties#

The determinant of a square matrix $A$, written $\det(A)$ or $|A|$, is a single number that packages invertibility, orientation, and volume scaling into one scalar. You do not need a formula first: three rules completely determine $\det$ on every $n \times n$ matrix, and every other property (including the cofactor formula and the product of pivots) follows from them.

Rule 1 — Identity. The determinant of the identity is one:

The standard basis has unit volume and the standard orientation, so the number assigned to $I$ is $1$.

Rule 2 — Row exchange. Swapping any two rows of $A$ multiplies the determinant by $-1$:

A single swap flips orientation; an even number of swaps leaves the sign unchanged, and an odd number flips it.

Rule 3 — Linearity in each row. If you fix every row except row $i$, the map that sends that row to $\det(A)$ is linear. Explicitly: if row $i$ is a sum $u + v$, then $\det(A)$ is the sum of the two determinants obtained by putting $u$ or $v$ in row $i$ (other rows fixed). Scaling row $i$ by a scalar $c$ multiplies $\det(A)$ by $c$.

These three rules are the whole axiomatic definition. From them you can derive further facts without a closed formula:

• If two rows are equal, $\det(A) = 0$ (swap them and Rule 2 says the determinant equals its own negative).
• Adding a multiple of one row to another leaves $\det(A)$ unchanged (linearity plus the equal-row fact).
• A zero row forces $\det(A) = 0$ (scale that row by $0$).
• $\det(A) = 0$ if and only if $A$ is singular (not invertible) — elimination will produce a zero pivot, which we connect next.

The same three rules can be stated for columns instead of rows: the determinant is multilinear and alternating in the columns, and $\det(I) = 1$. In practice we use whichever view is convenient.

Determinant from Pivots#

Week 3 turned $A$ into an upper triangular matrix $U$ by elimination. Each elementary step either leaves the determinant unchanged or multiplies it by a known factor, so you can read $\det(A)$ off the pivots.

Recall the operations and their effect on $\det$:

| Elimination step | Effect on $\det$ | | --- | --- | | Subtract a multiple of one row from another | No change | | Swap two rows | Multiply by $-1$ | | Scale a row by $c$ | Multiply by $c$ (we rarely scale rows during plain elimination) |

After elimination without scaling, $U$ is upper triangular with the pivots $p_1, p_2, \ldots, p_n$ on the diagonal. For a triangular matrix the determinant is the product of the diagonal entries (expand along the first column, or use multilinearity on the standard basis). Therefore

where the sign is $-1$ raised to the number of row exchanges performed during elimination. Equivalently, if $PA = LU$ with $P$ a permutation matrix, then

since $\det(L) = 1$ for unit lower triangular $L$, and $\det(P) = \pm 1$ according as $P$ is an even or odd permutation.

Worked $2 \times 2$. For $A = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$, elimination subtracts $2$ times row 1 from row 2 (no swap) and produces pivots $2$ and $1$. So

The classical $2 \times 2$ formula $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$ gives $2\cdot 3 - 1\cdot 4 = 2$, matching.

Zero pivot means zero determinant. If elimination cannot produce a full set of nonzero pivots (even after all useful row exchanges), some diagonal entry of $U$ is zero, the product of pivots is zero, and $\det(A) = 0$. That is the algebraic test for singularity: $A$ is invertible if and only if $\det(A) \neq 0$.

The pivot product is the fastest way to compute a determinant by hand for small dense systems you already eliminate. The next section gives a recursive formula that does not require running elimination first.

Cofactor Expansion#

The cofactor expansion (Laplace expansion) expresses $\det(A)$ as a weighted sum of determinants of smaller matrices. Fix the first row for concreteness. For an $n \times n$ matrix $A = (a_{ij})$,

where the cofactor of entry $a_{1j}$ is

and the minor $M_{1j}$ is the determinant of the $(n-1) \times (n-1)$ matrix obtained by deleting row $1$ and column $j$ of $A$. The sign pattern $(-1)^{i+j}$ is the checkerboard of $+$ and $-$ starting with $+$ in the top-left corner.

You may expand along any row $i$ or column $j$:

Choose a row or column with many zeros to shorten the arithmetic.

Full $3 \times 3$ example. Expand

along the first row:

The third term is zero because $a_{13} = 0$. The remaining cofactors are

So

(Equivalently, writing the expansion with the minors visible: $1\cdot\det\begin{pmatrix}1&1\\2&1\end{pmatrix} - 2\cdot\det\begin{pmatrix}3&1\\0&1\end{pmatrix} + 0 = -1 - 6 = -7$.)

Check via pivots. Eliminate $A$ without row swaps: subtract $3$ times row 1 from row 2 to get $\begin{pmatrix} 1 & 2 & 0 \\ 0 & -5 & 1 \\ 0 & 2 & 1 \end{pmatrix}$; then add $\frac{2}{5}$ times the new row 2 to row 3. The pivots are $1$, $-5$, and $\frac{7}{5}$, and $1 \cdot (-5) \cdot \frac{7}{5} = -7$, matching the cofactor result.

Cofactor expansion is $O(n!)$ if applied naively to large $n$; for computation prefer elimination (product of pivots). For theory and small hand calculations — and for the characteristic polynomial in Week 9 — expansion is the right tool.

Cramer's Rule#

Cramer's rule solves a square system $Ax = b$ when $A$ is invertible, by writing each unknown as a ratio of two determinants. For each index $i$, form the matrix $A_i$ by replacing column $i$ of $A$ with the right-hand side $b$. Then

The formula is a direct consequence of multilinearity in the columns: the solution identity $A x = b$ says $b$ is the linear combination $x_1 a_1 + \cdots + x_n a_n$ of the columns of $A$; expanding $\det(A_i)$ and using alternating multilinear properties isolates $x_i \det(A)$.

Worked $2 \times 2$. Solve

We already have $\det(A) = 2$. Replace column 1 with $b$:

so $x_1 = 4/2 = 2$. Replace column 2 with $b$:

so $x_2 = 2/2 = 1$. Check: $2\begin{pmatrix}2\\4\end{pmatrix} + 1\begin{pmatrix}1\\3\end{pmatrix} = \begin{pmatrix}5\\11\end{pmatrix}$.

When to use it. Cramer's rule is excellent for $2 \times 2$ systems and for theoretical arguments that need an explicit formula for $x_i$. For $n \ge 3$ it is usually slower than elimination: you compute $n+1$ determinants instead of one factorization. If $\det(A) = 0$, the rule does not apply — the system has either no solution or infinitely many, which Week 5 will classify by rank.

Determinant as Volume#

There is a geometric meaning that makes the algebraic rules intuitive. If the columns of an $n \times n$ matrix $A$ are vectors $a_1, \ldots, a_n$ in $\mathbb{R}^n$, they span a parallelepiped (a parallelogram when $n = 2$). The absolute value of the determinant is the volume of that solid:

(The same statement holds with rows instead of columns.) The sign of $\det(A)$ encodes orientation: positive means the ordered basis $(a_1,\ldots,a_n)$ has the same orientation as the standard basis; negative means it is mirrored (an odd number of reflections or row/column swaps).

Unit cube. The columns of $I$ are the standard unit vectors, which span the unit cube of volume $1$, matching $\det(I) = 1$.

Axis-aligned box. If

the columns are edge vectors of lengths $1$, $2$, and $3$ along the coordinate axes. The box volume is $1 \cdot 2 \cdot 3 = 6$, and $\det(A) = 6$.

Linear maps scale volume. Applying $A$ to any region multiplies $n$-dimensional volume by $|\det(A)|$. In particular, if $\det(A) = 0$, the image of the unit cube is flattened into a lower-dimensional set of volume zero — the columns are linearly dependent, and $A$ is singular. That is the geometric reason a zero determinant means “no inverse”: you cannot reverse a map that has crushed volume to zero.

This volume picture returns in change-of-variables formulas (Jacobians) and in continuous densities on configuration spaces. Algebraically, the same scalar $\det(A)$ will appear in Week 9 as the constant term structure of the characteristic polynomial $\det(A - \lambda I) = 0$.

Knowledge check#

Browser lab#

Compute a determinant, verify the row-swap sign flip, and use it as a volume.

import numpy as np

A = np.array([[1., 2., 0.],
              [3., 1., 1.],
              [0., 2., 1.]])

print("det(A) =", np.linalg.det(A))

A_swapped = A.copy()
A_swapped[[0, 1]] = A_swapped[[1, 0]]
print("det after row swap =", np.linalg.det(A_swapped))

edges = np.array([[1., 0., 0.],
                   [0., 2., 0.],
                   [0., 0., 3.]])
print("box volume via det =", abs(np.linalg.det(edges)))