Determinants

The determinant of an square matrix is a number which is computed from the entries of the matrix. If A is a square matrix, the determinant of A is denoted by $\det A$ or $|A|$ .

For $2 \times 2$ matrices, the formula is

$$\left|\matrix{a & b \cr c & d \cr}\right| = a d - b c.$$

Think of the following picture:

$$\hbox{\epsfysize=0.5in \epsffile{determinants-1.eps}}$$

Example. Compute the determinant of $\displaystyle \left[\matrix{3 & -2 \cr 5 & 4 \cr}\right]$ .

$$\left|\matrix{3 & -2 \cr 5 & 4 \cr}\right| = (3)(4) - (5)(-2) = 12 - (-10) = 22. \quad\halmos$$

For matrices which are $3 \times 3$ or larger, you can compute the determinant using a recursive algorithm called expansion by cofactors.

Example. Compute the determinant of

$$\left[\matrix{1 & -1 & 3 \cr 4 & 5 & 2 \cr 1 & 0 & 7 \cr}\right].$$

First, pick any row or column. It's usually good to pick a row or column with lots of zeros. I'll use column 2.

Go down column 2 one element at a time. For each element:

1. Cross out the row and column containing the element to leave a $2 \times 2$ matrix.

2. Find the product of the element, the determinant of the $2 \times 2$ matrix, and a plus or minus sign. The sign is determined by a "checkboard pattern":

$$\left[\matrix{ + & - & + \cr - & + & - \cr + & - & + \cr}\right]$$

The determinant is the sum of the products.

I'll work through the steps one element at a time. Cross out the row and column containing -1:

$$\left|\matrix{\cdot & {\bf -1} & \cdot \cr 4 & \cdot & 2 \cr 1 & \cdot & 7 \cr}\right|$$

Compute the $2 \times 2$ determinant:

$$\left|\matrix{4 & 2 \cr 1 & 7 \cr}\right| = 26.$$

Multiply the element, the $2 \times 2$ determinant, and a minus sign:

$$(-1)\cdot 26\cdot (-1) = 26.$$

Cross out the row and column containing 5:

$$\left|\matrix{1 & \cdot & 3 \cr \cdot & {\bf 5} & \cdot \cr 1 & \cdot & 7 \cr}\right|$$

Compute the $2 \times 2$ determinant:

$$\left|\matrix{1 & 3 \cr 1 & 7 \cr}\right| = 4.$$

Multiply the element, the $2 \times 2$ determinant, and a plus sign:

$$5\cdot 4\cdot (+1) = 20.$$

Cross out the row and column containing 0:

$$\left|\matrix{1 & \cdot & 3 \cr 4 & \cdot & 2 \cr \cdot & {\bf 0} & \cdot \cr}\right|$$

Compute the $2 \times 2$ determinant:

$$\left|\matrix{1 & 3 \cr 4 & 2 \cr}\right| = -10.$$

Multiply the element, the $2 \times 2$ determinant, and a minus sign:

$$0\cdot (-10)\cdot (-1) = 0.$$

The total is $26 + 20 + 0 = 46$ .

As an exericse, try expanding the determinant of this matrix using another row or column and see if you get the same answer.

Example. Compute $\displaystyle \left|\matrix{-3 & 1 & 1 \cr 1 & 2 & 1 \cr 5 & -1 & 3 \cr}\right]$ .

I'll expand by cofactors of the first row:

$$\left|\matrix{ -3 & 1 & 1 \cr 1 & 2 & 1 \cr 5 & -1 & 3 \cr}\right| = (-3) \left|\matrix{2 & 1 \cr -1 & 3 \cr}\right] - (1) \left|\matrix{1 & 1 \cr 5 & 3 \cr}\right| + (1) \left|\matrix{1 & 2 \cr 5 & -1 \cr}\right| =$$

$$(-3)(7) - (1)(-2) + (1)(-11) = -30.\quad\halmos$$

Notice that expansion by cofactors reduces the computation of an $n \times n$ determinant to the computation of n $(n - 1) \times (n - 1)$ determinants.

There are other approaches you can use to compute determinants. For instance, in linear algebra you learn about row reduction, which provides a fairly efficient way to compute determinants for larger matrices.

Here are some properties of determinants.

Proposition. (a) If A has two equal rows, then $\det A = 0$ .

(b) If two rows of A are swapped, the value of the determinant is multiplied by -1:

$$\det \left[\matrix{ \vdots \cr \leftarrow \vec{x} \rightarrow \cr \vdots \cr \leftarrow \vec{y} \rightarrow \cr \vdots \cr}\right] = -\det \left[\matrix{ \vdots \cr \leftarrow \vec{y} \rightarrow \cr \vdots \cr \leftarrow \vec{x} \rightarrow \cr \vdots \cr}\right].$$

(c) The determinant of a sum is the sum of the determinants one row at a time:

$$\det \left[\matrix{ \vbox{\vphantom{$\dfrac{x}{x}$}} (\hbox{FOO}) \cr \leftarrow \vec{x} + \vec{y} \rightarrow \cr \vbox{\vphantom{$\dfrac{x}{x}$}} (\hbox{BAR}) \cr}\right] = \det \left[\matrix{ \vbox{\vphantom{$\dfrac{x}{x}$}} (\hbox{FOO}) \cr \leftarrow \vec{x} \rightarrow \cr \vbox{\vphantom{$\dfrac{x}{x}$}} (\hbox{BAR}) \cr}\right] + \det \left[\matrix{ \vbox{\vphantom{$\dfrac{x}{x}$}} (\hbox{FOO}) \cr \leftarrow \vec{y} \rightarrow \cr \vbox{\vphantom{$\dfrac{x}{x}$}} (\hbox{BAR}) \cr}\right].$$

(The parts of the matrices labelled "FOO" and "BAR" are the same in all 3 matrices: They don't change. The sum occurs in a single row.)

(d) A number may be factored out of one row of a determinant at a time:

$$\det \left[\matrix{ \vbox{\vphantom{$\dfrac{x}{x}$}} (\hbox{FOO}) \cr \leftarrow k \cdot \vec{x} \rightarrow \cr \vbox{\vphantom{$\dfrac{x}{x}$}} (\hbox{BAR}) \cr}\right] = k \cdot \det \left[\matrix{ \vbox{\vphantom{$\dfrac{x}{x}$}} (\hbox{FOO}) \cr \leftarrow \vec{x} \rightarrow \cr \vbox{\vphantom{$\dfrac{x}{x}$}} (\hbox{BAR}) \cr}\right].$$

(e) The determinant of a product is the product of the determinants:

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

(f) The determinant of the $n \times n$ identity matrix is 1:

$$\det \left[\matrix{ 1 & 0 & \cdots & 0 \cr 0 & 1 & \cdots & 0 \cr \vdots & \vdots && \vdots \cr 0 & 0 & \cdots & 1 \cr}\right] = 1.$$

Proof. The proofs that these properties hold for arbitrary $n \times n$ matrices are fairly involved; you'd see them in a course in linear algebra.

I'll verify that a couple of the properties hold in some special cases.

As an example of (a), here's the determinant of a $3 \times 3$ matrix with two equal rows, which I'm computing by expanding by cofactors of row 1:

$$\det \left[\matrix{ a & b & c \cr d & e & f \cr d & e & f \cr}\right] = a \cdot \det \left[\matrix{e & f \cr e & f \cr}\right] - b \cdot \det \left[\matrix{d & f \cr d & f \cr}\right] + c \cdot \det \left[\matrix{d & e \cr d & e \cr}\right] =$$

$$a (e f - f e) - b (d f - f d) + c (d e - e d) = 0.$$

Here's an example of (e) with $2 \times 2$ matrices,

$$\left[\matrix{x + x' & y + y' \cr a & b \cr}\right] = (x + x') \cdot b - (y + y') \cdot a = (x b - y a) + x' b - y' a) = \det \left[\matrix{x & y \cr a & b \cr}\right] + \det \left[\matrix{x' & y' \cr a & b \cr}\right].\quad\halmos$$

Some of these properties are illustrated in the following examples.

Example. Suppose that

$$\left|\matrix{a & b \cr c & d \cr}\right| = 10.$$

Compute:

(a) $\displaystyle \left|\matrix{c & d \cr a & b \cr}\right|$ .

(b) $\displaystyle \left|\matrix{3 a & 3 b \cr 2 c & 2 d \cr}\right|$ .

(c) $\displaystyle \left|\matrix{4 a + c & 4 b + d \cr c & d \cr}\right|$ .

(a) Swapping two rows multiplies the determinant by -1:

$$\left|\matrix{c & d \cr a & b \cr}\right| = -\left|\matrix{a & b \cr c & d \cr}\right| = -10.\quad\halmos$$

(b) Factor 3 out of the first row, then factor 2 out of the second row:

$$\left|\matrix{3 a & 3 b \cr 2 c & 2 d \cr}\right| = 3 \cdot \left|\matrix{a & b \cr 2 c & 2 d \cr}\right| = 2 \cdot 3 \left|\matrix{a & b \cr c & d \cr}\right| = 6 \cdot 10 = 60.\quad\halmos$$

(c) Break up the determinant using a sum of $(4 a, 4 b)$ and $(c, d)$ in the first row. Factor 4 out of the first determinant; the second determinant is 0 because the matrix has two equal rows.

$$\left|\matrix{4 a + c & 4 b + d \cr c & d \cr}\right| = \left|\matrix{4 a & 4 b \cr c & d \cr}\right| + \left|\matrix{c & d \cr c & d \cr}\right| = 4 \cdot \left|\matrix{a & b \cr c & d \cr}\right| + 0 = 4 \cdot 10 = 40.\quad\halmos$$

Example. Give specific $2 \times 2$ matrices A and B for which

$$\det (A + B) \ne \det A + \det B.$$

$$\det \left(\left[\matrix{1 & 0 \cr 0 & 1 \cr}\right] + \left[\matrix{-1 & 0 \cr 0 & -1 \cr}\right]\right) = \det \left[\matrix{0 & 0 \cr 0 & 0 \cr}\right] = 0.$$

$$\det \left[\matrix{1 & 0 \cr 0 & 1 \cr}\right] + \det \left[\matrix{-1 & 0 \cr 0 & -1 \cr}\right] = 1 + 1 = 2.\quad\halmos$$

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