Determinants

3.4. Determinants#

For a square nxn matrix A, the determinant is denoted in a few different ways:

(3.57)#\[\begin{equation} \det \vv{A} =|\vv{A}|=\begin{vmatrix} A_{11} & \cdots & A_{1n} \\ \vdots & & \vdots \\ A_{n1} & \cdots & A_{nn} \end{vmatrix} \end{equation}\]

The determinant is defined in a recursive way.For \(n=1\), \(|\vv{A}|= A_{11}\) ( the matrix element) For \(n \ge 2\), \(|\vv{A}|\) is defined as:

(3.58)#\[\begin{equation} |\vv{A}|= \sum_{j=1}^n A_{ij} C_{ij} = \sum_{j=1}^n (-1)^{i+j} A_{ij} M_{ij} \end{equation}\]

where i is any row of A, \(C_{ij}\) is the cofactor of A:

(3.59)#\[\begin{equation} C_{ij} = (-1)^{i+j} M_{ij}, \end{equation}\]

and \(M_{ij}\) is the minor of A. The minor is the determinant of the matrix obtained by removing row i and column j from A. Equivalently,

(3.60)#\[\begin{equation} |\vv{A}| = \sum_{i=1}^n A_{ij} C_{ij} = \sum_{i=1}^n (-1)^{i+j} A_{ij} M_{ij} \end{equation}\]

where now j is any column of A.

3.4.1. 2x2 matrix#

We will use the definition of the determinant to show this must be the case! Let’s use the first definition and pick the first row \(i=1\):

(3.62)#\[\begin{align} \begin{vmatrix}a & b \\ c & d\end{vmatrix} &= (-1)^{1+1} \cdot A_{11} M_{11} + (-1)^{1+2} A_{12} M_{12} \\ &= a \begin{vmatrix} \phantom{a} & \phantom{b} \\ \phantom{c} & d \end{vmatrix} - b \begin{vmatrix} \phantom{a} & \phantom{b} \\ c & \phantom{d} \end{vmatrix} \\ &= a d - b c \end{align}\]

3.4.2. Larger matrices#

The determinants of larger matrices can be computed by reducing them to sums of 2x2 determinants. To do this quickly, it can be helpful to envision \((-1)^{i+j}\) as a checkerboard of signs, then visualize the minors.

For example, to evaluate

(3.63)#\[\begin{equation} \begin{vmatrix}1 & 3 & 0 \\ 2 & 6 & 4 \\ -1 & 0 & 2 \end{vmatrix} \end{equation}\]

The sign matrix is:

(3.64)#\[\begin{equation} \begin{bmatrix}+ & - & + \\ - & + & - \\ + & - & + \end{bmatrix} \end{equation}\]

Even faster, start from the plus sign in the upper left corner, then alternate until you get to your chosen row or column! Let’s use row 3:

(3.65)#\[\begin{align} |\vv{A}| &= + (-1) \cdot \begin{vmatrix} 3 & 0 \\ 6 & 4 \end{vmatrix} - 0 \cdot \begin{vmatrix}1 & 0 \\ 2 & 4 \end{vmatrix} + 2 \cdot \begin{vmatrix}1 & 3 \\ 2 & 6 \end{vmatrix} \\ &= -(3 \cdot 4 - 0 \cdot 6) + 2(1 \cdot 6 - 2 \cdot 3) \\ &= -12 \end{align}\]

Note that the same result could be achieved using any row or column. For example, column 3 gives:

(3.66)#\[\begin{align} |\vv{A}|&= +(0) \cdot \begin{vmatrix}2 & 6 \\ -1 & 0 \end{vmatrix} - 4 \cdot \begin{vmatrix}1 & 3 \\ -1 & 0 \end{vmatrix} + 2 \cdot \begin{vmatrix}1 & 3 \\ 2 & 6 \end{vmatrix} \\ &= -4 \cdot (0+3) + 2 \cdot (6-6)\\ &= -12 \end{align}\]

It’s usually a good idea to expand along the row or column with the most zeros! For example, let’s evaluate

(3.67)#\[\begin{align} \begin{vmatrix}1 & -2 & 0 & 0 \\ 4 & 3 & 5 & 0 \\ 0 &2 & 7 & 5 \\ 0 & 0 & 2 & 0 \end{vmatrix} &= -2 \cdot \begin{vmatrix}1 & -2 & 0 \\ 4 & 3 & 0 \\ 0 &2 & 5 \end{vmatrix} \\ &= -2 \cdot 5 \cdot \begin{vmatrix}1 & -2 \\ 4 & 3 \end{vmatrix} \\ &= -10 \cdot (3+8) \\ &= -110 \end{align}\]

where we chose row 4, then column 3 to do the calculation faster!

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