Determinants - Essential Mathematics for Chemical Engineers

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

\det \vv{A} =|\vv{A}|=\begin{vmatrix} A_{11} & \cdots & A_{1n} \\ \vdots & & \vdots \\ A_{n1} & \cdots & A_{nn} \end{vmatrix}

(1)

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:

|\vv{A}|= \sum_{j=1}^n A_{ij} C_{ij} = \sum_{j=1}^n (-1)^{i+j} A_{ij} M_{ij}

(2)

where i is any row of A, $C_{ij}$ is the cofactor of A:

C_{ij} = (-1)^{i+j} M_{ij},

(3)

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,

|\vv{A}| = \sum_{i=1}^n A_{ij} C_{ij} = \sum_{i=1}^n (-1)^{i+j} A_{ij} M_{ij}

(4)

where now j is any column of A.

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 $i=1$ :

\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}

(6)

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

\begin{vmatrix}1 & 3 & 0 \\ 2 & 6 & 4 \\ -1 & 0 & 2 \end{vmatrix}

(7)

The sign matrix is:

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

(8)

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:

\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}

(9)

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

\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}

(10)

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

\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}

(11)

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

3.4 Determinants

2x2 matrix¶

Larger matrices¶