Determinants, Area and Cofactors

One num­ber for a whole matrix#

Every square matrix car­ries one spe­cial num­ber, its deter­mi­nant. That sin­gle num­ber tells you whether the matrix can be inverted, and it also mea­sures how the matrix stretches areas. Here you will learn to eval­u­ate $2 \times 2$ and $3 \times 3$ deter­mi­nants, use them to find the area of a tri­an­gle and to test collinear­ity, and meet minors and cofac­tors, which we will need for inverses in the next les­son.

Deter­mi­nants#

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

For a $3 \times 3$ matrix, expand along the first row, like this:

$$\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2\begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3\begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix}.$$

In fact you may expand along any row or col­umn, using the sign pat­tern $\begin{smallmatrix} + & - & + \\ - & + & - \\ + & - & + \end{smallmatrix}$. A teacher's tip: always pick the row or col­umn with the most zeros. It saves both time and mis­takes.

Facts worth remem­ber­ing. $\lvert A^T \rvert = \lvert A \rvert$. Swap­ping two rows changes the sign. Two equal rows give $0$. For $A$ of order $n$, $\lvert kA \rvert = k^n\lvert A \rvert$. And $\lvert AB \rvert = \lvert A \rvert\lvert B \rvert$.

Area of a tri­an­gle#

$$\displaystyle \Delta = \frac{1}{2}\left\lvert \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right\rvert.$$

If the three points lie on one line, there is no tri­an­gle at all, so the points are collinear exactly when the deter­mi­nant is $0$.

Minors and cofac­tors#

The minor $M_{ij}$ is the deter­mi­nant left after delet­ing row $i$ and col­umn $j$. The cofac­tor attaches a sign to it: $A_{ij} = (-1)^{i+j}M_{ij}$. Mul­ti­ply the entries of any row by their own cofac­tors and add: you get the deter­mi­nant. Mul­ti­ply them by the cofac­tors of another row instead, and the sum is always $0$.

Try these your­self#

1. Eval­u­ate $\begin{vmatrix} 7 & 3 \\ -2 & 5 \end{vmatrix}$; $\begin{vmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{vmatrix}$.
2. Eval­u­ate $\begin{vmatrix} 1 & -1 & 2 \\ 3 & 0 & 1 \\ -2 & 4 & 5 \end{vmatrix}$ along the sec­ond row.
3. Find $x$ if $\begin{vmatrix} x & 4 \\ 3 & x \end{vmatrix} = \begin{vmatrix} 2 & 3 \\ -1 & 5 \end{vmatrix}$.
4. If $A$ is $3 \times 3$ with $\lvert A \rvert = 4$, find $\lvert 2A \rvert$ and $\lvert A^T \rvert$.
5. Find the area of the tri­an­gle $(-2, -1)$, $(3, 4)$, $(5, -2)$.
6. Show that $(1, -1)$, $(3, 5)$, $(-1, -7)$ are collinear.
7. Use deter­mi­nants to find the line through $(-1, 4)$ and $(3, -2)$.
8. Find the minors and cofac­tors of the sec­ond row of $\begin{vmatrix} 2 & -1 & 0 \\ 1 & 3 & 5 \\ 4 & 0 & -2 \end{vmatrix}$ and eval­u­ate the deter­mi­nant with them.
9. Show with­out expand­ing that $\begin{vmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{vmatrix} = 0$ (add col­umn $2$ to col­umn $3$).
10. Find $k$ if the area of the tri­an­gle $(1, k)$, $(4, 0)$, $(0, 2)$ is $5$.

Answers to check against#