How One Determinant Tells You Whether Three Points Lie on a Line

Every square matrix has a determinant — one number, written $|A|$ or $\det A$, that reveals a lot: whether the matrix can be inverted, how it scales areas, and whether equations built from it have a unique solution.

This part covers evaluating determinants, minors and cofactors, and using determinants for areas and collinearity.

**A $2\times 2$ determinant is $ad-bc$, and a $3\times 3$ determinant is found by expanding along any row or column: multiply each entry by the $2\times 2$ determinant left after deleting its row and column, using the sign pattern $+,-,+$.

Order 2:**

$$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc\left|\begin{array}{cc}5& 2\\ 3& 4\end{array}\right|=20-6=14$$

Sign pattern for order 3:

$$\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}$$

Worked example — along row 1.

$$\Delta =\left|\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 1& 0& 6\end{array}\right|=1(24-0)-2(0-5)+3(0-4)=24+10-12=22$$

Same determinant, along column 1, which contains a zero:

$$\Delta =1(24-0)-0+1(10-12)=24-2=22$$

Choose wisely. Expanding along the row or column with the most zeros saves work.

An everyday example. Checking a shop bill by adding down the columns instead of across the rows gives the same total — just as every row or column expansion gives the same determinant.

The substance. Only square matrices have determinants; a $2\times 3$ matrix has none.

**The minor $M_{ij}$ of an entry is the determinant left after deleting its row and column, its cofactor is $A_{ij}=(-1{)}^{i+j}{M}_{ij}$, and a determinant equals the sum of the entries of any row or column each multiplied by its own cofactor.

Worked example — minors and cofactors of row 1** of $\Delta$ above:

$$M_{11}=\left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|=24,A_{11}=24$$

$$M_{12}=\left|\begin{array}{cc}0& 5\\ 1& 6\end{array}\right|=-5,A_{12}=(-1{)}^3(-5)=5$$

$$M_{13}=\left|\begin{array}{cc}0& 4\\ 1& 0\end{array}\right|=-4,A_{13}=-4$$

Expanding with cofactors:

$$\Delta =a_{11}{A}_{11}+a_{12}{A}_{12}+a_{13}{A}_{13}=1(24)+2(5)+3(-4)=22$$

A useful fact. Entries of one row multiplied by the cofactors of a different row always add up to zero. Row 2 with row 1's cofactors gives $0(24)+4(5)+5(-4)=0$.

An everyday example. Covering one row and one column of a classroom seating chart with two rulers leaves a smaller chart — that is how a minor is formed.

The substance. A minor and its cofactor differ only in sign, and they are equal exactly when $i+j$ is even.

**The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is half the absolute value of the determinant of its coordinates, and the three points are collinear exactly when that determinant is zero.

Formula:**

$$\text{Area}=\frac{1}{2}\left|\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\right|$$

Worked example — area. Vertices $(1,0)$, $(6,0)$ and $(4,3)$, in centimetres:

$$\left|\begin{array}{ccc}1& 0& 1\\ 6& 0& 1\\ 4& 3& 1\end{array}\right|=1(0-3)-0+1(18-0)=15\text{Area}=\frac{15}{2}=7.5{\text{ cm}}^2$$

Check: the base is $5$ cm and the height $3$ cm, so the area is $\frac{1}{2}\times 5\times 3=7.5$ cm${}^2$.

Worked example — collinearity. Are $(1,2)$, $(3,6)$ and $(5,10)$ collinear?

$$\left|\begin{array}{ccc}1& 2& 1\\ 3& 6& 1\\ 5& 10& 1\end{array}\right|=1(6-10)-2(3-5)+1(30-30)=0$$

The area is zero, so the points are collinear.

Finding an unknown vertex. Set the determinant equal to **$\pm$ twice the given area and solve both equations.

An everyday example. A surveyor who knows the coordinates of three corners of a triangular plot can compute its area directly, without measuring any height.

The substance. Area is never negative** — the determinant's sign only reflects the order in which the vertices are listed.