How a Determinant Finds the Area of a Triangle in One Line

Determinants measure areas, test whether points lie on a line, and — through the adjoint — give a formula for the inverse of a matrix. These uses connect coordinate geometry with the algebra of matrices.

This part covers area and collinearity, the adjoint and its properties, and the inverse by the adjoint method.

**The area of the triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is half the absolute value of $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$, and the three points are collinear exactly when this determinant is zero.

Worked example (area).** Find the area of the triangle with vertices $(1,0)$, $(6,0)$ and $(4,3)$.

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

The area is 7.5 square units, matching $\frac{1}{2}\times 5\times 3$.

Worked example (collinearity). For $(2,5)$, $(4,9)$ and $(7,15)$, the determinant is $2(9-15)-5(4-7)+1(60-63)=-12+15-3=0$, so the points are collinear.

Worked example (unknown vertex). If the triangle with vertices $(k,0)$, $(4,0)$ and $(0,2)$ has area 4, the determinant is $8-2k$, so $\frac{1}{2}|8-2k|=4$ gives $k=0$ or $k=8$.

An everyday example. A surveyor finding the area of a triangular plot from its corner coordinates on a village map uses exactly this determinant.

The substance. Take the absolute value — the determinant is negative when the vertices are listed clockwise, but an area cannot be negative.

**The adjoint of a square matrix A is the transpose of its matrix of cofactors, and its central property is $A(\text{adj}A)=(\text{adj}A)A=|A|I$.

For order 2.** If $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, then $\text{adj}A=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ — swap the diagonal entries and change the signs of the other two.

Properties for a matrix of order n:

- $A(\text{adj}A)=|A|I$
- $|\text{adj}A|=|A{|}^{n-1}$
- $\text{adj}(AB)=(\text{adj}B)(\text{adj}A)$

Worked example. For $A=\left[\begin{array}{ccc}1& 2& 0\\ 0& 1& 3\\ 2& 0& 1\end{array}\right]$, $|A|=1(1-0)-2(0-6)+0=13$. The cofactors are

- Row 1: $C_{11}=1$, $C_{12}=6$, $C_{13}=-2$
- Row 2: $C_{21}=-2$, $C_{22}=1$, $C_{23}=4$
- Row 3: $C_{31}=6$, $C_{32}=-3$, $C_{33}=1$

$$\text{adj}A=\left[\begin{array}{ccc}1& -2& 6\\ 6& 1& -3\\ -2& 4& 1\end{array}\right]$$

Check one entry of $A(\text{adj}A)$: row 1 of A times column 1 of the adjoint is $1(1)+2(6)+0(-2)=13=|A|$.

An everyday example. A spreadsheet that fills every cell of a new table from a small calculation on the old one builds its result entry by entry, just as the adjoint is built from cofactors.

The substance. Forgetting the transpose is the classic error — the cofactor from row i and column j goes into row j and column i of the adjoint.

**If $|A|\ne 0$, the inverse is $A^{-1}=\frac{1}{|A|}\text{adj}A$, which follows directly from $A(\text{adj}A)=|A|I$; if $|A|=0$, no inverse exists.

Worked example (order 2).** For $A=\left[\begin{array}{cc}3& 4\\ 2& 5\end{array}\right]$, $|A|=15-8=7$, so

$$A^{-1}=\frac{1}{7}\left[\begin{array}{cc}5& -4\\ -2& 3\end{array}\right]$$

Check: the top-left entry of $A{A}^{-1}$ is $\frac{1}{7}(15-8)=1$.

Worked example (order 3). Using the adjoint found above, with $|A|=13$:

$$A^{-1}=\frac{1}{13}\left[\begin{array}{ccc}1& -2& 6\\ 6& 1& -3\\ -2& 4& 1\end{array}\right]$$

Check row 2 of A times column 2 of the adjoint: $0(-2)+1(1)+3(4)=13$, which becomes 1 after dividing by 13.

Useful properties: $(AB{)}^{-1}=B^{-1}{A}^{-1}$, $(A^T{)}^{-1}=(A^{-1}{)}^T$ and $|A^{-1}|=\frac{1}{|A|}$.

An everyday example. **Decoding a message scrambled by a known $2\times 2$ matrix means multiplying by its inverse, found in one line from the adjoint.

The substance. Always check $|A|$ first** — computing nine cofactors for a singular matrix is wasted effort.