4.1 Determinants and the Area of a Triangle

Solve the pair of equations $a_1x+b_1y=c_1$, $a_2x+b_2y=c_2$ by elimination, as you did in Class 10. Multiply the first by $b_2$, the second by $b_1$ and subtract: $(a_1b_2-a_2b_1)\,x=c_1b_2-c_2b_1$. Doing the same for $y$ gives $(a_1b_2-a_2b_1)\,y=a_1c_2-a_2c_1$. Everything hangs on one number, $a_1b_2-a_2b_1$, built from the four coefficients. If it is not zero, you can divide and there is exactly one solution. If it is zero, the two lines are parallel or the same line, and there is no unique solution. The number determines the fate of the system — which is where its name comes from.

That number is the determinant of the coefficient array $\begin{bmatrix}a_1&b_1\\a_2&b_2\end{bmatrix}$. In this section you will meet determinants of order 1, 2 and 3, learn to expand a $3\times3$ determinant along whichever row or column makes the arithmetic shortest, and see a beautiful geometric use: the area of a triangle, and a one-line test for three points lying on a straight line. Sections 4.2 and 4.3 then use determinants to invert matrices and solve systems of equations.

Determinants of Order 1 and 2

Determinants are defined only for square matrices (same number of rows as columns; see Section 3.1 for orders of matrices). A determinant is a number, while a matrix is an array of numbers — keep the two ideas apart.

The chapter teaser showed the geometry behind Definition 4.1: if the columns $(a_{11},a_{21})$ and $(a_{12},a_{22})$ are drawn as arrows from the origin, $|A|$ is the area of the parallelogram they span, with a $+$ sign when the turn from the first column to the second is anticlockwise and a $-$ sign when it is clockwise. In particular $|A|=0$ exactly when the two columns lie along one line.

Determinants of Order 3: Expansion along Any Row or Column

A $3\times3$ determinant is built from three $2\times2$ determinants. Pick an entry of the first row, delete the row and the column through it, and a $2\times2$ determinant is left over. Multiply the entry by that leftover determinant, attach a sign, and add the three results.

Why the minus sign in the middle? Each entry $a_{ij}$ carries the sign $(-1)^{i+j}$: plus when $i+j$ is even, minus when it is odd. Laid out over the matrix, the signs form a chessboard that always starts with $+$ in the top-left corner: $$\begin{matrix}+&-&+\\-&+&-\\+&-&+\end{matrix}$$ Multiplying out Definition 4.2 gives six products, three with each sign: $$|A|=a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}$$ $$\qquad+\,a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}.$$ Every product takes exactly one entry from each row and exactly one from each column. That symmetry is the reason behind the most useful fact about determinants.

Why it is true. Multiply out the second-column expansion above: $$-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{22}(a_{11}a_{33}-a_{13}a_{31})$$ $$\qquad-\,a_{32}(a_{11}a_{23}-a_{13}a_{21}).$$ You get exactly the six products listed before Theorem 4.1, with the same signs. The same check works for each of the six rows and columns (try one in Exercise II). $\square$

Area of a Triangle and Collinearity

In Class 10 you found the area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ from the formula $\tfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$. That expression is hard to remember — but it is exactly a $3\times3$ determinant in disguise.

Proof. Expand $D$ along its third column, whose entries are all $1$ with signs $+,-,+$: $$D=\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}-\begin{vmatrix}x_1&y_1\\x_3&y_3\end{vmatrix}+\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}$$ $$=(x_2y_3-x_3y_2)-(x_1y_3-x_3y_1)$$ $$\qquad+\,(x_1y_2-x_2y_1).$$ Collecting the terms in $x_1$, $x_2$, $x_3$ gives $x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$ — the Class 10 expression. So (a) is the Class 10 formula. Three points are collinear exactly when the “triangle” they form is flat, i.e. has area $0$, which by (a) means $D=0$: that is (b). For (c), a point $P(x,y)$ lies on the line $P_1P_2$ if and only if $P$, $P_1$, $P_2$ are collinear, i.e. (by (b)) the determinant in (c) is $0$. Expanding it along the first row gives $x(y_1-y_2)-y(x_1-x_2)+(x_1y_2-x_2y_1)=0$, a linear equation in $x$ and $y$ — a straight line. $\square$

The sign of $D$ carries extra information: $D\gt0$ when $P_1\to P_2\to P_3$ goes round the triangle anticlockwise and $D\lt0$ when it goes clockwise. Because areas are never negative, we take the absolute value — and this is why an area question with an unknown vertex usually has two answers ($D=+2\Delta$ or $D=-2\Delta$).

Exercises