determinantlinear algebra
I am having a hard time with the following question
Let $P_0,P_1,P_2$ be points in $\mathbb{R}^2$. We denote $r_i = P_0P_i, i =1,2$. Let P be the 2 by 2 matrix with rows $r_1$ and $r_2$. Let A denote the area of the triangle with vertices $P_0,P_1,P_2$. Explain why $A=\frac{1}{2}|\det(P)|$
Any hints or tips would be greatly appreciated.
There are quite a few ways to derive this formula, but thinking of the determinant as a volume has a straightforward geometric appeal.
Place the triangle $\triangle{ABC}$ on the plane $x=1$ in $\mathbb R^3$ and consider the pyramid with this triangle as its base and the origin for its apex:
This pyramid’s altitude is $1$, so its volume is equal to the area of its base, $\triangle{ABC}$. If we glue a reflected copy of this pyramid to its base, we get a parallelepiped whose volume is given by the triple product $$\overrightarrow{OA}\times\overrightarrow{OB}\cdot\overrightarrow{OC} = \det\begin{bmatrix}\overrightarrow{OA} & \overrightarrow{OB} & \overrightarrow{OC}\end{bmatrix} = \det\begin{bmatrix}1&1&1 \\ A(1)&B(1)&C(1) \\ A(2)&B(2)&C(2) \end{bmatrix}.$$ The volume of the pyramid, and hence also the area of the triangle, is half of this value.
We can also construct this geometrically in $\mathbb R^2$, but it requires some extra work to pass to a $3\times3$ determinant. Similarly to the above construction, we adjoin a reflected version of the triangle to form a parallelogram:
The area of this paralellogram is equal to the value of the determinant $$\det\begin{bmatrix}\overrightarrow{CA}&\overrightarrow{CB}\end{bmatrix} = \det\begin{bmatrix}A(1)-C(1) & B(1)-C(1) \\ A(2)-C(2) & B(2)-C(2)\end{bmatrix}.$$ Embedding this into a $3\times3$ matrix and using some properties of determinants, $$\begin{align} \begin{vmatrix}A(1)-C(1) & B(1)-C(1) \\ A(2)-C(2) & B(2)-C(2)\end{vmatrix} &= \begin{vmatrix}0&0&1 \\ A(1)-C(1) & B(1)-C(1) & 0 \\ A(2)-C(2) & B(2)-C(2) & 0\end{vmatrix} \\ &= \begin{vmatrix}0&0&1 \\ A(1)-C(1) & B(1)-C(1) & C(1) \\ A(2)-C(2) & B(2)-C(2) & C(2)\end{vmatrix} \\ &= \begin{vmatrix}1&1&1 \\ A(1) & B(1) & C(1) \\ A(2) & B(2) & C(2)\end{vmatrix} \end{align}.$$ The second step is valid because if we expand the determinant along the top row, only the lower-left $2\times2$ submatrix contributes to its value, so the entries below the $1$ in the last column can be anything. I prefer the derivation as a volume, though, because it doesn’t require any algebraic trickery.
Hint 1: What happens to new matrix if you do $R_1 \to R_1 - R_2 - R_3 -R_4 -R_5$ where $R$ represents the rows of the matrix in part (i).
Hint 2: The answer for the determinant is zero.
Best Answer
We can choose a coordinate system so that the origin is at one angle of the triangle and the x-axis lies upon one side of the triangle.
In that coordinate system, the vertices of the triangle are (0, 0), (b, 0), and (a, h). The are of the triangle is 1/2 height times base= bh/2.
Taking P0= (0, 0), P1P0= (b, 0) and P2P0= (a, h). The determinant is $\left|\begin{array}{cc}b & 0 \\ a & h\end{array}\right|= bh$