Chapter 7.4, Problem 84E

To determine

To calculate:

The equation of the parabola $y=a{x}^2+bx+c$ that passes through the points $\left(-2,2\right),\left(1,-1\right),\left(2,-6\right)$ using matrices.

Answer to Problem 84E

  $y=-x^2-2x+2$ is the equation of the parabola.

Explanation of Solution

Given information:

The given parabola equation $y=a{x}^2+bx+c$ passes through the points $\left(-2,2\right),\left(1,-1\right),\left(2,-6\right)$

Calculation:

Let the equation of the parabola that passes through the points be $y=a{x}^2+bx+c$ .

Since the points lie on the parabola, they satisfies the parabola equation when we substitute them.

Substituting each of the three points in the equations gives us three equations in three variables.

  $\begin{array}{l}4a-2b+c=2\to \left(1\right)\\ a+b+c=-1\to \left(2\right)\\ 4a+2b+c=-6\to \left(3\right)\end{array}$

The given system of equations can be written in matrix notation as shown below,

  $\left(\begin{array}{ccc}4& -2& 1\\ 1& 1& 1\\ 4& 2& 1\end{array}\right)\left(\begin{array}{c}a\\ b\\ c\end{array}\right)=\left(\begin{array}{c}2\\ -1\\ -6\end{array}\right)$

Multiply both sides by the inverse of the coefficient matrix.

Note that the product of a matrix with its inverse is an identity matrix of the same order.

Hence, we will get two column matrix on both sides, comparing them will give us the values of a,b,c.

  $\left(\begin{array}{c}a\\ b\\ c\end{array}\right)={\left(\begin{array}{ccc}4& -2& 1\\ 1& 1& 1\\ 4& 2& 1\end{array}\right)}^{-1}\left(\begin{array}{c}11\\ 2\\ 16\end{array}\right)$

Consider, ${\left(\begin{array}{ccc}4& -2& 1\\ 1& 1& 1\\ 4& 2& 1\end{array}\right)}^{-1}$

Augment with a $3\times 3$ identity matrix,

  $=\left[\begin{array}{ccccccc}4& -2& 1& \mid & 1& 0& 0\\ 1& 1& 1& \mid & 0& 1& 0\\ 4& 2& 1& \mid & 0& 0& 1\end{array}\right]$

Reduce the matrix to row echelon form,

  $=\left[\begin{array}{ccccccc}4& -2& 1& \mid & 1& 0& 0\\ 0& 4& 0& \mid & -1& 0& 1\\ 0& 0& \frac{3}{4}& \mid & \frac{1}{8}& 1& -\frac{3}{8}\end{array}\right]$

Reduce further,

  $=\left[\begin{array}{ccccccc}1& 0& 0& \mid & \frac{1}{12}& -\frac{1}{3}& \frac{1}{4}\\ 0& 1& 0& \mid & -\frac{1}{4}& 0& \frac{1}{4}\\ 0& 0& 1& \mid & \frac{1}{6}& \frac{4}{3}& -\frac{1}{2}\end{array}\right]$

Hence, we get

  $=\left(\begin{array}{ccc}\frac{1}{12}& -\frac{1}{3}& \frac{1}{4}\\ -\frac{1}{4}& 0& \frac{1}{4}\\ \frac{1}{6}& \frac{4}{3}& -\frac{1}{2}\end{array}\right)$

Now, $\left(\begin{array}{c}a\\ b\\ c\end{array}\right)={\left(\begin{array}{ccc}4& -2& 1\\ 1& 1& 1\\ 4& 2& 1\end{array}\right)}^{-1}\left(\begin{array}{c}11\\ 2\\ 16\end{array}\right)$

  $=\left(\begin{array}{ccc}\frac{1}{12}& -\frac{1}{3}& \frac{1}{4}\\ -\frac{1}{4}& 0& \frac{1}{4}\\ \frac{1}{6}& \frac{4}{3}& -\frac{1}{2}\end{array}\right)\left(\begin{array}{c}2\\ -1\\ -6\end{array}\right)$

  $=\left(\begin{array}{c}-1\\ -2\\ 2\end{array}\right)$

Therefore,

  $a=-1,b=-2,c=2$

Hence, the equation of the parabola is $y=-x^2-2x+2$

Graph:

  

Interpretation:

Thus, the graph is verified.