Answer to Question #230374 in Linear Algebra for moe

Question

A parabola $y=ax^{2}+bx+c$  passes through the points (-2, 10); (1,4) and (2,6).

Which of the following are correct? More than one answer may be correct

1. The augmented matrix for the system of equations connecting a, b and c is $\left(\begin{array}{cccc}4&-2&1&:&10\\1&1&1&:&4\\4&2&1&: &6\end{array}\right)$
2. The determinant of the coefficient matrix for the system of equations connecting a, b and c is -12.
3. The parabola above passes through the origin.
4. By applying Gauss elimination, the normal form of the augmented matrix for the system of equations connecting a, b and c is$\left(\begin{array}{cccc}1&0&0&:&1\\0&1&0&:&-1\\0&0&1&: &4\end{array}\right). ​$

Answer 1

$(-2, 10)$

$$a(-2)^2+b(-2)+c=10$$

$(1,4)$

$$a(1)^2+b(1)+c=4$$

$(2,6)$

$$a(2)^2+b(2)+c=6$$

$$\begin{matrix} 4a-2b+c=10\\ a+b+c=4\\ 4a+2b+c=6\\ \end{matrix}$$

1. The augmented matrix for the system of equations connecting a, b and c is

$$A=\begin{pmatrix} 4 & -2 & 1 & : & 10 \\ 1 & 1 & 1 & : & 4 \\ 4 & 2 & 1 & : & 6 \\ \end{pmatrix}$$

True.

2.

$$\begin{vmatrix} 4 & -2 & 1 \\ 1 & 1 & 1 \\ 4 & 2 & 1 \\ \end{vmatrix}=4\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}-(-2)\begin{vmatrix} 1 & 1 \\ 4 & 1 \end{vmatrix}+1\begin{vmatrix} 1 & 1 \\ 4 & 2 \end{vmatrix}$$

$$=-4-6-2=-12$$

True.

3.

$(0,0)$

$$a(0)^2+b(0)+c=0=>c=0$$

$$\begin{matrix} 4a-2b+0=10\\ a+b+0=4\\ 4a+2b+0=6\\ \end{matrix}$$

$$\begin{matrix} 8a=16\\ a+b=4\\ 4a+2b=6\\ \end{matrix}$$

$$\begin{matrix} a=2\\ b=2\\ 4a+2b=6\\ \end{matrix}$$

No solution.

False.

4.

$$A=\begin{pmatrix} 4 & -2 & 1 & : & 10 \\ 1 & 1 & 1 & : & 4 \\ 4 & 2 & 1 & : & 6 \\ \end{pmatrix}$$

$R_1=R_1/4$

$$\begin{pmatrix} 1 & -1/2 & 1/4 & : & 5/2 \\ 1 & 1 & 1 & : & 4 \\ 4 & 2 & 1 & : & 6 \\ \end{pmatrix}$$

$R_2=R_2-R_1$

$$\begin{pmatrix} 1 & -1/2 & 1/4 & : & 5/2 \\ 0 & 3/2 & 3/4 & : & 3/2 \\ 4 & 2 & 1 & : & 6 \\ \end{pmatrix}$$

$R_3=R_3-4R_1$

$$\begin{pmatrix} 1 & -1/2 & 1/4 & : & 5/2 \\ 0 & 3/2 & 3/4 & : & 3/2 \\ 0 & 4 & 0 & : & -4 \\ \end{pmatrix}$$

$R_2=2R_2/3$

$$\begin{pmatrix} 1 & -1/2 & 1/4 & : & 5/2 \\ 0 & 1 & 1/2 & : & 1 \\ 0 & 4 & 0 & : & -4 \\ \end{pmatrix}$$

$R_1=R_1+R_2/2$

$$\begin{pmatrix} 1 & 0 & 1/2 & : & 3 \\ 0 & 1 & 1/2 & : & 1 \\ 0 & 4 & 0 & : & -4 \\ \end{pmatrix}$$

$R_3=R_3-4R_2$

$$\begin{pmatrix} 1 & 0 & 1/2 & : & 3 \\ 0 & 1 & 1/2 & : & 1 \\ 0 & 0 & -2 & : & -8 \\ \end{pmatrix}$$

$R_3=-R_3/2$

$$\begin{pmatrix} 1 & 0 & 1/2 & : & 3 \\ 0 & 1 & 1/2 & : & 1 \\ 0 & 0 & 1 & : & 4 \\ \end{pmatrix}$$

$R_1=R_1-R_3/2$

$$\begin{pmatrix} 1 & 0 & 0 & : & 1 \\ 0 & 1 & 1/2 & : & 1 \\ 0 & 0 & 1 & : & 4 \\ \end{pmatrix}$$

$R_2=R_2-R_3/2$

$$\begin{pmatrix} 1 & 0 & 0 & : & 1 \\ 0 & 1 & 0 & : & -1 \\ 0 & 0 & 1 & : & 4 \\ \end{pmatrix}$$

The normal form of the augmented matrix for the system of equations connecting a, b and c is

$$\begin{pmatrix} 1 & 0 & 0 & : & 1 \\ 0 & 1 & 0 & : & -1 \\ 0 & 0 & 1 & : & 4 \\ \end{pmatrix}$$

True.