Question

Change the following equations in to augmented matrix

x-y+2z=1

3x-y+5z=-2

4x+2y+(x²-8)z=(x+2)

And determine values of x where:-

There is no solution

And where there is exactly one solution

And where there infinitely many solutions

Answer 1

Augmented matrix 

$$A=\begin{bmatrix} 1 & -1 & 2 & 1 \\ 3 & -1 & 5 & -2 \\ 4 & 2 & x^2-8 & x+2 \end{bmatrix}$$

$R_2=R_2-3R_1$

$$\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 2 & -1 & -5 \\ 4 & 2 & x^2-8 & x+2 \end{bmatrix}$$

$R_3=R_3-4R_1$

$$\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 2 & -1 & -5 \\ 0 & 6 & x^2-16 & x-2 \end{bmatrix}$$

$R_2=R_2/2$

$$\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & -1/2 & -5/2 \\ 0 &6 & x^2-16 & x-2 \end{bmatrix}$$

$R_1=R_1+R_2$

$$\begin{bmatrix} 1 & 0 & 3/2 & -3/2 \\ 0 & 1 & -1/2 & -5/2 \\ 0 & 6 & x^2-16 & x-2 \end{bmatrix}$$

$R_3=R_3-6R_2$

$$\begin{bmatrix} 1 & 0 & 3/2 & -3/2 \\ 0 & 1 & -1/2 & -5/2 \\ 0 & 0 & x^2-13 & x+13 \end{bmatrix}$$

I) no solution

$$\begin{matrix} x^2-13=0 \\ x+13\not=0 \end{matrix}=>x=\pm\sqrt{13}$$

Ii) exactly one solution

$$x^2-13\not=0=>x\not=\pm\sqrt{13}$$

Iii) infinitely many solutions

$$\begin{matrix} x^2-13=0 \\ x+13=0 \end{matrix}=>x\in \varnothing$$