Answer to Question #198921 in Linear Algebra for FELICIA NDANGUMUNI

Question

Consider the following augmented matrix 1 -1 2 1

3 -1 5 -2

-4 2 2x^2-8 x+2

Determine the values of x for which the system has

(i) no solution,

(ii) exactly one solution,

(iii) infinitely many solutions

Answer 1

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

$R_2=R_2-3R_1$

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

$R_3=R_3+4R_1$

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

$R_2=R_2/2$

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

$R_1=R_1+R_2$

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

$R_3=R_3+2R_2$

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

I) no solution

$\begin{matrix} x^2-1=0 \\ x+1\not=0 \end{matrix}=>x=1$

Ii) exactly one solution

$x^2-1\not=0=>x\not=\pm1$

Iii) infinitely many solutions

$\begin{matrix} x^2-1=0 \\ x+1=0 \end{matrix}=>x=-1$