Question

Find dimensions and basis of solution space W for a homogeneous system.
x^2+2y+2z-s+3t=0
x+2y+3z+s+t=0
3x+6y+8z+s+5t=0

Answer 1

The augmented matrix is

$$\begin{bmatrix} 1 & 2& 2 & -1& 3 \\ 1 & 2& 3 & 1& 1\\ 3 & 6& 8 & 1& 5 \end{bmatrix}$$

Reduce the system to echelon form:

$$R_3\to R_3-3R_1\implies \begin{bmatrix} 1 & 2& 2 & -1& 3 \\ 1 & 2& 3 & 1& 1\\ 0 & 0& 2 & 4& -4 \end{bmatrix}$$

$$R_2\to R_2-R_1\implies \begin{bmatrix} 1 & 2& 2 & -1& 3 \\ 0 & 0& 1 & 2& -2\\ 0 & 0& 2 & 4& -4 \end{bmatrix}$$

$$R_3\to R_3-2R_2\implies \begin{bmatrix} 1 & 2& 2 & -1& 3 \\ 0 & 0& 1 & 2& -2\\ 0 & 0& 0 & 0& 0 \end{bmatrix}$$

This corresponds to the system:

$$z+2s-2t=0,x+2y+2z-s+3t=0$$

The system in echelon form has 2 nonzero equations in five unknowns. Hence the system has $5-2=3$ free variables which are $y,s,t$. Thus,

$$\dim W=3$$

We obtain the basis for W:

i. $y=1,s=0,t=0$

$$v_1=(-2,1,0,0,0)$$

ii. $y=0,s=1,t=0$

$$v_2=(5,0,-2,1,0)$$

iii.$y=0,s=0,t=1$

$$v_3=(-7,0,2,0,1)$$

The set $\{v_1,v_2,v_3\}$ is a basis of the solution space W.