find general solution from linear equations

Question

reduced to row-echelon form $$\begin{pmatrix} 1&2&3&-2&4\\ 2&4&8&1&9\\ 3&6&13&4&14\\ \end{pmatrix}$$

i got $$\begin{pmatrix} 1&2&0&-19/2&5/2\\ 0&0&1&5/2&1/2\\ 0&0&0&0&0\\ \end{pmatrix}$$

so i have two equations $$X_1+X_2-19/2X_4+5/2X_5$$ and $$X_3+5/2X_4+X_5/2$$

but im not sure how to make these two into one general solution

Answer 1

Supposing you are dealing with the homogeneous system associated to the row-echelon reduced matrix $$A\textbf X=\textbf 0,$$ we have to solve: $$\begin{cases} X_1+2X_2-\frac{19}2X_4+\frac52X_5=0\\ X_3+\frac 52 X_4+\frac 12 X_5=0 \end{cases} $$ Therefore we can put $X_1, X_2, X_3$ as parameters, $X_1=t, X_2=s, X_3=v$ and solve the system in the two variables remaining $X_4, X_5$ by standards $2\times 2$ linear systems resolving methods.

Answer 2

Since $\dim\ker\begin{pmatrix} 1&2&3&-2&4\\ 0&0&2&5&1\\ \end{pmatrix}=3$, personally I'd like to give some zeros and then calculate the others:

• finding $x=\begin{pmatrix} a\\ b\\ 0\\ 0\\ 0\\ \end{pmatrix}$, we get $a=2,b=-1$
• finding $x=\begin{pmatrix} 0\\ 0\\ a\\ b\\ c\\ \end{pmatrix}$, we get $a=-22,b=5,c=19$
• finding $x=\begin{pmatrix} 0\\ a\\ b\\ c\\ 0\\ \end{pmatrix}$, we get $a=19,b=-10,c=4$

Then$$\ker\begin{pmatrix} 1&2&3&-2&4\\ 0&0&2&5&1\\ \end{pmatrix}=<\begin{pmatrix} 2\\ -1\\ 0\\ 0\\ 0\\ \end{pmatrix},\begin{pmatrix} 0\\ 0\\ -22\\ 5\\ 19\\ \end{pmatrix},\begin{pmatrix} 0\\ 19\\ -10\\ 4\\ 0\\ \end{pmatrix}>$$