$$3a+2b+c+3d+5e=0\\ 6a+4b+3c+5d+7e=0\\ 9a+6b+5c+7d+9e=0\\ 3a+2b+4d+8e=0$$
After using Gauss elimination method, I have \begin{bmatrix} 3&2&1&3&5\\ 0&0&1&-1&-3\\ 0&0&0&0&0\\ 0&0&0&0&0 \end{bmatrix}=\begin{bmatrix} 0\\0\\0\\0\end{bmatrix}
Even though I have 1 free variable(5varibales-4equations), I don't know how to solve to get all of the a,b,c,d,e. Please help me how to continue. Thank you all in advance .
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COMMENT.- There are $5$ unknowns and $4$ equations so there are a lot of solutions (an obvious one is the zero solution). But doing $x = 3a + 2b$ we obtain the following system of $4$ equations with $4$ unknowns: $$x+c+3d+5e=0\\ 2x+3c+5d+7e=0\\ 3x+5c+7d+9e=0\\ x+4d+8e=0$$ which can be solved with any of the usual methods.
On
Using Gauss-Jordan elimination:
First, set up the matrix: $$\begin{bmatrix}3&2&1&3&5\\6&4&3&5&7\\9&6&5&7&9\\3&2&0&4&8\end{bmatrix}$$
Then, replace $R_2-2R_2\rightarrow R_2, R_3-3R_1\rightarrow R_3, R_4-R_1\rightarrow R_4$: $$\begin{bmatrix}3&2&1&3&5\\0&0&1&-1&-3\\0&0&2&-2&-6\\0&0&-1&1&3\end{bmatrix}$$ The, replace $R_1-R_2\rightarrow R_1, R_3-2R_2\rightarrow R_3, R_2+R_4\rightarrow R_4$: $$\begin{bmatrix}3&2&0&4&8\\0&0&1&-1&-3\\0&0&0&0&0\\0&0&0&0&0\end{bmatrix}$$
You are left with the following system of equations: $$3a+2b+4d+8e=0$$$$c-d-3e=0$$
This can be rearranged to form: $$4d+8e=-3a-2b$$$$d+3e=c$$
Now you can choose any values you'd like for $a,b,c$. Based on these values, there will be one possible set of values for $d$ and $e$: $$d=-\frac{8c+9a+6b}{4}$$ $$e=\frac{3a+2b+4c}{4}$$
At this point, you can delete the last two rows of your matrix; you can also "mark" the columns. The ones that have a leading nonzero entry for some row are marked "D" for "dependent", the others are marked "I". So in your case, we get to $$ \begin{bmatrix} D&I&D&I&I\\ 3&2&1&3&5\\ 0&0&1&-1&-3\\ \end{bmatrix} $$ That means that the variables $a$ and $c$ are dependent, and $b, d, e$ are independent. In other words, you can pick ANY value you like for $b,d,e$, and then find $a$ and $c$ from the two equations. For instance, if we pick $(b, d, e) = (1,0,0)$, we get that $$ 3a + 2b + c + 3d + 5e = 3a + 2 + c + 0 + 0 = 3a + 2 + c = 0 $$ and $$ c - d - 3e = c - 0 = 3\cdot 0 = c = 0$$
which we can now solve, bottom up, to get that $c = 0$ and $3a + 2 = 0$ so $a = -2/3$.
Now you could do this for any $b,d,e$ triple and get a solution, and in fact this gives all possible solutions, but you can get them all at once by rewriting the two equations a little differently. First, through, I want to do one more elimination step, which is to use the "1" in row 2, col 3, to remove the numbers above it as well as below, so we subtract the second row from the first to get
$$ \begin{bmatrix} D&I&D&I&I\\ 3&2&0&4&8\\ 0&0&1&-1&-3\\ \end{bmatrix} $$ so that our equations are now $$ c - d - 3e = 0 \\ a + 2b + 4d + 8e = 0 $$ which I'm going to rewrite by bringing everything except $a$ and$c$ to the other side: $$ c = d + 3e\\ a = -2b - 4d - 8e $$ Now you can see that any solution looks like $$ \pmatrix{a\\b\\c\\d\\e} = \pmatrix{-2b-4d-8e\\b\\d + 3e \\ d \\ e}. $$ And that's it --- that's the solution to the whole problem.
NB: All this was done assuming that your original row-reduced matrix was correct; it appears not to be. The good news is that you can now carry out the analogous computations yourself, and test that you understand the general method.