I am just creating random sets of vectors to try and practice solving systems of equations. I thought I had been doing well, but I came up with a set of vectors that keep stumping me. In fact, I've been trying to figure out where I went wrong for almost two hours, yet I still have no clue.
I tried the following:
$$x_1\begin{bmatrix}
-1 \\
-2 \\
-3
\end{bmatrix}+x_2\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}+x_3\begin{bmatrix}
4 \\
2 \\
0
\end{bmatrix}
=
\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}$$
Every time I try to solve this I end up stuck at a similar place.
Here is one example of what I have tried:
$$\begin{eqnarray}
-x_1&+x_2&+4x_3&=a\\
-2x_1&+3x_2&+2x_3&=b\\
-3x_1&+5x_2&&=c
\end{eqnarray}$$
Then for example, I try to eliminate the $x_3$ constant from equations 1 and 2.
$$\begin{eqnarray}
-x_1&+x_2&+4x_3&=a\\
4x_1&-6x_2&-4x_3&=-2b\\
3x_1&-5x_2&&=a-2b
\end{eqnarray}$$
Then, I try to eliminate the $x_2$ constant from the new equation and the original equation 3.
$$\begin{eqnarray}
3x_1&-5x_2&=a-2b\\
-3x_1&+5x_2&=c\\
\end{eqnarray}$$
but I end up with:
$$0+0=a-2b+c$$
I have tried eliminating different factors, but I cannot figure out what I'm doing wrong.
Best Answer
Guide:
Not every system has a unique solution. For your system to have a solution, we need $a-2b+c=0$.
If we have $a-2b+c=0$, now we can let $x_1=t$, from $-3x_1+5x_2=c$, we can solve for $x_2$. Now, having $x_1$ and $x_2$, we can use the equation to solve for $x_3$. It always have infinitely many solutions if $a-2b+c=0$.