# Thread: Basic matrix problem with finding general solutions

1. ## Basic matrix problem with finding general solutions

So basically, I have these two: Imgur: The most awesome images on the Internet examples in my book and i have no idea how they came to that conclusion. I'm obviously missing something fundamental here but I cannot see how you get those columns from those systems.

2. ## Re: Basic matrix problem with finding general solutions

The problem is "Find the general solution of the system $\displaystyle x_1+ 2x_2- 3x_3= 5$".

They write the solution as $\displaystyle \vec{x}= \begin{bmatrix}5 \\ 0 \\ 0\end{bmatrix}+ s\begin{bmatrix}-2 \\ 1 \\ 0 \end{bmatrix}+ t\begin{bmatrix}3 \\ 0 \\ 1 \end{bmatrix}$

Clearly, they are writing $\displaystyle x_1$, $\displaystyle x_2$, $\displaystyle x_3$ as components of the vector $\displaystyle \vec{x}= \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}$. Since $\displaystyle x_1+ 2x_2- 3x_3= 5$ we have immediately $\displaystyle x_1= 5- 2x_2+ 3x_3$.

So $\displaystyle \vec{x}= \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}= \begin{bmatrix}5- 2x_2+ 3x_3 \\ x_2 \\ x_3\end{bmatrix}$$\displaystyle = \begin{bmatrix}5 \\ 0 \\ 0 \end{bmatrix}+ \begin{bmatrix}-2x_2 \\ x_2 \\ 0 \end{bmatrix}+ \begin{bmatrix}3x_3 \\ 0 \\ x_3\end{bmatrix}$$\displaystyle = \begin{bmatrix}5 \\ 0 \\ 0 \end{bmatrix}+ x_2\begin{bmatrix}-2 \\ 1 \\ 0 \end{bmatrix}+ x_3\begin{bmatrix}3 \\ 0 \\ 1\end{bmatrix}$.

They have chosen to replace "$\displaystyle x_2$" and "$\displaystyle x_3$" with "s" and "t".

The second problem is exactly the same thing except that they have "= 0" rather than "= 5" so there is no $\displaystyle \begin{bmatrix}5 \\ 0 \\ 0 \end{bmatrix}$. They are using the phrase "solution space" rather than "solution set" because, since this is a plane that includes the origin, it forms a vector space where the previous solution set does not. (Often a plane that does not include the origin is called a "linear manifold".)