For which vectors $(b_1 , b_2 , b_3)$ do these systems have a solution?
$\begin{bmatrix}
1 & 1 & 1 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix}$ $\times$ $\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}$=$\begin{bmatrix}
b_1 \\
b_2 \\
b_3 \\
\end{bmatrix}$
I have to solve this without solving the system (how could you even solve the system, it says this times something equals something?) but I don't know how? Does this have something to do with column space? The column space of the first matrix would be all of $R^3$ but I don't know how that helps me figure out what the vector $(b_1, b_2, b_3)$ is?
Is the question just asking for $b_1=x_1+x_2+x_3$,
$b_2=x_2+x_3$,
$b_3=x_3$?
Best Answer
You have $$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{bmatrix} \times \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix}=\begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ \end{bmatrix},$$ let $h_1 = \begin{bmatrix}1 \\0 \\0 \\\end{bmatrix},$ $h_2 = \begin{bmatrix}1 \\1 \\0 \\\end{bmatrix},$ $h_3 = \begin{bmatrix}1 \\1 \\1 \\\end{bmatrix}$,$b = \begin{bmatrix}b_1 \\b_2 \\b_3 \\\end{bmatrix}$ rewrite the initial system in the following form $$x_1h_1 + x_2h_2 + x_3h_3 = b,$$ so you have a linear combination of $h_1, h_2, h_3$, moreover it is not hard to notice that $h_1, h_2, h_3$ are linearly independent.
Here comes the question: having $3$ linearly independent vectors in $\mathbb{R}^3$ how many vectors in $\mathbb{R}^3$ you can represent as a linear combination of given independent vectors?
P.S. There is no need to inverse your matrix, since the question is to find all $b$ for which the solution exists.