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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 $b_1=x_1+x_2+x_3$,
$b_2=x_2+x_3$,
$b_3=x_3$?

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    $\begingroup$ Your matrix has rank 3, use it. $\endgroup$ – echzhen Feb 26 '16 at 22:54
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    $\begingroup$ given $b_1, b_2, b_3,$ all you need to do is find out what $x_3, x_2, x_1$ must be. $\endgroup$ – Will Jagy Feb 26 '16 at 22:58
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    $\begingroup$ @WillJagy so you mean $x_3=b_3$, $x_2=b_2-b_3$, and $x_1=b_1-b_2$? $\endgroup$ – idknuttin Feb 26 '16 at 23:00
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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.

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  • $\begingroup$ if$h_1, h_2, and h_3$ are linearly independent then doesn't the mean there are an infinite number of solutions? $\endgroup$ – idknuttin Feb 26 '16 at 23:16
  • $\begingroup$ @idknuttin No, solution is one and only one. But it's not your question. Look, solution is the vector $x$. Vector $b$ is given, so the question is: for which given $b$ you can find $x$? $\endgroup$ – echzhen Feb 26 '16 at 23:18
  • $\begingroup$ Is the answer to my original question what I wrote in my question about what $x_1, x_2$ and $x_3$ equal? $\endgroup$ – idknuttin Feb 26 '16 at 23:22
  • $\begingroup$ Your teacher may disagree with you since it wasn't your problem to find $x$. You have to use some fundamental properties of $\mathbb{R}^n$ to state that: it doesn't matter which $b$ I receive, I can always find $x$. $\endgroup$ – echzhen Feb 26 '16 at 23:23
  • $\begingroup$ I'm sorry I meant $b_1=x_1+x_2+x_3$, $b_2=x_2+x_3$, $b_3=x_3$ $\endgroup$ – idknuttin Feb 26 '16 at 23:26
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The system has solution iff $$\begin{bmatrix} 1 & 1 & 1 & b_1\\ 0 & 1 & 1 & b_2 \\ 0 & 0 & 1 & b_3 \\ \end{bmatrix} $$ this matrix has rank $3$ for the Rouché-Capelli's theorem

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You can rewrite these equations as:

$ b_3 = x_3 ,\\$ $ b_2 - b_3 = x_2, b_1 - b_2 = x_1$

It's the same as inverting the given matrix!!

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  • $\begingroup$ So I state the solution as this system only has solutions for vectors $x_3=b_3$, $x_2=b_2-b_3$, and $x_1=b_1-b_2$? $\endgroup$ – idknuttin Feb 26 '16 at 23:04
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If $A$ is matrix $n \times n$, $x$ is the $n \times 1$ matrix variable and $b$ a $n \times 1$ matrix given, then the system $$ A . x = b$$ has solution ever that $A$ is invertible. Since, your problem has ever solution.

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