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Given the system of linear equations:

$\hspace{20pt}6x_2 + 2x_3 + 10x_4 = b_1$

$x_1 +x_2+4x_3- 2x_4 \hspace{5pt}=b_2$

$x_1 - 2x_2 + 3x_3 - 7x_4 = b_3$

(a) Find all possible Values of $b_1$,$b_2$,and $b_3$ for which this system has solutions;

(b)Find all possible solutions of this system if $b_1=6$,$b_2=7$, and $b_3 = 4$.

I was able to solve (a), but not sure what the solution should look like for (b)?

I set up an augmented matrix obtain rref of the matrix:(edit after comment)

$\begin{bmatrix} 0&6&2&10&b_1\\ 1&1&4&-2&b_2\\ 1&-2&3&-7&b_3\\ \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0&\frac{11}{3}&-\frac{11}{3}&b_2-\frac{1}{6}b_1\\ 0&1&\frac{1}{3}&\frac{5}{3}&\frac{1}{6}b_1\\ 0&0&0&0&b_3-b_2+\frac{1}{2}b_1\\ \end{bmatrix}$

So for (a): $b_2 = \frac{1}{6}b_1$; $b_3=b_2-\frac{1}{2}b_1=-\frac{1}{3}b_1$.

We get, $\begin{bmatrix} \frac{1}{6}b_1\\ \frac{1}{6}b_1\\ -\frac{1}{3}b_1 \end{bmatrix}$ the column space of matrix is the line containing vector $\begin{bmatrix} \frac{1}{6}\\ \frac{1}{6}\\ -\frac{1}{3} \end{bmatrix}$.

For part (b), the rank of the matrix is $2 +$ nullspace $= 4$ (number of columns). Thus $x_3$, and $x_4$ will be scalars so let $x_3=s$,$x_4=t$.

For given values of $b_1=6$,$b_2=7$,$b_3=4$ we have the augmented matrix: \begin{bmatrix} 1&0&\frac{11}{3}&-\frac{11}{3}&6\\ 0&1&\frac{1}{3}&\frac{5}{3}&1\\ 0&0&0&0&0\\ \end{bmatrix}

$\Rightarrow \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}x_1+\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}x_2+\begin{bmatrix} \frac{1}{3}\\ \frac{1}{3}\\ 0 \end{bmatrix}s+\begin{bmatrix} \frac{-11}{3}\\ \frac{5}{3}\\ 0 \end{bmatrix}t=\begin{bmatrix} 6\\ 1\\ 0 \end{bmatrix}$

$x_1+0x_2+\frac{11}{3}s - \frac{11}{3}t = 6$; $x_1 = 6-\frac{11}{3}s+\frac{11}{3}t$

$0x_1+x_2+\frac{1}{3}s+\frac{5}{3}t = 1$; $x_2= 1- \frac{1}{3}s-\frac{5}{3}t$

$x_3=s$

$x_4=t$

Not sure, does this answer (b)?

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    $\begingroup$ From the first part, you should have found $b_3-b_2+\dfrac{1}{2}b_1$ (sign error?) in that last entry, which must be zero for there to be a solution, so we need $b_3-b_2+\dfrac{1}{2}b_1 = 0 \implies b_1 = 2 (b_2 - b_3)$ $\endgroup$ Commented Mar 3, 2017 at 6:53
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    $\begingroup$ Now I am sure you have a sign error as I show above, cause for the second part, you should get $$ \begin{bmatrix} 1 & 0 & \dfrac{11}{3} & -\dfrac{11}{3} & 6 \\ 0 & 1 & \dfrac{1}{3} & \dfrac{5}{3} & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ This makes $x_3$ and $x_4$ free variables. $\endgroup$ Commented Mar 3, 2017 at 6:59

1 Answer 1

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Row reduction of matrix

Find the fundamental columns.

Reorder rows to simplify arithmetic.

$$ % E \left[ \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{array} \right] % A \left[ \begin{array}{crrr} 0 & 6 & 2 & 10 \\ 1 & 1 & 4 & -2 \\ 1 & -2 & 3 & -7 \\ \end{array} \right] % = % PA \left[ \begin{array}{crrr} 1 & -2 & 3 & -7 \\ 1 & 1 & 4 & -2 \\ 0 & 6 & 2 & 10 \\ \end{array} \right] % $$

Clear column 1: $$ % E \left[ \begin{array}{rcc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] % PA \left[ \begin{array}{crrr} 1 & -2 & 3 & -7 \\ 1 & 1 & 4 & -2 \\ 0 & 6 & 2 & 10 \\ \end{array} \right] % = % A \left[ \begin{array}{crcr} 1 & -2 & 3 & -7 \\ 0 & 3 & 1 & 5 \\ 0 & 6 & 2 & 10 \\ \end{array} \right] % $$

Clear column 2: $$ % E \left[ \begin{array}{cc c} 1 & \frac{2}{3} & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & -2 & 1 \\ \end{array} \right] % PA \left[ \begin{array}{crcr} 1 & -2 & 3 & -7 \\ 0 & 3 & 1 & 5 \\ 0 & 6 & 2 & 10 \\ \end{array} \right] % = % A \left[ \begin{array}{cccc} 1 & 0 & \frac{11}{3} & -\frac{11}{3} \\ 0 & 1 & \frac{1}{3} & \frac{5}{3} \\ 0 & 0 & 0 & 0 \\ \end{array} \right] % $$ Result $$ % \begin{align} \mathbf{A} &\mapsto \mathbf{E_{A}} \\ % \left[ \begin{array}{crrr} 0 & 6 & 2 & 10 \\ 1 & 1 & 4 & -2 \\ 1 & -2 & 3 & -7 \\ \end{array} \right] % &\mapsto % \left[ \begin{array}{cccc} \boxed{1} & 0 & \frac{11}{3} & -\frac{11}{3} \\ 0 & \boxed{1} & \frac{1}{3} & \frac{5}{3} \\ 0 & 0 & 0 & 0 \\ \end{array} \right] % \end{align} % $$ The boxed pivots reveal the fundamental columns: $1$ and $2$.

(a) For a data vector $b$ to reside in the range space of the columns, it must be a linear combination of the fundamental columns:

$$ \boxed{ b\in\mathcal{R}\left( \mathbf{A} \right) \quad \iff \quad b = \alpha \left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ \end{array} \right] + \beta \left[ \begin{array}{r} 6 \\ 1 \\ -2 \\ \end{array} \right] } % \tag{1} $$ where $\alpha$ and $\beta$ are arbitrary constants.

(b) What are the restriction on $b_{1}$ so that the following vector resides in the range space of the columns? $$ b= \left[ \begin{array}{c} b_{1} \\ \color{blue}{7} \\ \color{blue}{4} \\ \end{array} \right] = \alpha \left[ \begin{array}{c} 0 \\ \color{blue}{1} \\ \color{blue}{1} \\ \end{array} \right] + \beta \left[ \begin{array}{r} 6 \\ \color{blue}{1} \\ \color{blue}{-2} \\ \end{array} \right] $$ There are a couple of ways to attack this problem, one of which is to solve the matrix equation $$ % \left[ \begin{array}{cr} \color{blue}{1} & \color{blue}{1} \\ \color{blue}{1} & \color{blue}{-2} \\ \end{array} \right] %% \left[ \begin{array}{r} \alpha \\ \beta \\ \end{array} \right] %% = \left[ \begin{array}{c} \color{blue}{7} \\ \color{blue}{4} \\ \end{array} \right] %% $$

The solution is $$ \left[ \begin{array}{r} \alpha \\ \beta \\ \end{array} \right] = \frac{1}{3} \left[ \begin{array}{cr} 2 & 1 \\ 1 & -1 \\ \end{array} \right] \left[ \begin{array}{c} 7 \\ 4 \\ \end{array} \right] % = \left[ \begin{array}{c} 6 \\ 1 \\ \end{array} \right] $$ Put the mixing constants into $(1)$ $$ 6 \left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ \end{array} \right] + 1 \left[ \begin{array}{r} 6 \\ 1 \\ -2 \\ \end{array} \right] % = \left[ \begin{array}{r} 6 \\ 7 \\ 4 \\ \end{array} \right] % $$ This shows that $$ \boxed{ b_{1} = 6 } $$

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