Suppose that $x_1 = -1$, $x_2 = 2$, $x_3 = 4$, $x_4 = -3$ is a solution of a non-homogeneous linear system $A\mathbf{x} = \mathbf{b}$ and that the solution set of the homogeneous system $A\mathbf{x} =\mathbf{0}$ is given by the formulas:
$$\begin{align*}
x_1 &= -3r + 4s,\\
x_2 &= r – s,\\
x_3 &= r,\\
x_4 &= s.
\end{align*}$$
Find the vector form of the general solutions of Ax = 0 and Ax = b
I ended up with something like:
( -3 4)
( 1 -1)
( 1 0)
( 0 1)
where I separated the $r$ and $s$ values, I haven't tried to actually solve though because I'm kinda confused about what I'm suppose to do with this.
Best Answer
You answer of the solutions to $A\vec{x}=0$ being equal to the span of $\begin{bmatrix} -3\\ 1\\ 1\\ 0 \end{bmatrix}$ and $ \begin{bmatrix} 4\\ -1\\ 0\\ 1 \end{bmatrix}$ is correct; now let's find the solutions to $A\vec{x}=\vec{b}$. If the kernel of a matrix is not equal to only the zero vector (as in our case, it actually has a dimension of two), then the matrix is non-invertible, and has infinitely many solutions for $\vec{b}$. By the property of linear transformations, if $A\vec{x}=0$, then $A(\vec{p}+\vec{x})=\vec{b}$ (where $\vec{p}$ is a vector such that $A\vec{p}=\vec{b}$) because $A(\vec{p}+\vec{x})=A\vec{p}+A\vec{x}=A\vec{p}+\vec{0}=\vec{b}+\vec{0}=\vec{b}$. Since you know that $ \begin{bmatrix} -1\\ 2\\ 4\\ -3 \end{bmatrix}$ is a solution to $A\vec{x}=\vec{b}$, the set of solutions to this equations will consist of $\vec{p}$+any solution to $A\vec{x}=0$, which you found to be the span of $\begin{bmatrix} -3\\ 1\\ 1\\ 0 \end{bmatrix}$ and $ \begin{bmatrix} 4\\ -1\\ 0\\ 1 \end{bmatrix}$. Or, in other words, the set of $\left \{\vec{p}+\vec{x}| A\vec{x}=0 \right \}$ is a set of solutions to $A\vec{x}=\vec{b}$. Or, represented numerically, solutions to $A\vec{x}=\vec{b}$ are of form $ \underset{\text{A given solution to } A\vec{x}=\vec{b}}{\underbrace{\begin{bmatrix} -1\\ 2\\ 4\\ -3 \end{bmatrix}}}$ $+$ $\underset{\text{span of Kernel } A} {\underbrace{r\begin{bmatrix} -3\\ 1\\ 1\\ 0 \end{bmatrix} + s \begin{bmatrix} 4\\ -1\\ 0\\ 1 \end{bmatrix}}}$, as Arturo pointed out.