08 Matrix Equation

$$ A\vec{v}=\vec{b} $$

If $A$ is an $m\times n$ matrix and $\vec{x}$ is a vector in $\mathbb{R}^n,$ then

$$A \vec{x} = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \cdots & \vec{a}_i \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_n \end{bmatrix},$$

where $\vec{a}_{i}$ is the $i$th column of $A$.

$$= x_{1}\vec{a}_{1}+x_{2}\vec{a}_{2} +\cdots+x_{n}\vec{a}_{n},$$

$$ A \in \mathbb{R}^{m\times n}, \vec{x} \in \mathbb{R}^n, \vec{b} \in \mathbb{R}^m. $$ which is a linear combination of the columns of $A$ using the coefficients of $A$ as weights.

What this means is that $A\vec{x} = \vec{b}$ only has a solution if, and only if, $\vec{b}$ is a linear combination of the columns of $A$.

Given $A \in \mathbb{R}^{m\times n}$,
$\forall \vec{b} \in \mathbb{R}^n, A\vec{x}=\vec{b}$ has a solution.
each $\vec{b} \in R^m$ is a linear combination of the columns of $A$.
The columns of $A$ span $\mathbb{R}^m$
$A$ has a pivot in every row.

Example 1

$$\begin{bmatrix}-2 & 3 & 5 \\0 & 1 & 7\end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 5 \end{bmatrix} = -1 \begin{bmatrix} -2 \\ 0 \end{bmatrix} +0 \begin{bmatrix} 3 \\ 2 \end{bmatrix} + 5 \begin{bmatrix} 5 \\ 7 \end{bmatrix}$$

A Real Example

Given

$$ \left[\begin{array}{ccc|c} -2 & 1 & 0 & b_{1} \\ 1 & 1 & 3 & b_{2} \\ -1 & 0 & -1 & b_{3} \end{array}\right], $$ for what values of $b_{1},b_{2},b_{3}$ does this system have a solution?
After $\text{rref}$ing the matrix, you get

$$ \left[\begin{array}{ccc|c} 1 & 0 & 1 & -b_{3} \\ 0 & 1 & 2 & b_{3}+b_{2} \\ 0 & 0 & 0 & b_{1}-b_{2}-3b_{3} \end{array}\right]. $$ This means that there is a solution to this system so long as $b_{1}-b_{2}-3b_{3} = 0$, If $b_{1}-b_{2}-3b_{3} \neq 0$, then the system is inconsistent.

So we can guarantee that we always have a solution when there is a pivot in every row.