## Linear combination of columns

Consider the system given by $$Ax=b$$ where $$A= \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$$, $$x= \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$, and $$b= \begin{bmatrix} 7\\8 \end{bmatrix}$$.

Recall that $$Ax$$ in this case denotes the tuple $$\begin{bmatrix} x_1 + 2x_2 + 3x_3\\4x_1+5x_2+6x_3\end{bmatrix}$$.

Using tuple arithmetic, this tuple can be written as $$x_1\begin{bmatrix} 1\\4\end{bmatrix} + x_2\begin{bmatrix} 2\\5\end{bmatrix} + x_3\begin{bmatrix} 3\\6\end{bmatrix}$$.

What we have here is called a linear combination of the tuples $$\begin{bmatrix}1\\4\end{bmatrix}$$, $$\begin{bmatrix}2\\5\end{bmatrix}$$, and $$\begin{bmatrix}3\\6\end{bmatrix}$$. (In general, a linear combination of the tuples $$t_1,\ldots, t_k$$ has the form $$a_1t_1+a_2t_2+\cdots +a_kt_k$$ where $$a_i$$ is a scalar for each $$i= 1,\ldots,k$$.)

The question of whether or not $$Ax=b$$ has a solution can be interpreted as follows: Is $$b$$ a linear combination of the columns of $$A$$?

In general, if $$A= [A_1 \cdots A_n]$$ where $$A_i$$ is the $$i$$th column of $$A$$ and $$x = \begin{bmatrix}x_1\\ \vdots \\ x_n\end{bmatrix}$$, then $$Ax = x_1A_1 + x_2A_2 + \cdots + x_nA_n$$. Solving $$Ax=b$$ means finding a linear combination of $$A_1,\ldots,A_n$$ that gives $$b$$.

## Exercises

For each of the following, write it as a linear combination of tuples with $$x,y,z$$ as scalars.

1. $$\begin{bmatrix} x+2y-z \\ y +z \\ 2x\end{bmatrix}$$

2. $$\begin{bmatrix} 2z-y+x \\ x-y+z \\ y+2x\end{bmatrix}$$