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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 \(b\).

Quick Quiz


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