Section 4 Examples

Let A = $\begin{bmatrix} a_1,&a_2,&...,&a_n \end{bmatrix} \text{ and } \vec{x} = \begin{bmatrix} x_1\\x_2\\x_3\\...\\x_n \end{bmatrix}$
$A \vec{x} = x_1a_a + x_2a_2+...+ x_na_n$

Thus,

(1)
\begin{align} \begin{bmatrix} 3&0&1\\1&1&0\\2&-1&1 \end{bmatrix} \begin{bmatrix} 1\\2\\-1 \end{bmatrix} = \begin{bmatrix} 3(1)+0(2) + 1(-1) \\ 1(1) + 2(1) + 0(-1) \\ 2(1) + -1(2) + 1(-1) \end{bmatrix} = \begin{bmatrix} 2\\3\\-1 \end{bmatrix}. \end{align}

which is equivalent to

(2)
\begin{align} 1\begin{bmatrix} 3\\1\\2 \end{bmatrix} + 2\begin{bmatrix} 0\\1\\-1 \end{bmatrix} + (-1)\begin{bmatrix} 1\\0\\1 \end{bmatrix} = \begin{bmatrix} 2\\3\\-1 \end{bmatrix} \end{align}

The first way of multiplying takes each row in the first matrix (A) and multiplies it by the corresponding column entry in the vector ($\vec{x}$) and adding the products to make the new entry in the matrix of dimensions defined by the number of rows in the first matrix by the number of columns in the second matrix. If it's a matrix of nxm dimensions times a vector of mx1 dimensions, the end result will always be nx1. An easy mnemonic for remembering this is the inner two dimensions must match.

If we have the matrix equation $A\vec{x} = \vec{b}$, where $A \text{ and } \vec{b}$ are given, $A = \begin{bmatrix}1&2&3\\0&1&1\\-1&-1&0\\ \end{bmatrix} \text{ and } \vec{b} = \begin{bmatrix} 1\\2\\2 \end{bmatrix}$ and are asked to find a vector that satisfies $A\vec{x} = \vec{b}$, we simply augment A with $\vec{b}$ and solve as we have learned in previous sections. These three values are the conditions under which the augmented matrix is consistent and can be represented as a vector $\begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}$.

In this case, we find that matrix to be $\begin{bmatrix} -\frac{7}{2}\\ \frac{3}{2} \\ \frac{1}{2} \end{bmatrix}$. It doesn't turn out too nicely, but try the reduction yourself and see if the logic makes sense.

page revision: 2, last edited: 08 Feb 2015 02:19