MATH 240

Fri. September 6th, 2019

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Matrix Product

Given a vector $x$ and a matrix $A$, we want to define their product $Ax$. (This operation can be used to represent things like rotations in video games.)

$\begin{gathered} A=\begin{pmatrix} 3&2&1\\4&5&6\\1&4&3 \end{pmatrix}, \ x=\begin{pmatrix} 1\\2\\3 \end{pmatrix}\\ Ax=\begin{pmatrix} 3&2&1\\4&5&6\\1&4&3 \end{pmatrix} \begin{pmatrix} 1\\2\\3 \end{pmatrix} \end{gathered}$
Rotate each row of the matrix so it becomes a column, then multiply that column element-wise with the vector and add the results.
$\begin{aligned} Ax&=\begin{pmatrix} 3(1) + 2(2) + 1(3)\\ 4(1) + 5(2) + 6(3)\\ 1(1) + 4(2) + 3(3) \end{pmatrix}\\ &=\begin{pmatrix} 3+4+3\\4+10+18\\1+8+9 \end{pmatrix}\\ &=\begin{pmatrix} 10\\32\\18 \end{pmatrix} \end{aligned}$
The number of columns in the matrix must be the same as the number of rows in the vector for this operation to be defined.

If $A$ is a $n$ by $m$ matrix with columns $a_1... a_n$ and $x$ is a column vector with length $m$, then
$Ax=x_1a_1+...+x_na_n=b,$

whose solution set is the same as the solution set of the equations whose augmented matrix is $\begin{bmatrix} a_1...a_n,b \end{bmatrix}$.

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Ex. Let $A=\begin{pmatrix} 1&3&4\\-4&2&-6\\-3&-2&-7 \end{pmatrix}$.

Can we choose $b$ in $\R^3$ in an arbitrary manner and still have $Ax=b$ be consistent? (This is logically equivalent to the question, "Do the columns of $A$ span $\R^3$?")

$\begin{aligned} b=&\begin{pmatrix} b_1\\b_2\\b_3 \end{pmatrix}\\ \text{Augmented matrix: }&\begin{pmatrix} 1&3&4&b_1\\-4&2&-6&b_2\\-3&-2&-7&b_3 \end{pmatrix}\\ =&\begin{pmatrix} 1&3&4&b_1\\0&14&10&b_2+4b_1\\0&7&5&b_3+3b_1 \end{pmatrix}\\ =&\begin{pmatrix} 1&3&4&b_1\\0&14&10&b_2+4b_1\\0&0&0&b_3+3b_1-\frac{1}{2}(b_2+4b_1) \end{pmatrix} \end{aligned}$

Based on the bottom row, we can see that for some choices of $b$, this system will be inconsistent; therefore the columns of $A$ do not span $\R^3$.