MATH 240

Fri. September 6th, 2019


Matrix Product

Given a vector xx and a matrix AA, we want to define their product AxAx. (This operation can be used to represent things like rotations in video games.)

A=(321456143), x=(123)Ax=(321456143)(123)\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.
Ax=(3(1)+2(2)+1(3)4(1)+5(2)+6(3)1(1)+4(2)+3(3))=(3+4+34+10+181+8+9)=(103218)\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 AA is a nn by mm matrix with columns a1...ana_1... a_n and xx is a column vector with length mm, then
Ax=x1a1+...+xnan=b,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 [a1...an,b]\begin{bmatrix} a_1...a_n,b \end{bmatrix}.


Ex. Let A=(134426327)A=\begin{pmatrix} 1&3&4\\-4&2&-6\\-3&-2&-7 \end{pmatrix}.

Can we choose bb in R3\R^3 in an arbitrary manner and still have Ax=bAx=b be consistent? (This is logically equivalent to the question, "Do the columns of AA span R3\R^3?")

b=(b1b2b3)Augmented matrix: (134b1426b2327b3)=(134b101410b2+4b1075b3+3b1)=(134b101410b2+4b1000b3+3b112(b2+4b1))\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 bb, this system will be inconsistent; therefore the columns of AA do not span R3\R^3.