Question

Express Aa as a linear combination of the columns of A

Answer 1

A= \[\left[\begin{matrix}1 & 2 \\ 3 & 4 \\ 5 & 6\end{matrix}\right]\]

Answer 2

\[\alpha = \left[\begin{matrix}7 \\ -8\end{matrix}\right]\]

Answer 3

would it be like Aa=0

Answer 4

2 R^3 vectors do not produce an R^2 vector

Answer 5

What do you mean?!

Answer 6

if alpha had a third component of 0 it would be workable

Answer 7

i misread it

Answer 8

the question is
\[A\alpha=b\]

Answer 9

7 -8
----
1 2
3 4
5 6

distribute the alpha down the columns and add the rows

Answer 10

\[\begin{pmatrix}1 & 2 \\ 3 & 4 \\ 5 & 6\end{pmatrix} \begin{pmatrix}7 \\ -8\end{pmatrix}\]

\[ -7 \begin{pmatrix} 1\\3\\5 \end{pmatrix} -8 \begin{pmatrix} 2\\4\\6 \end{pmatrix} \]

Answer 11

not -7, but 7

Answer 12

7 -8
----
1 2
3 4
5 6


7(1) - 8(2)
7(3) - 8(4)
7(5) - 8(6)

Answer 13

so then it would be \[\left[\begin{matrix}-7 \\-21 \\ -35\end{matrix}\right] - \left(\begin{matrix}16 \\ 32 \\ 48\end{matrix}\right)\]

Answer 14

yes, but i typoed -7, brain working faster than the fingers is all

Answer 15

oh yeah haha I realized that too and didn't fix it, silly. Okay thank you!!

Answer 16

youre welcome