Question

given the vectors a=[-4,1,7] b=[2,0,-3], and c=[1,-1,5] find 3a-2b+4c

Answer 1

Do you know how to do scalar multiplication?

Answer 2

\[
d\begin{bmatrix}
a \\ b \\ c\end{bmatrix} = \begin{bmatrix}da\\db\\dc\end{bmatrix}
\]

Answer 3

Then there is vector addition: \[
\begin{bmatrix}a\\b\\c\end{bmatrix} + \begin{bmatrix}d\\e\\c\end{bmatrix}=\begin{bmatrix}a+d\\b+e\\c+f\end{bmatrix}
\]

Answer 4

that would be a [-12, 4, 21]

Answer 5

First find \(3\mathbf{a}\)\[
3\mathbf{a}=3\begin{bmatrix}-12\\4\\21\end{bmatrix} =\begin{bmatrix}-36\\12\\63\end{bmatrix}
\]

Answer 6

Now can you find \(-2\mathbf{b}\) and \(4\mathbf{c}\)?

Answer 7

why would u multiply -12 by 3, isnt it 3a= s(-4, 1, 70 which equals -12, 4, 21 ?

Answer 8

3(-4, 1, 7) **

Answer 9

Ummm \(1\cdot 3=3\).

Answer 10

lol my bad, ok let me show u what ive done: 3[-4, 1, 7] = [-12, 3, 21] then 2[2,0,-3] = [4,0,-6] and then 4[1,-1,5] = [4, -4, 20]

Answer 11

Okay, now just subtract/add them

Answer 12

so no wits [-12, 3, 21] - [4, o, -6] + [ 4, -4, 20]

Answer 13

do i add all them together, or add the x's then y's then z's and have [x, y, z]

Answer 14

Look at what I said before.

Answer 15

\[
\begin{bmatrix}a\\b\\c\end{bmatrix} + \begin{bmatrix}d\\e\\f\end{bmatrix}=\begin{bmatrix}a+d\\b+e\\c+f\end{bmatrix}
\]

Answer 16

ok my answet [-12, -1, 47]

Answer 17

That looks correct.

Answer 18

A Mathematica v9 solution is attached.