- #1

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u = (1,1,1)

v = (1,2,3)

w = (2,-1,1)

We follow the method here:

[itex] x \ \begin{bmatrix}

1 \\

1 \\

1 \end{bmatrix} \ + \ y \

\begin{bmatrix}

1 \\

2 \\

3 \end{bmatrix} \ + \ z \

\begin{bmatrix}

2 \\

- 1 \\

1 \end{bmatrix} \ = \ \begin{bmatrix}

1 \\

-2 \\

5 \end{bmatrix} \ =\ \begin{bmatrix}

x \\

x \\

x \end{bmatrix} \ + \

\begin{bmatrix}

y \\

2y \\

3y \end{bmatrix} \ + \

\begin{bmatrix}

2z \\

- z \\

z \end{bmatrix} \ = \ \begin{bmatrix}

1 \\

-2 \\

5 \end{bmatrix}[/itex]

and continue, but I don't understand this fully.

The way I understand a 3-tuple is that (1,2,3) is

1 in the x axis, 2 in the y-axis & 3 in the z-axis.

I can't help but return to thinking this way & wanting

to write

u = (1,1,1)

v = (1,2,3)

w = (2,-1,1)

as

[itex]

x \ \begin{bmatrix}

1 \\

1 \\

2 \end{bmatrix} \ + \ y \

\begin{bmatrix}

1 \\

2 \\

-1 \end{bmatrix} \ + \ z \

\begin{bmatrix}

1 \\

3 \\

1 \end{bmatrix} \ = \ \begin{bmatrix}

1 \\

-2 \\

5 \end{bmatrix} \ =\ \begin{bmatrix}

x \\

x \\

2x \end{bmatrix} \ + \

\begin{bmatrix}

y \\

2y \\

- y \end{bmatrix} \ + \

\begin{bmatrix}

z \\

3z \\

z \end{bmatrix} \ = \ \begin{bmatrix}

1 \\

-2 \\

5 \end{bmatrix}[/itex]

Needless to say it's because I don't understand the reason why we do it

one way and not the other, I mean it doesn't make sense because in my

underdeveloped and confused understanding of linear algebra we can

transpose the vector and switch between the two ways I've done the

matrices here & none of it makes sense.

It would help a lot if anyone could clear this up with a good

explanation!