- Thread starter Mondo
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Yes, that is what I thought you were saying but then I thought that the matrix multiplication was wrong. I am not sure why you switching addition confused me so much but I was wrong? You were right? I hope that SK has not been following this post.No. What I am saying is that

3x + 4y = 11

-3y + 2x = 22

is the same system as

3x + 4y = 11

2x - 3y = 22

And if linear algebra says that those are different, then linear algebra is nonsense.

I have this vague recollection that you need to order the variables consistently for purposes of linear algebra, but there is no point in relying on my half-baked recollections.

Seriously, your proof was nice (until I thought it was wrong).

ROFLMAOYes, that is what I thought you were saying but then I thought that the matrix multiplication was wrong. I am not sure why you switching addition confused me so much but I was wrong? You were right? I hope that SK has not been following this post.

Seriously, your proof was nice (until I thought it was wrong).

All good, my friend.

The moment linear algebra makes its appearance, I try to crawl under the sofa. I asked here once what its utility was and never got an answer that satisfied me. I picked up some of it in a course on abstract algebra, but I took that course in 1968. The amount I remember about linear algebra is paltry enough to be potentially fatal.

"A little knowledge is a dangerous thing;

Drink deep or touch not the Empyrean spring"

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\(\displaystyle

\begin{pmatrix}x'\\y'\end{pmatrix} = \begin{pmatrix}cos A & -sin A\\sin A & cos A\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}\)

translates to:

\(\displaystyle x' = x cos A - y sin A\)

\(\displaystyle y' = x sin A + y cos A\)

which is what JeffM said back in post #2.

PS. Still haven't solved my lining up sin A issue. Anyone??

\begin{pmatrix}x'\\y'\end{pmatrix} = \begin{pmatrix}cos A & -sin A\\sin A & cos A\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}\)

translates to:

\(\displaystyle x' = x cos A - y sin A\)

\(\displaystyle y' = x sin A + y cos A\)

which is what JeffM said back in post #2.

PS. Still haven't solved my lining up sin A issue. Anyone??

Last edited:

Thanks for resolving that.\(\displaystyle

\begin{pmatrix}x'\\y'\end{pmatrix} = \begin{pmatrix}cos A & -sin A\\sin A & cos A\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}\)

translates to:

\(\displaystyle x' = x cos A - y sin A\)

\(\displaystyle y' = x sin A + y cos A\)

which is what JeffM said back in post #2.

PS. Still haven't solved my lining up sin A issue. Anyone??

I am almost as vague on doing alignment in LaTeX as I am about linear algebra, but I suspect you need to establish a column for signs to get everything neat.