Question

Linear algebra question

Answer 1

do you recall cramers rule? since were going to need to apply it

Answer 2

Yeah I know Cramers rule.

Answer 3

But I am confused with all those transposes.

Answer 4

They are just trying to tell you that those are column vectors, not row vectors; a transpose just swaps from row to column notation.

Answer 5

I don't think they matter though. Because the determinant of a transpose does not change the determinant.

Answer 6

Right?

Answer 7

\[\begin{vmatrix}
1&0&0&3\\
2&-5&0&6\\
0&1&-1&0\\
-1&2&3&1
\end{vmatrix}~~~\begin{vmatrix}x_1\\x_2\\x_3\\x_4\end{vmatrix}=\begin{vmatrix}1\\0\\0\\-1 \end{vmatrix}\]
\[Ax=b\]

Answer 8

now, what is this cramer rule cause I cant remember it to clearly

Answer 9

Ahh thanks. I needed to see that vizaulization.

Answer 10

Basically you swap whatever column of x you are trying to solve for with b and then that x value is

x=det(a)/(det(ai) where ai is the replaced row.

Answer 11

column sorry not row.

Answer 12

lol, i never would have remembered that

Answer 13

so in this case I would replace r3 with b and find the determinant of that. Then I would go det det(ai)/det(a) .

Answer 14

It's det (ai)/det(a) sorry. I mixed it up at first.

Answer 15

Thanks a lot!

Answer 16

have fun with it :) that transpose stuff was just so that they could fit the x and b parts in a line and take up less room on the page