Question

Please help need help urgently with this matrix question

Answer 1

ad determinant of a matrix is ad-bc so here's a link that can help you out http://www.mathsisfun.com/algebra/matrix-determinant.html

Answer 2

*a determinant

Answer 3

For the first determinant\[x(3v-2w)-y(3u-w)+z(2u-v) = 3\]The second determinant equation is\[x(-6w+9v)-y(-3w+9u)+z(-3v+6u)\]
That probably won't be ovely helpful actually sorry. I'll keep on looking

Answer 4

However discarding the values for x,y,z (they are the same in both) we can simplify to find some sort of ratio?
\[3v-2w-3u+w+2u-v=2v-w-u=3\]Let's see if there is a similar ratio in the next equation
\[-6w+9v+3w+9u-3v+6u=6v-3w+15u=unkwown\]

Answer 5

So far we have \(2v-w-u = 3\) and \(6v-3w+15u = X\) we can divide our second equation through by 3 which will line us up with the ratios, giving\[2v-w+5u = X/3\]

Answer 6

We know from \(2v-w-u=3\) that \(2v-w = 3+u\) so we can sub that in to our equation. Giving \[3+u+5u = X/3\]

Answer 7

Actually I think everything I've just done might be bogus... sorry people

Answer 8

Two row operations have been performed...

Answer 9

Unkle? You stopped... don't let me do that :D

Answer 10

\[\begin{vmatrix}x&y&z\\u&v&w\\1&2&3\end{vmatrix}\]\[=-\begin{vmatrix}x&y&z\\1&2&3\\u&v&w\end{vmatrix}\]

Answer 11

\[=-\frac13\begin{vmatrix}x&y&z\\3&6&9\\u&v&w\end{vmatrix}\]

Answer 12

\[\begin{vmatrix}x&y&z\\\color{red}u&\color{red}v&\color{red}w\\\color{blue}1&\color{blue}2&\color{blue}3\end{vmatrix}=-\begin{vmatrix}x&y&z\\\color{blue}1&\color{blue}2&\color{blue}3\\\color{red}u&\color{red}v&\color{red}w\end{vmatrix}\]

What Unkle means is that if you switch a pair of rows of a matrix, the determinant of that new matrix becomes the negative of the original determinant...

Answer 13

also, scaling one row scales the determinant accordingly

Answer 14

So does that mean that the answer is -1/3 *3 = -1

Answer 15

What did we do? We switched a pair of rows, and after that, we multiplied one row by -3, right?

Answer 16

\[\begin{vmatrix}x&y&z\\\color{red}u&\color{red}v&\color{red}w\\\color{blue}1&\color{blue}2&\color{blue}3\end{vmatrix}=D\]\[\begin{vmatrix}x&y&z\\\color{blue}1&\color{blue}2&\color{blue}3\\\color{red}u&\color{red}v&\color{red}w\end{vmatrix}=-D\]
And by switching rows, the determinant becomes negative...

Answer 17

Now, multiplying ONE row by a scalar, would also change the determinant, also multiplying it by THAT same scalar
\[\begin{vmatrix}x&y&z\\\color{violet}{-3}\cdot\color{blue}1&\color{violet}{-3}\cdot\color{blue}2&\color{violet}{-3}\cdot\color{blue}3\\\color{red}u&\color{red}v&\color{red}w\end{vmatrix}=\color{violet}{-3}\cdot-D\]\[\large \begin{vmatrix}x&y&z\\\color{blueviolet}{-3}&\color{blueviolet}{-6}&\color{blueviolet}{-9}\\\color{red}u&\color{red}v&\color{red}w\end{vmatrix}=\color{green}{3D}\]

Answer 18

thanks i get it now

Answer 19

That is good.

Answer 20

@terenzreignz i have another question and i really need help with it because it is due today do you think you could help me solve that as well

Answer 21

That had better not be a quiz, though. Post it :)