# Matrix determinant

• Mar 16th 2008, 12:14 PM
mathsstudent8
Matrix determinant
How is the best way to find a determinant for matrices of rows and columns more than 3?
The question i need to solve is this:
A=
1211
2210
361u
00uv

Find the determinant of the matrix. Sketch in a u-v plane the set of points for which A has no inverse.

What is a u-v plane? Please somebody help.
• Mar 16th 2008, 01:40 PM
galactus
You can use cofactors to find the determinant, but I would run it through a calculator that does them. My opinion about doing determinants by hand is kind of like Plato's about partial fractions and integrals. A calculator will do the grunt work in a second.

The deteminant is $\displaystyle 2(u^{2}-3u+2v)$

The matrix does not have an inverse if the determinant is 0.
• Mar 16th 2008, 04:29 PM
mathsstudent8
Thanks for your help, but i need to know how to do it for an exam. How did you figure it out?
• Mar 16th 2008, 05:01 PM
galactus
I ran it thorugh the calculator, but here's one way to do it the ol-fashioned way.

It's not too bad. Interchange rows 3 and 4, then perform 2 elementary row operations to turn it into an upper trianguar matrix.

Then multiply the diagonal.

Interchange row 3 and 4:

$\displaystyle \begin{bmatrix}1&2&1&1\\2&2&1&0\\0&0&u&v\\3&6&1&u\ end{bmatrix}$

Multiply row 1 by -3 and add to row 4:

$\displaystyle \begin{bmatrix}1&2&1&1\\2&2&1&0\\0&0&u&v\\0&0&-2&u-3\end{bmatrix}$

2/u times row 3 added to row 4:

$\displaystyle \begin{bmatrix}1&2&1&1\\2&2&1&0\\0&0&u&v\\0&0&0&u+ \frac{2v}{u}-3\end{bmatrix}$

Now, multiply the diagonal:

$\displaystyle (1)(2)(u)(u+\frac{2v}{u}-3)=2(u^{2}-3u+2v)$
• Mar 17th 2008, 02:34 AM
mathsstudent8
Do you happen to know how to sketch the u-v plane? I don't know what that is :S
• Mar 17th 2008, 02:57 AM
mr fantastic
Quote:

Originally Posted by mathsstudent8
Do you happen to know how to sketch the u-v plane? I don't know what that is :S

What if you were asked to sketch $\displaystyle x^{2}-3x+2y = 0$ in the x-y plane ...... Capisce?
• Mar 17th 2008, 05:03 AM
mathsstudent8
Man i'm so going to fail this exam.

What if you were asked to sketch $\displaystyle x^{2}-3x+2y = 0$ in the x-y plane ...... Capisce?
$\displaystyle x^{2}-3x+2y = 0 \Rightarrow y = -\frac{x^2}{2} + \frac{3x}{2}$ is a parabola.
Now just put back x --> u and y --> v to get that in the u-v plane the set of points you want lie on the parabola $\displaystyle v = -\frac{u^2}{2} + \frac{3u}{2}$.