Matrix determinant

**mathsstudent8** · March 16th 2008, 12:14 PM

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.

**galactus** · March 16th 2008, 01:40 PM

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 $2(u^{2}-3u+2v)$

The matrix does not have an inverse if the determinant is 0.

**mathsstudent8** · March 16th 2008, 04:29 PM

Thanks for your help, but i need to know how to do it for an exam. How did you figure it out?

**galactus** · March 16th 2008, 05:01 PM

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:

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

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

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

2/u times row 3 added to row 4:

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

Now, multiply the diagonal:

$(1)(2)(u)(u+\frac{2v}{u}-3)=2(u^{2}-3u+2v)$

**mathsstudent8** · March 17th 2008, 02:34 AM

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

**mr fantastic** · March 17th 2008, 02:57 AM

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 $x^{2}-3x+2y = 0$ in the x-y plane ...... Capisce?

**mathsstudent8** · March 17th 2008, 05:03 AM

Man i'm so going to fail this exam.

Thanks for your help.

**mr fantastic** · March 17th 2008, 02:35 PM

Originally Posted by mr fantastic

What if you were asked to sketch $x^{2}-3x+2y = 0$ in the x-y plane ...... Capisce?

$x^{2}-3x+2y = 0 \Rightarrow y = -\frac{x^2}{2} + \frac{3x}{2}$ 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 $v = -\frac{u^2}{2} + \frac{3u}{2}$ .

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