# Proving a matrix is non-singular

• Jan 16th 2009, 10:05 AM
craig
Proving a matrix is non-singular
Hi again, came across this matrix question which has really got me stuck.

A = $\left(\begin{array}{cc}q&3\\-2&q-1\end{array}\right)$

1. Find det A, worked this out to be $q(q-1)+6$.

2. Show that A is non-singular for all values of q.

If it is non singular, then the det does not equal 0, I think? But how would I go about showing this, solve the quadratic equation in terms of q?

• Jan 16th 2009, 10:27 AM
Plato
Quote:

Originally Posted by craig
A = $\left(\begin{array}{cc}q&3\\-2&q-1\end{array}\right)$
1. Find det A, worked this out to be $q(q-1)+6$.
2. Show that A is non-singular for all values of q.

Are there any real solutions for $q^2 -q + 6=0$?
• Jan 16th 2009, 10:31 AM
craig
Quote:

Originally Posted by Plato
Are there any real solutions for $q^2 -q + 6=0$?

Just found the answers, in there they have :

$q^2 -q + 6=(q-\frac{1}{2})^2 + 5\frac{3}{4}$

>0 for all real q.

No idea where they have got this from though?
• Jan 16th 2009, 10:49 AM
o_O
I'm sure you're familiar with the method of "completing the square"?
• Jan 16th 2009, 11:01 AM
craig
I think I've got it, I knew it was completing the square, but wasn't sure why he was doing it exactly. But the $(q-\frac{1}{2})^2 + 5\frac{3}{4}$ is basically saying that whatever number q is, wether it is +ve or -ve,once it is squared and then added to $5\frac{3}{4}$, will always be greater than 0, therefore non singular.

Is this right?

Sorry it took this long to work this rather basic thing out, been revising since 11 this morning, mind just went blank.

Thanks for the help and patience :)
• Jan 16th 2009, 12:23 PM
HallsofIvy
Quote:

Originally Posted by craig
I think I've got it, I knew it was completing the square, but wasn't sure why he was doing it exactly. But the $(q-\frac{1}{2})^2 + 5\frac{3}{4}$ is basically saying that whatever number q is, wether it is +ve or -ve,once it is squared and then added to $5\frac{3}{4}$, will always be greater than 0, therefore non singular.

Is this right?

Yes, that is right. You could also have solve the equation using the quadratic formula and observed that it has two complex roots, no real root.

Quote:

Sorry it took this long to work this rather basic thing out, been revising since 11 this morning, mind just went blank.

Thanks for the help and patience :)