PocketMatrix

As it's Friday afternoon, I got one of my colleagues to give me a maths lesson to solve this problem (I've never been a maths geek so if this is all wrong then tell me and I'll fire him).

He says that you need to work out the determinant of A. If that's non-zero then there's only one answer. Then you can solve the equations to get the actual answer.

I got a bit lost in the scribbling in the first part, but he started working out the determinant by replacing the first column with the null result and putting it over A.

So for x it's:

<pre>x=[ 0 7 1]
[ 0 4 -1]
[ 0 3 0]
----------
A</pre>

Then do the same for y and z (didn't get that bit myself).

Then the algebra bit went:

<pre>2x+7y+z=0
x+4y-z=0
x+3y =0</pre>

The bottom two give you:

<pre>y-z=0 , so y=z</pre>

Then substitute that in the first equation to get:

<pre>2x+8y=0</pre>

Then double the third equation above and subtract:

<pre>2y=0</pre>

And lo and behold:

<pre>x=0
y=0
z=0</pre>

Ok, his job is safe.