Question 1164525
<pre>A square matrix is Invertible if and only if its determinant is non-zero.

{{{(matrix(3,3,
1, 2, x,
3, 0, y,
1, 1, z))}}}

We find its determinant

{{{abs(matrix(3,3,
1, 2, x,
3, 0, y,
1, 1, z))}}}{{{""=""}}}{{{1*0*z+2*y*1+x*3*1-x*0*1-2*3*z-1*y*1}}}{{{""=""}}}{{{0+2y+3x-0-6z-y}}}{{{""=""}}}{{{3x+y-6z}}}

We can choose any values for x, y, z such that 

{{{3x+y-6z<>0}}}

I'll arbitrary choose x=3, y=-2, z=1 (to make the determinant 1, so its
inverse will have all integer elements:

{{{(matrix(3,3,
1, 2, 3,
3, 0, -2,
1, 1, 1))}}}

Then its inverse is

{{{(matrix(3,3,
2, 1, -4,
-5, -2, 1,
3, 1, -6))}}}

Edwin</pre>