Question 341598
{{{A=(matrix(2,2,2,-5,3,1))}}}    

Evaluate the square of A and find x,y,z not all zero such that the matrix 
    
xI+yA+zAA  is the zero matrix.
<pre><b>
We will find those three terms separately:

{{{xI=x(matrix(2,2,1,0,0,1))=(matrix(2,2,x,0,0,x))}}}

{{{yA=y(matrix(2,2,2,-5,3,1))=(matrix(2,2,2y,-5y,3y,y))}}}

To find zAA we first find the square of A, which is AA

{{{AA=(matrix(2,2,2,-5,3,1))(matrix(2,2,2,-5,3,1))=
(matrix(2,2,(2)(2)+(-5)(3),(2)(-5)+(-5)(1), (3)(2)+(1)(3),(3)(-5)+(1)(1)))= 
(matrix(2,2,4-15,-10-5, 6+3,-15+1))=(matrix(2,2,-11,-15,9,-14))}}}

(That was the first thing asked for, the square of A)

Now we multiply that by z to get zAA:

{{{zAA = z(matrix(2,2,-11,-15,9,-14))= (matrix(2,2,-11z,-15z,9z,-14z))}}}

Next we find their sum:

{{{xI+yA+zAA=(matrix(2,2,x,0,0,x))+(matrix(2,2,2y,-5y,3y,y))+(matrix(2,2,-11z,-15z,9z,-14z))=(matrix(2,2,x+2y-11z,-5y-15z, 3y+9z, x+y-14z))}}}

This must equal to the zero 2×2 matrix, so,

{{{(matrix(2,2,x+2y-11z,-5y-15z, 3y+9z, x+y-14z))=(matrix(2,2,0,0,0,0))}}}

We form this system by setting each element = 0:

{{{system(x+2y-11z=0,-5y-15z=0, 3y+9z=0, x+y-14z=0)}}}

We solve that system by Gaussian elimination:

{{{(matrix(4,5,

1,2,-11,"|",0,
0,-5,-15,"|",0,
0,3,9,"|",0, 
1,1,-14,"|",0))}}}

By using row operations we find the row reduced echelon form
of that augmented matrix.

{{{(matrix(4,5,

1,0,-17,"|",0,
0,1,3,"|",0,
0,0,0,"|",0, 
0,0,0,"|",0))}}}

The first two rows of that matrix gives us
the system:

{{{system(x-17z=0,y+3z=0)}}}

or

{{{system(x=17z,y=-3z)}}}

To find all solutions, we choose a constant k for z, then

{{{system(x=17k,y=-3k,z=k)}}}

Edwin</pre>