Question 1062852
<pre>
Just the (a) part:

Let A = {{{(matrix(2,2,w,x,y,z))}}}

be idempotent.  Then

   {{{A^2}}}{{{""=""}}}{{{A}}}

 {{{(matrix(2,2,w,x,y,z))^2}}}{{{""=""}}}{{{(matrix(2,2,w,x,y,z))}}}

{{{(matrix(2,2,w,x,y,z))*(matrix(2,2,w,x,y,z))}}}{{{""=""}}}{{{(matrix(2,2,w,x,y,z))}}}

{{{(matrix(2,2,w^2+xy,wx+xz,wy+yz,xy+z^2))}}}{{{""=""}}}{{{(matrix(2,2,w,x,y,z))}}}

So we set each elements on the left equal to the corresponding
element on the right:

{{{system(w^2+xy=w, wx+xz=x, wy+yz=y, xy+z^2=z)}}}

Divide the second one through by x or the third one through
by y:

w + z = 1 

So we pick, say w = 1/4 and z = 3/4 since they have sum 1 

Substituting in

{{{w^2+xy=w}}}

{{{(1/4)^2+xy=1/4}}}

{{{1/16+xy=1/4}}}

{{{1+16xy=4}}}
 {{{16xy=3}}}
 {{{xy=3/16}}}

Choose x=1 and y=3/16, because their product is 3/16.

Substitute in

A = {{{(matrix(2,2,w,x,y,z))}}}

A = {{{(matrix(2,4,1/4,"","",1,3/16,"","",3/4)))}}}

Edwin</pre>