Question 1131091
<pre>{{{(matrix(2,2,1,1,1,2))X+(matrix(2,2,2,-1,-1,1))X(matrix(2,2,1,5,1,2))=(matrix(2,2,1,1,1,1))}}}

Let {{{X=(matrix(2,2,a,b,c,d))}}}

{{{(matrix(2,2,1,1,1,2))(matrix(2,2,a,b,c,d))+(matrix(2,2,2,-1,-1,1))(matrix(2,2,a,b,c,d))(matrix(2,2,1,5,1,2))=(matrix(2,2,1,1,1,1))}}}

{{{(matrix(2,2,a+c,b+d,a+2c,b+2d))+(matrix(2,2,2a-c,2b-d,-a+c,-b+d))(matrix(2,2,1,5,1,2))=(matrix(2,2,1,1,1,1))}}}

{{{(matrix(2,2,a+c,b+d,a+2c,b+2d))+(matrix(2,2,(2a-c)+(2b-d),5(2a-c)+2(2b-d),(-a+c)+(-b+d),5(-a+c)+2(-b+d)))=(matrix(2,2,1,1,1,1))}}}


{{{(matrix(2,2,a+c,b+d,a+2c,b+2d))+(matrix(2,2,2a+2b-c-d,10a+4b-5c-2d,-a-b+c+d,-5a-2b+5c+2d))=(matrix(2,2,1,1,1,1))}}}

{{{(matrix(2,2,3a+2b-d,10a+5b-5c-d,-b+3c+d,-5a-b+3c+4d))=(matrix(2,2,1,1,1,1))}}}

Equate corresponding elements on both sides:

{{{system(3a+2b-d=1,10a+5b-5c-d=1,-b+3c+d=1,-5a-b+3c+4d=1)}}} 

Solve that system and get {{{matrix(1,4,a=9/80,b=17/40,c=33/80,d=3/16)}}}

Therefore,

{{{X=(matrix(2,2,a,b,c,d))=(matrix(2,2,9/80,17/40,33/80,3/16))}}}

Note: this is so tedious, I could have made an error or errors. 
So although this is the correct procedure, you'd better check it. 
I didn't.

Edwin</pre>