Let A=[100−1100−11]=I+C
where I=[100010001],C=[000−1000−10]
Then,
C2=[000000100],
C3=[000000000]=C4=C5=…….
Now,
B=7A20−20A7+2I
=7(I+C)20−20(I+C)7+2I
=7(I+20C+C220C2+...)−20(I+7C+C27C2+..)+2I
=–11I+910C2
=[−11000−1109100−11]
So,
b13=910