Given,
A=[0a110c230]
⇒A2=[0a110c230][0a110c230]
⇒A2=[a+23ca2ca+3c132a2+3c]
Now finding,
A3=[a+23ca2ca+3c132a2+3c][0a110c230]
⇒A3=[2ca+3a2+3ca+2aa+2+3cα+2+3c3+2acca+2c+3c22a+4+6c6+3a+9c2ca+3]
Now equating A3=A and comparing both side we get,
2ca+3=0⇒c=2a−3
And a+2+3c=1
⇒a+2+3(2a−3)=1
⇒a+1−2a9=0
⇒2a2+2a−9=0
⇒a=4−2+4+4×9×2=4−2+76≈46.7≈1.4
Hence, a∈(1,2]
So, on comparing with a∈(n−1,n] we get n=2