The correct option is B more than 2
Let x be the number of 1's on the main diagonal and y be the number of 1's above the main diagonal, then
x+2y=5
⇒x=1,y=2orx=3,y=1.
When x=1, the main diagonal can be chosen in 3 ways, and the elements above the main diagonal in 3 ways. Therefore, there are 9 such matrices.These are
A1=⎡⎢⎣111100100⎤⎥⎦, A2=⎡⎢⎣011110100⎤⎥⎦
A3=⎡⎢⎣011100101⎤⎥⎦, A4=⎡⎢⎣101001110⎤⎥⎦
A5=⎡⎢⎣001011110⎤⎥⎦, A6=⎡⎢⎣001001111⎤⎥⎦
A7=⎡⎢⎣110101010⎤⎥⎦, A8=⎡⎢⎣010111010⎤⎥⎦
A9=⎡⎢⎣100011011⎤⎥⎦
When x=3, the main diagonal can be chosen in 3 way; and the element above the main diagonal in 3 ways. Therefore, there are 3 such matrices.
A10=⎡⎢⎣110110001⎤⎥⎦, A11=⎡⎢⎣101010101⎤⎥⎦,
A12=⎡⎢⎣100011011⎤⎥⎦
Let X=⎡⎢⎣xyz⎤⎥⎦ and B=⎡⎢⎣100⎤⎥⎦, and note that
A1X=B and A12X=B have infinite number of solutions.
AX=B has no solution when
A=A6,A8,A10,A11