# The number of $3\times 3$ matrices A whose entries are either 0 or 1 and for which the system $A=\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}$ has exactly two distinct solution is

$(a)\;0\qquad(b)\;2^9-1\qquad(c)\;168\qquad(d)\;2$

Let $A=\begin{bmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{bmatrix}$
Where $a_i,b_i,c_i$ have values 0 or 1 for $i$=1,2,3
Then the given system is equivalent to
$a_1x+b_1y+c_1z=1$
$a_2x+b_2y+c_2z=0$
$a_3x+b_3y+c_3z=0$
Which represent three distinct planes .But three planes cannot intersect at two distinct points therefore no such system exists.
Hence (a) is the correct answer.