Question

The number of $3\times 3$ metrices $A$ whose entries are either $0$ or $1$ and for which the system $A=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$ has exactly two distinct solution is?

• A. $0$
• B. $2^9-1$
• C. $168$
• D. $2$

Answer 1

The correct option is A $0$

Let $A=\left[\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_3\\ a_3& b_3& c_3\end{array}\right]$

where $a_i,b_i$ and $c_i$ have values $0$ or $1$

for $i=1,2,3$

Then, the given system is equivalent to

$a_{1}x+b_{1}y+c_{1}z=1$

$a_{2}x+b_{2}y+c_{2}z=0$

$a_{3}x+b_{3}y+c_{3}z=0$

which represent three distinct planes. But three planes cannot intersect at two distinct points, therefore no such system exists.

Hence, the correct answer.