multiple choice matrices problem If $M$ is a $3 \times 3$ matrix such that
$$
\begin{aligned}
\begin{pmatrix} 0 & 1 &2\end{pmatrix}M &= \begin{pmatrix} 1 & 0 &0\end{pmatrix} \text{ and}\\
\begin{pmatrix} 3 & 4 &5\end{pmatrix}M &= \begin{pmatrix} 0 & 1 &0\end{pmatrix} \text{ ,}
\end{aligned}
$$
then $\begin{pmatrix} 6 &7 &8\end{pmatrix}M$ is equal to  
(A) $\begin{pmatrix} 2 &1 &−2\end{pmatrix}$
(B) $\begin{pmatrix} 0 &0 &1\end{pmatrix}$
(C) $\begin{pmatrix} -1 &2 &0\end{pmatrix}$
(D) $\begin{pmatrix} 9 &10 &8\end{pmatrix}$  
i know that $M$ is a $3 \times 3$ matrix so we have 9 unknown and from these two equation we get six unknown so i can't solve it

Considering the Options,we have $\begin{pmatrix} 6 &7 &8\end{pmatrix}M=\begin{pmatrix} -1 &2 &0\end{pmatrix}$. Since If $A$ be an $m\times n$ matrix and we partition $A$ into rows then $$A = \begin{bmatrix}
a(1,:) \\ a(2,:) \\ a(3,:)
\end{bmatrix}$$  If $B$ an $n\times r$ matrix, then the $i$th row of product $AB$ is determind by multiplying the $i$th row of $A$ times $B$. Thus the $i$th row of $AB$ is $a(i,:)B$. In general the product $AB$ can be partitioned into rows as follows: $$AB = \begin{bmatrix}
a(1,:)B \\ a(2,:)B \\ a(3,:)B
\end{bmatrix}$$ Now in this question if we let $A = \begin{bmatrix}
0&1&2 \\ 3&4&5 \\ 6&7&8
\end{bmatrix}$ then have an matrix equation as follows:$$AM = \begin{bmatrix}
1&0&0 \\ 0&1&0 \\ x&y&z
\end{bmatrix}$$Now $$\det (AM)=\det(A)\det(M)=\det(\begin{bmatrix}
1&0&0 \\ 0&1&0 \\ x&y&z
\end{bmatrix})$$ and since $\det(A)=0$ we have $\det(\begin{bmatrix}
1&0&0 \\ 0&1&0 \\ x&y&z
\end{bmatrix})=0$. Therefore $z=0$. And only option is (C).
