# 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

-
Whatever you might mean with your notation, note that $2 \cdot [3, 4, 5] - [0,1,2] = [6,7,8]$. –  Henry T. Horton Jul 17 '12 at 15:24
@HenryT.Horton Thanhs –  Argha Jul 17 '12 at 15:30
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).