Let M be a 33 matrix such that Mpmatrix100pmatrix=pmatrix123pmatrix, Mpmatrix010pmatrix=pm…

Question

Let $M$ be a $3	imes3$ matrix such that $M\begin{pmatrix}1\0\0\end{pmatrix}=\begin{pmatrix}1\2\3\end{pmatrix}$, $M\begin{pmatrix}0\1\0\end{pmatrix}=\begin{pmatrix}0\1\2\end{pmatrix}$, $M\begin{pmatrix}0\0\1\end{pmatrix}=\begin{pmatrix}-1\1\1\end{pmatrix}$. If $M\begin{pmatrix}x\y\z\end{pmatrix}=\begin{pmatrix}1\7\11\end{pmatrix}$, then $x+y+z$ equals:

Accepted Answer

The columns of $M$ are the given images, so $M=\begin{pmatrix}1&0&-1\2&1&1\3&2&1\end{pmatrix}$.

The system $M\begin{pmatrix}x\y\z\end{pmatrix}=\begin{pmatrix}1\7\11\end{pmatrix}$ gives:
$$x-z=1,\qquad 2x+y+z=7,\qquad 3x+2y+z=11.$$

From the first, $x=1+z$. Substituting into the second: $2(1+z)+y+z=7\Rightarrow y+3z=5\Rightarrow y=5-3z$.

Substituting into the third: $3(1+z)+2(5-3z)+z=11\Rightarrow13-2z=11\Rightarrow z=1$.

So $x=2,\ y=2,\ z=1$, and $x+y+z=5$.

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