The problem involves a $3 \times 3$ matrix $M$ and its action on standard basis vectors. We need to find the sum $x+y+z$ where $M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$.

Finding the Matrix M

The columns of matrix $M$ are determined by its action on the standard basis vectors:

• $M \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ gives the first column of $M$.
• $M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}$ gives the second column of $M$.
• $M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$ gives the third column of $M$.

Therefore, the matrix $M$ is:

$ M = \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix} $

Solving the Vector Equation

We are given the equation $M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$. Substituting the matrix $M$ we found:

$ \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix} $

This matrix equation corresponds to the following system of linear equations:

1. $x - z = 1$
2. $2x + y + z = 7$
3. $3x + 2y + z = 11$

Calculating x, y, and z

We can solve this system. From equation (1), we get $x = 1 + z$. Substitute this into equations (2) and (3):

• Equation (2) becomes: $2(1+z) + y + z = 7 \implies 2 + 2z + y + z = 7 \implies y + 3z = 5$. (Equation 4)
• Equation (3) becomes: $3(1+z) + 2y + z = 11 \implies 3 + 3z + 2y + z = 11 \implies 2y + 4z = 8 \implies y + 2z = 4$. (Equation 5)

Now, solve the system formed by equations (4) and (5):

1. $y + 3z = 5$
2. $y + 2z = 4$

Subtracting equation (5) from equation (4) gives:

$ (y + 3z) - (y + 2z) = 5 - 4 $ $ z = 1 $

Substitute $z=1$ back into equation (5):

$ y + 2(1) = 4 \implies y + 2 = 4 \implies y = 2 $

Substitute $z=1$ back into the expression for $x$ ($x = 1+z$):

$ x = 1 + 1 \implies x = 2 $

So, we have $x=2$, $y=2$, and $z=1$. Let's check this with the original vector equation:

$ M \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(2)+0(2)-1(1) \\ 2(2)+1(2)+1(1) \\ 3(2)+2(2)+1(1) \end{pmatrix} = \begin{pmatrix} 2-1 \\ 4+2+1 \\ 6+4+1 \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix} $

This matches the given condition.

Calculating the Sum x+y+z

Finally, calculate the sum $x+y+z$:

$ x+y+z = 2 + 2 + 1 = 5 $