Find Sum of Variables x y z for Matrix Transformation

Given that $M$ is a $3 \times 3$ matrix and its action on the standard basis vectors is given by:
$M \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \quad M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}, \quad \text{and} \quad M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$

The columns of matrix $M$ correspond directly to the transformations of the standard basis vectors. Thus, the matrix $M$ is formed as: $M = \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix}$

We are given that: $M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$

Substituting the matrix $M$ gives: $\begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$

This matrix equation yields the following system of linear equations:

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

From equation (1), we express $x$ in terms of $z$: $x = z + 1$

Substituting $x = z + 1$ into equations (2) and (3):

• From equation (2): $2(z + 1) + y + z = 7 \implies y + 3z + 2 = 7 \implies y + 3z = 5 \quad \text{--- (4)}$

• From equation (3): $3(z + 1) + 2y + z = 11 \implies 2y + 4z + 3 = 11 \implies 2y + 4z = 8 \implies y + 2z = 4 \quad \text{--- (5)}$

Subtracting equation (5) from equation (4): $(y + 3z) - (y + 2z) = 5 - 4 \implies z = 1$

Substituting $z = 1$ into equation (5) to find $y$: $y + 2(1) = 4 \implies y = 2$

Substituting $z = 1$ to find $x$: $x = 1 + 1 = 2$

Thus, the values of the variables are $x = 2$, $y = 2$, and $z = 1$.

Finally, the sum $x + y + z$ is: $x + y + z = 2 + 2 + 1 = 5$