I need to solve this sytem for $u,v,w$. I´ve tried basic algebra, but my answer does not mach the one from the book.

I need to solve this sytem for $u,v,w$. I´ve tried basic algebra, but my answer does not mach the one from the book.

On
You can write it as a matrix multiplication
$$ \begin{bmatrix}\vec{a} & \vec{b} & \vec{c} \end{bmatrix}^\top = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & -1 \\ 0 & 1 & -2 \end{bmatrix} \begin{bmatrix}\vec{u} & \vec{v} & \vec{w} \end{bmatrix}^\top $$
$$ \begin{bmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{bmatrix}= \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & -1 \\ 0 & 1 & -2 \end{bmatrix} \begin{bmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{bmatrix} $$
$$ \begin{bmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{bmatrix}= \begin{bmatrix} u_x+w_x & u_y+w_y & u_z + w_z \\ 2 u_x+v_x-w_x & 2 u_y+v_y-w_y &2 u_z+v_z-w_z \\ v_x -2 w_x &v_y -2 w_y & v_z -2 w_z \end{bmatrix} \checkmark$$
$$ \begin{bmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & -1 \\ 0 & 1 & -2 \end{bmatrix}^{-1} \begin{bmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{bmatrix} $$
$$ \begin{bmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{bmatrix} = \begin{bmatrix}-1 & 1 & -1 \\ 4 & -2 & 3 \\ 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{bmatrix} $$
$$\begin{bmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{bmatrix} = \begin{bmatrix} -a_x+b_x-c_x & -a_y+b_y-c_y & -a_z+b_z-c_z \\ 4 a_x -2 b_x+3 c_x &4 a_y -2 b_y+3 c_y & 4 a_z -2 b_z+3 c_z \\ 2 a_x - b_x+ c_x & 2 a_y - b_y+ c_y & 2 a_z - b_z+ c_z \end{bmatrix} $$
An easy way (unless this is a course on using matrix methods) is to sequentially substitute: $$ \vec{u} = \vec{a} - \vec{w} $$ $$ \vec{b} = 2\vec{u} + \vec{v} - \vec{w} = 2\vec{a} + \vec{v} - 3\vec{w} \longrightarrow \vec{v} = \vec{b} - 2\vec{a} + 3\vec{w} $$ $$ \vec{c} = \vec{v} - 2\vec{w} = \vec{b} - 2\vec{a} + \vec{w} \longrightarrow \vec{w} = \vec{c} - \vec{b} + 2\vec{a} $$ And having found $\vec{w}$ back-substitute: $$ \vec{w} = 2\vec{a} - \vec{b} + \vec{c} $$ $$ \vec{v} = \vec{b} - 2\vec{a} + 3\vec{w} = 4\vec{a} -2\vec{b} + 3\vec{c} $$ $$ \vec{u} = \vec{a} - \vec{w} = -\vec{a} + \vec{b} -\vec{c} $$
You should check that this solves the equations by plugging this in.