# How do you solve the system x+2y+z=2, 2x+3y+3z=-3, and 2x+3y+2z=2?

Aug 14, 2017

$x = 3 , y = 2 , z = - 5$

#### Explanation:

We have the equation:

$\left\{\begin{matrix}x + 2 y + z = 2 \\ 2 x + 3 y + 3 z = - 3 \\ 2 x + 3 y + 2 z = 2\end{matrix}\right.$

In vector matrix form we can write this as:

$\left(\begin{matrix}1 & 2 & 1 \\ 2 & 3 & 3 \\ 2 & 3 & 2\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}2 \\ - 3 \\ 2\end{matrix}\right)$

Or as:

$\boldsymbol{A} \boldsymbol{\underline{x}} = \boldsymbol{\underline{b}}$

Where:

 bb(A) = ( (1,2,1), (2,3,3), (2,3,2) ); bb(ul x) = ( (x), (y), (z) ); bb(ul b) = ( (2), (-3), (2) )

So then, the solution can be found by inverting the matrix $\boldsymbol{A}$, to get:

$\boldsymbol{\underline{x}} = {\boldsymbol{A}}^{-} 1 \boldsymbol{\underline{b}}$

To invert the matrix A, first we compute the matrix of cofactors:

$C o f \left(\boldsymbol{A}\right) = \left(\begin{matrix}+ | \left(3 3\right) & \left(3 2\right) | & - | \left(2 3\right) & \left(2 2\right) | & + | \left(2 3\right) & \left(2 3\right) | \\ - | \left(2 1\right) & \left(3 2\right) | & + | \left(1 1\right) & \left(2 2\right) | & - | \left(1 2\right) & \left(2 3\right) | \\ + | \left(2 1\right) & \left(3 3\right) | & - | \left(1 1\right) & \left(2 3\right) | & + | \left(1 2\right) & \left(2 3\right) |\end{matrix}\right)$

$\text{ } = \left(\begin{matrix}6 - 9 & - \left(4 - 6\right) & 6 - 6 \\ - \left(4 - 3\right) & 2 - 2 & - \left(3 - 4\right) \\ 6 - 3 & - \left(3 - 2\right) & 3 - 4\end{matrix}\right)$

$\text{ } = \left(\begin{matrix}- 3 & 2 & 0 \\ - 1 & 0 & 1 \\ 3 & - 1 & - 1\end{matrix}\right)$

We then compute the adjoint of $A$ or ${\boldsymbol{C}}^{T}$

$a \mathrm{dj} \left(\boldsymbol{A}\right) = \left(\begin{matrix}- 3 & - 1 & 3 \\ 2 & 0 & - 1 \\ 0 & 1 & - 1\end{matrix}\right)$

We must also compute the determinant of $\boldsymbol{A}$:

$\det \left(\boldsymbol{A}\right) = \left(1\right) | \left(3 , 3\right) , \left(3 , 2\right) | - \left(2\right) | \left(2 , 3\right) , \left(2 , 2\right) | + \left(1\right) | \left(2 , 3\right) , \left(2 , 3\right) |$
$\text{ } = \left(6 - 9\right) - \left(2\right) \left(4 - 6\right) + \left(6 - 6\right)$
$\text{ } = - 3 + 4$
$\text{ } = 1$

And so we get the inverse using:

${\boldsymbol{A}}^{-} 1 = \frac{1}{\det} \left(\boldsymbol{A}\right) a \mathrm{dj} \left(\boldsymbol{A}\right)$
$\setminus \setminus \setminus = 1 \cdot a \mathrm{dj} \left(\boldsymbol{A}\right)$
$\setminus \setminus \setminus = \left(\begin{matrix}- 3 & - 1 & - 3 \\ 2 & 0 & - 1 \\ 0 & 1 & - 1\end{matrix}\right)$

Thus, the solution of the linear system is:

$\boldsymbol{\underline{x}} = \left(\begin{matrix}- 3 & - 1 & 3 \\ 2 & 0 & - 1 \\ 0 & 1 & - 1\end{matrix}\right) \left(\begin{matrix}2 \\ - 3 \\ 2\end{matrix}\right)$

$\setminus \setminus \setminus = \left(\begin{matrix}- 6 + 3 + 6 \\ 4 + 0 - 2 \\ 0 - 3 - 2\end{matrix}\right)$
$\setminus \setminus \setminus = \left(\begin{matrix}3 \\ 2 \\ - 5\end{matrix}\right)$

Hence the solution is:

$x = 3 , y = 2 , z = - 5$