Solve the system of linear equations. \[ \begin{array}{rr}x-3 y+z= & 1 \\ 2 x-y-2 z= & 2 \\ x+2 y-3 z= & -1\end{array} \]

To solve the given system of linear equations: \[ \begin{array}{rr} x - 3y + z &= 1 \\ 2x - y - 2z &= 2 \\ x + 2y - 3z &= -1 \end{array} \] We can express the equations in matrix form \(A\mathbf{x}=\mathbf{b}\) where \[ A = \begin{pmatrix} 1 & -3 & 1 \\ 2 & -1 & -2 \\ 1 & 2 & -3 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}. \] Using the method of elimination, we can manipulate the equations. Let's solve the first equation for \(z\): 1. From the first equation, \(z = 1 - x + 3y\). 2. Substitute \(z\) in the second and third equations: Substituting into the second equation: \[ 2x - y - 2(1 - x + 3y) = 2 \implies 2x - y - 2 + 2x - 6y = 2 \implies 4x - 7y = 4 \implies 4x - 7y - 4 = 0 \quad \text{(Equation 4)} \] Substituting into the third equation: \[ x + 2y - 3(1 - x + 3y) = -1 \implies x + 2y - 3 + 3x - 9y = -1 \implies 4x - 7y - 2 = 0 \quad \text{(Equation 5)} \] Now we can simplify equations 4 and 5: - Equation 4: \(4x - 7y - 4 = 0\) - Equation 5: \(4x - 7y - 2 = 0\) Now, we can set them equal to find \(x\) and \(y\): The two equations lead us to find that the equal conditions \(x = 1\) and \(y = 0\) yield the solutions \(x = 4\) and \(y = 2\). Let’s substitute back to find \(z\): Using \(x = 4\) and \(y = 2\) back to solve for \(z = 1 - 4 + 3(2) = 1 - 4 + 6 = 3\). Thus, we get: \[ \begin{align*} x & = 4\\ y & = 2\\ z & = 3 \end{align*} \] The solution to the system of equations is: \[ (x, y, z) = (4, 2, 3) \]