Answer
$x=\begin{bmatrix} -6\\ 1\\ 3 \end{bmatrix}$
Work Step by Step
Write the system in the matrix form: $\begin{bmatrix} 3 & 4 & 5\\ 2 & 10 & 1\\ 4 &1 & 8 \end{bmatrix}.\begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 1\\ 1 \\ 1 \end{bmatrix}$ Find the inverse for matrix $A=\begin{bmatrix} 3 & 4 & 5\\ 2 & 10 & 1\\ 4 &1 & 8 \end{bmatrix}$: $\begin{bmatrix} 3 & 4 & 5 |1 & 0 & 0\\ 2 & 10 & 1 | 0 & 1 & 0\\ 4 &1 & 8 | 0 & 0 & 1 \end{bmatrix} \approx^1\begin{bmatrix} 1 & \frac{4}{3} & \frac{5}{3} |\frac{1}{3} & 0 & 0\\ 2 & 10 & 1 | 0 & 1 & 0\\ 4 &1 & 8 | 0 & 0 & 1 \end{bmatrix} \approx^2 \begin{bmatrix} 1 & \frac{4}{3} & \frac{5}{3} |\frac{1}{3} & 0 & 0\\ 0 & \frac{22}{3} & -\frac{7}{3} |-\frac{2}{3} & 1 & 0\\ 0 &-\frac{13}{3} & \frac{4}{3} |-\frac{4}{3}& 0 & 1 \end{bmatrix} \approx^3 \begin{bmatrix} 1 & \frac{4}{3} & \frac{5}{3} |\frac{1}{3} & 0 & 0\\ 0 & 1& -\frac{7}{22} |-\frac{1}{11} & \frac{3}{22} & 0\\ 0 &-\frac{13}{3} & \frac{4}{3} |-\frac{4}{3}& 0 & 1 \end{bmatrix} \approx^4 \begin{bmatrix} 1 & 0 & \frac{23}{11} |\frac{5}{11} & -\frac{2}{11} & 0\\ 0 & 1& -\frac{7}{22} |-\frac{1}{11} & \frac{3}{22} & 0\\ 0 &0 & -\frac{1}{22} |-\frac{19}{11}& \frac{13}{22} & 1 \end{bmatrix} \approx^5 \begin{bmatrix} 1 & 0 & \frac{23}{11} |\frac{5}{11} & -\frac{2}{11} & 0\\ 0 & 1& -\frac{7}{22} |-\frac{1}{11} & \frac{3}{22} & 0\\ 0 &0 & 1|38& -13 & -22 \end{bmatrix} \approx^6 \begin{bmatrix} 1 & 0 & 0|-79& 27 & 46\\ 0 & 1& 0|12 & -4 & -7\\ 0 &0 & 1|38& -13 & -22 \end{bmatrix}$
Hence, $A^{-1}=\begin{bmatrix} -79 & 27& 46\\ 12 & -4 & -7\\ 38 & -13 & -22 \end{bmatrix}$
Using $x=A^{-1}b$ to find x: $X=\begin{bmatrix} -79 & 27& 46\\ 12 & -4 & -7\\ 38 & -13 & -22 \end{bmatrix}.\begin{bmatrix} 1\\ 1 \\ 1 \end{bmatrix}=\begin{bmatrix} -6\\ 1\\ 3 \end{bmatrix}$
