Chapter 2.5, Problem 6P

Explanation of Solution

Solving the given linear system:

Consider the given system of linear equations,

$\begin{array}{l}{x}_1+x_3=4\\ 4x_1+x_2-2x_3=0\\ 3x_1+x_2-x_3=2\end{array}$

For a system of the form $Ax=b$, it can be written as follows:

$x=A^{-1}b$

For the given system of linear equations, the inverse obtained in Problem 2 can be used, which is given below,

$A^{-1}=\left[\begin{array}{l}\frac{1}{2}\frac{1}{2}-\frac{1}{2}\\ -1-23\\ \frac{1}{2}-\frac{1}{2}\frac{1}{2}\end{array}\right]$

Thus, it can be written as follows:

$\begin{array}{l}{A}^{-1}\end{array}$