Chapter 8 - Test - Page 951: 2

The solution is, $x=t,y=t-1,z=t;t\in \mathbb{R}$.

Consider the given system of equations $\begin{align} & x-2y+z=2 \\ & 2x-y-z=1 \\ \end{align}$ Therefore, the system of equations can be expressed as below: $\left[ \begin{matrix} 1 & -2 & 1 \\ 2 & -1 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 2 \\ 1 \\ \end{matrix} \right]$ Consider, $z=t,t\in \mathbb{R}$. So, $\begin{align} & \left[ \begin{matrix} 1 & -2 \\ 2 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]+t\left[ \begin{matrix} 1 \\ -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 \\ 1 \\ \end{matrix} \right] \\ & \left[ \begin{matrix} 1 & -2 \\ 2 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]=\left[ \begin{matrix} 2 \\ 1 \\ \end{matrix} \right]-t\left[ \begin{matrix} 1 \\ -1 \\ \end{matrix} \right] \\ & \left[ \begin{matrix} 1 & -2 \\ 2 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]=\left[ \begin{matrix} 2-t \\ 1+t \\ \end{matrix} \right] \end{align}$ Therefore, this system can be expressed as $AX=B$ Where $A=\left[ \begin{matrix} 1 & -2 \\ 2 & -1 \\ \end{matrix} \right],B=\left[ \begin{matrix} 2-t \\ 1+t \\ \end{matrix} \right],X=\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ Therefore, the solution of the equation is given by: $X={{A}^{-1}}B$ Consider the determinant of the matrix $\begin{align} & \left| A \right|=\left| \begin{array}{*{35}{r}} 1 & -2 \\ 2 & -1 \\ \end{array} \right| \\ & =-1+4 \\ & =3 \end{align}$ Consider the adjoint of the matrix $\begin{align} & \text{adj}\left( A \right)={{\left[ \begin{matrix} +\left( -1 \right) & -\left( 2 \right) \\ -\left( -2 \right) & +\left( 1 \right) \\ \end{matrix} \right]}^{t}} \\ & ={{\left[ \begin{array}{*{35}{r}} -1 & -2 \\ 2 & 1 \\ \end{array} \right]}^{t}} \\ & =\left[ \begin{array}{*{35}{r}} -1 & 2 \\ -2 & 1 \\ \end{array} \right] \end{align}$ Therefore, the inverse of the matrix $A$ is ${{A}^{-1}}=\frac{1}{3}\left[ \begin{array}{*{35}{r}} -1 & 2 \\ -2 & 1 \\ \end{array} \right]$ Therefore $\begin{align} & {{A}^{-1}}B=\frac{1}{3}\left[ \begin{array}{*{35}{r}} -1 & 2 \\ -2 & 1 \\ \end{array} \right]\left[ \begin{matrix} 2-t \\ 1+t \\ \end{matrix} \right] \\ & =\frac{1}{3}\left[ \begin{matrix} -\left( 2-t \right)+2\left( 1+t \right) \\ -2\left( 2-t \right)+\left( 1+t \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{-2+t+2+2t}{3} \\ \frac{-4+2t+1+t}{3} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} t \\ t-1 \\ \end{matrix} \right] \end{align}$ Therefore, the solution of the system of equations is $x=t,y=t-1,z=t$