Solve the following equation using Cramer's rule: x + y 2z = 1 , 2x 7z = 3 , x + y z = 5 .

Solution Using Cramer’s Rule

Given the system of linear equations: $$\{\begin{array}{l}x+y-2z=1\text{(1)}\\ 2x-7z=3\text{(2)}\\ x+y-z=5\text{(3)}\end{array}$$

Step 1: Write the system in matrix form $AX=B$

The coefficient matrix $A$, variable matrix $X$, and constant matrix $B$ are:

$$A=\left[\begin{array}{ccc}1& 1& -2\\ 2& 0& -7\\ 1& 1& -1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}1\\ 3\\ 5\end{array}\right]$$

Step 2: Compute the determinant of $A$, $\Delta$

$$\Delta =\left|\begin{array}{ccc}1& 1& -2\\ 2& 0& -7\\ 1& 1& -1\end{array}\right|=1\cdot (0\cdot (-1)-(-7)\cdot 1)-1\cdot (2\cdot (-1)-(-7)\cdot 1)+(-2)\cdot (2\cdot 1-0\cdot 1)$$ $$=1\cdot (0+7)-1\cdot (-2+7)+(-2)\cdot (2-0)$$ $$=7-5-4=-2$$

Step 3: Compute ${\Delta }_x$, ${\Delta }_y$, and ${\Delta }_z$

Replace the first column of $A$ with $B$ to find ${\Delta }_x$: $${\Delta }_x=\left|\begin{array}{ccc}1& 1& -2\\ 3& 0& -7\\ 5& 1& -1\end{array}\right|=1\cdot (0\cdot (-1)-(-7)\cdot 1)-1\cdot (3\cdot (-1)-(-7)\cdot 5)+(-2)\cdot (3\cdot 1-0\cdot 5)$$ $$=7-1\cdot (-3+35)-2\cdot 3=7-32-6=-31$$

Replace the second column of $A$ with $B$ to find ${\Delta }_y$: $${\Delta }_y=\left|\begin{array}{ccc}1& 1& -2\\ 2& 3& -7\\ 1& 5& -1\end{array}\right|=1\cdot (3\cdot (-1)-(-7)\cdot 5)-1\cdot (2\cdot (-1)-(-7)\cdot 1)+(-2)\cdot (2\cdot 5-3\cdot 1)$$ $$=1\cdot (-3+35)-1\cdot (-2+7)-2\cdot (10-3)=32-5-14=13$$

Replace the third column of $A$ with $B$ to find ${\Delta }_z$: $${\Delta }_z=\left|\begin{array}{ccc}1& 1& 1\\ 2& 0& 3\\ 1& 1& 5\end{array}\right|=1\cdot (0\cdot 5-3\cdot 1)-1\cdot (2\cdot 5-3\cdot 1)+1\cdot (2\cdot 1-0\cdot 1)$$ $$=1\cdot (0-3)-1\cdot (10-3)+1\cdot (2-0)=-3-7+2=-8$$

Step 4: Solve for $x$, $y$, and $z$

Using Cramer’s rule: $$x=\frac{{\Delta }_x}{\Delta }=\frac{-31}{-2}=\frac{31}{2},y=\frac{{\Delta }_y}{\Delta }=\frac{13}{-2}=-\frac{13}{2},z=\frac{{\Delta }_z}{\Delta }=\frac{-8}{-2}=4$$

Final Answer

The solution to the system is: $$\boxed{x=\frac{31}{2},y=-\frac{13}{2},z=4}$$