Question

3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.

Answer 1

Given: 3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20

$$\begin{gathered} D=\left|\begin{array}{ccc}3& -1& 2\\ 2& -1& 1\\ 3& 6& 5\end{array}\right| \\ 3\left(-5-6\right)+1\left(10-3\right)+2\left(12+3\right)=4 \end{gathered}$$

Since D is non-zero, the system of linear equations is consistent and has a unique solution.

$$\begin{gathered} {D}_1=\left|\begin{array}{ccc}6& -1& 2\\ 2& -1& 1\\ 20& 6& 5\end{array}\right| \\ =6\left(-5-6\right)+1\left(10-20\right)+2\left(12+20\right) \\ =-66-10+64 \\ =-12 \\ {D}_2=\left|\begin{array}{ccc}3& 6& 2\\ 2& 2& 1\\ 3& 20& 5\end{array}\right| \\ =3\left(10-20\right)-6\left(10-3\right)+2\left(40-6\right) \\ =-30-42+68 \\ =-4 \\ {D}_3=\left|\begin{array}{ccc}3& -1& 6\\ 2& -1& 2\\ 3& 6& 20\end{array}\right| \\ =3\left(-20-12\right)+1\left(40-6\right)+6\left(12+3\right) \\ =-96+34+90 \\ =28 \\ \mathrm{Now}, \\ x=\frac{D_1}{D}=\frac{-12}{4}=-3 \\ y=\frac{D_2}{D}=\frac{-4}{4}=-1 \\ z=\frac{D_3}{D}=\frac{28}{4}=7 \\ \therefore x=-3,y=-1\mathrm{and}\mathrm{z}=7 \end{gathered}$$