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

\[D = \begin{vmatrix}3 & - 1 & 2 \\ 2 & - 1 & 1 \\ 3 & 6 & 5\end{vmatrix}\] 
\[3\left( - 5 - 6 \right) + 1\left( 10 - 3 \right) + 2\left( 12 + 3 \right) = 4\] 
Since D is non-zero, the system of linear equations is consistent and has a unique solution.
\[ D_1 = \begin{vmatrix}6 & - 1 & 2 \\ 2 & - 1 & 1 \\ 20 & 6 & 5\end{vmatrix}\] 
\[ = 6\left( - 5 - 6 \right) + 1\left( 10 - 20 \right) + 2\left( 12 + 20 \right)\] 
\[ = - 66 - 10 + 64\] 
\[ = - 12\] 
\[ D_2 = \begin{vmatrix}3 & 6 & 2 \\ 2 & 2 & 1 \\ 3 & 20 & 5\end{vmatrix}\] 
\[ = 3\left( 10 - 20 \right) - 6\left( 10 - 3 \right) + 2\left( 40 - 6 \right)\] 
\[ = - 30 - 42 + 68\] 
\[ = - 4\] 
\[ D_3 = \begin{vmatrix}3 & - 1 & 6 \\ 2 & - 1 & 2 \\ 3 & 6 & 20\end{vmatrix}\] 
\[ = 3\left( - 20 - 12 \right) + 1\left( 40 - 6 \right) + 6\left( 12 + 3 \right)\] 
\[ = - 96 + 34 + 90\] 
\[ = 28\] 
Now, 
\[x = \frac{D_1}{D} = \frac{- 12}{4} = - 3\] 
\[y = \frac{D_2}{D} = \frac{- 4}{4} = - 1\] 
\[z = \frac{D_3}{D} = \frac{28}{4} = 7\] 
\[ \therefore x = - 3, y = - 1\text{ and }z = 7\]