Question

5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7

Answer 1

Given: 5x − 7y + z = 11
           6x − 8y − z = 15
           3x + 2y − 6z = 7

\[D = \begin{vmatrix}5 & - 7 & 1 \\ 6 & - 8 & - 1 \\ 3 & 2 & - 6\end{vmatrix}\] 
\[ = 5(48 + 2) + 7( - 36 + 3) + 1(12 + 24)\] 
\[ = 5(50) + 7( - 33) + 1(36)\] 
\[ = 55\] 
\[ D_1 = \begin{vmatrix}11 & - 7 & 1 \\ 15 & - 8 & - 1 \\ 7 & 2 & - 6\end{vmatrix}\] 
\[ = 11(48 + 2) + 7( - 90 + 7) + 1(30 + 56)\] 
\[ = 11(50) + 7( - 83) + 1(86)\] 
\[ = 55\] 
\[ D_2 = \begin{vmatrix}5 & 11 & 1 \\ 6 & 15 & - 1 \\ 3 & 7 & - 6\end{vmatrix}\] 
\[ = 5( - 90 + 7) - 11( - 36 + 3) + 1(42 - 45)\] 
\[ = 5( - 83) - 11( - 33) + 1( - 3)\] 
\[ = - 55\] 
\[ D_3 = \begin{vmatrix}5 & - 7 & 11 \\ 6 & - 8 & 15 \\ 3 & 2 & 7\end{vmatrix}\] 
\[ = 5( - 56 - 30) + 7(42 - 45) + 11(12 + 24)\] 
\[ = 5( - 86) + 7( - 3) + 11(36)\] 
\[ = - 55\] 
Now, 
\[x = \frac{D_1}{D} = \frac{55}{55} = 1\] 
\[y = \frac{D_2}{D} = \frac{- 55}{55} = - 1\] 
\[z = \frac{D_3}{D} = \frac{- 55}{55} = - 1\] 
\[ \therefore x = 1, y = - 1\text{ and }z = - 1\]