Question

6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8

Answer 1

Given: 6x + y − 3z = 5
            x + 3y − 2z = 5
           2x + y + 4z = 8

\[D = \begin{vmatrix}6 & 1 & - 3 \\ 1 & 3 & - 2 \\ 2 & 1 & 4\end{vmatrix}\] 
\[ = 6(12 + 2) - 1(4 + 4) - 3(1 - 6)\] 
\[ = 6(14) - 1(8) - 3( - 5)\] 
\[ = 91\] 
\[ D_1 = \begin{vmatrix}5 & 1 & - 3 \\ 5 & 3 & - 2 \\ 8 & 1 & 4\end{vmatrix}\] 
\[ = 5(12 + 2) - 1(20 + 16) - 3(5 - 24)\] 
\[ = 5(14) - 1(36) - 3( - 19)\] 
\[ = 91\] 
\[ D_2 = \begin{vmatrix}6 & 5 & - 3 \\ 1 & 5 & - 2 \\ 2 & 8 & 4\end{vmatrix}\] 
\[ = 6(20 + 16) - 5(4 + 4) - 3(8 - 10)\] 
\[ = 6(36) - 5(8) - 3( - 2)\] 
\[ = 182\] 
\[ D_3 = \begin{vmatrix}6 & 1 & 5 \\ 1 & 3 & 5 \\ 2 & 1 & 8\end{vmatrix}\] 
\[ = 6(24 - 5) - 1(8 - 10) + 5(1 - 6)\] 
\[ = 6(19) - 1( - 2) + 5( - 5)\] 
\[ = 91\] 
Now, 
\[x = \frac{D_1}{D} = \frac{91}{91} = 1\] 
\[y = \frac{D_2}{D} = \frac{182}{91} = 2\] 
\[z = \frac{D_3}{D} = \frac{91}{91} = 1\] 
\[ \therefore x = 1, y = 2\text{ and }z = 1\]