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Question
6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8
Solution
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,
\[x = \frac{D_1}{D} = \frac{91}{91} = 1\]
\[y = \frac{D_2}{D} = \frac{182}{91} = 2\]
\[z = \frac{D_3}{D} = \frac{91}{91} = 1\]
\[ \therefore x = 1, y = 2\text{ and }z = 1\]
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