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प्रश्न
2x − 3z + w = 1
x − y + 2w = 1
− 3y + z + w = 1
x + y + z = 1
उत्तर
\[D = \begin{vmatrix}2 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0\end{vmatrix}\]
\[2\begin{vmatrix}- 1 & 0 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} - 0 - 3\begin{vmatrix}1 & - 1 & 2 \\ 0 & - 3 & 1 \\ 1 & 1 & 0\end{vmatrix} - 1\begin{vmatrix}1 & - 1 & 0 \\ 0 & - 3 & 1 \\ 1 & 1 & 1\end{vmatrix}\]
\[ = 2\left[ - 1\left( 0 - 1 \right) - 0\left( 0 - 1 \right) + 2\left( - 3 - 1 \right) \right] - 3\left[ 1\left( 0 - 1 \right) + 1\left( 0 - 1 \right) + 2\left( 0 + 3 \right) \right] - 1\left[ 1\left( - 3 - 1 \right) + 1\left( 0 - 1 \right) + 0\left( 0 + 3 \right) \right]\]
\[ = - 21\]
\[ D_1 = \begin{vmatrix}1 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 2 \\ 1 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0\end{vmatrix}\]
\[1\begin{vmatrix}- 1 & 0 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} - 0 - 3\begin{vmatrix}1 & - 1 & 2 \\ 1 & - 3 & 1 \\ 1 & 1 & 0\end{vmatrix} - 1\begin{vmatrix}1 & - 1 & 0 \\ 1 & - 3 & 1 \\ 1 & 1 & 1\end{vmatrix}\]
\[ = 1\left[ - 1\left( 0 - 1 \right) - 0\left( 0 - 1 \right) + 2\left( - 3 - 1 \right) \right] - 3\left[ 1\left( 0 - 1 \right) + 1\left( 0 - 1 \right) + 2\left( 1 + 3 \right) \right] - 1\left[ 1\left( - 3 - 1 \right) + 1\left( 0 - 1 \right) + 2\left( 1 + 3 \right) \right]\]
\[ = - 21\]
\[ D_2 = \begin{vmatrix}2 & 1 & - 3 & 1 \\ 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0\end{vmatrix}\]
\[ = 2\begin{vmatrix}1 & 0 & 2 \\ 1 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} - 1\begin{vmatrix}1 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} + ( - 3)\begin{vmatrix}1 & 1 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} - 1\begin{vmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1\end{vmatrix}\]
\[2\left[ 1\left( 0 - 1 \right) + 2\left( 1 - 1 \right) \right] - 1\left[ 1\left( 0 - 1 \right) + 2\left( 0 - 1 \right) \right] - 3\left[ 1\left( 0 - 1 \right) - 1\left( 0 - 1 \right) + 2\left( 0 - 1 \right) \right] - 1\left[ 1\left( 1 - 1 \right) - 1\left( 0 - 1 \right) \right]\]
\[ = 6\]
\[ D_3 = \begin{vmatrix}2 & 0 & 1 & 1 \\ 1 & - 1 & 1 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0\end{vmatrix}\]
\[ = 2\begin{vmatrix}- 1 & 1 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} - 0 + 1\begin{vmatrix}1 & - 1 & 2 \\ 0 & - 3 & 1 \\ 1 & 1 & 0\end{vmatrix} - 1\begin{vmatrix}1 & - 1 & 1 \\ 0 & - 3 & 1 \\ 1 & 1 & 1\end{vmatrix}\]
\[ = 2\left[ - 1\left( 0 - 1 \right) - 1\left( 0 - 1 \right) + 2\left( - 3 - 1 \right) \right] + 1\left[ 1\left( 0 - 1 \right) + 1\left( 0 - 1 \right) + 2\left( 0 + 3 \right) \right] - 1\left[ 1\left( - 3 - 1 \right) + 1\left( 0 - 1 \right) + 1\left( 0 + 3 \right) \right]\]
\[ = - 6\]
\[ D_4 = \begin{vmatrix}2 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 1 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 1\end{vmatrix}\]
\[ = 2\begin{vmatrix}- 1 & 0 & 1 \\ - 3 & 1 & 1 \\ 1 & 1 & 1\end{vmatrix} - 0 - 3\begin{vmatrix}1 & - 1 & 1 \\ 0 & - 3 & 1 \\ 1 & 1 & 1\end{vmatrix} - 1\begin{vmatrix}1 & - 1 & 0 \\ 0 & - 3 & 1 \\ 1 & 1 & 1\end{vmatrix}\]
\[ = 2\left[ - 1\left( 1 - 1 \right) + 1\left( - 3 - 1 \right) \right] - 3\left[ 1\left( - 3 - 1 \right) + 1\left( 0 - 1 \right) + 1\left( 0 + 3 \right) \right] - 1\left[ 1\left( - 3 - 1 \right) + 1\left( 0 - 1 \right) \right]\]
\[ = 3\]
So, by Cramer's rule , we obtain
\[x = \frac{D_1}{D} = \frac{21}{21} = 1\]
\[y = \frac{D_2}{D} = \frac{6}{- 21} = - \frac{2}{7}\]
\[z = \frac{D_3}{D} = \frac{- 6}{- 21} = \frac{2}{7}\]
\[w = \frac{D_4}{D} = \frac{3}{- 21} = - \frac{1}{7}\]
\[\text{ Hence,} x = 1, y = - \frac{2}{7}, z = \frac{2}{7}, w = - \frac{1}{7}\]
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