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प्रश्न
Show that the following systems of linear equations is consistent and also find their solutions:
6x + 4y = 2
9x + 6y = 3
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उत्तर
Here,
\[6x + 4y = 2 . . . (1) \]
\[9x + 6y = 3 . . . (2)\]
\[AX = B \]
Here,
\[A = \begin{bmatrix}6 & 4 \\ 9 & 6\end{bmatrix}, X = \binom{x}{y}\text{ and }B = \binom{2}{3}\]
\[\begin{bmatrix}6 & 4 \\ 9 & 6\end{bmatrix}\binom{x}{y} = \binom{2}{3}\]
\[\left| A \right| = \begin{vmatrix}6 & 4 \\ 9 & 6\end{vmatrix}\]
\[ = 36 - 36\]
\[ = 0\]
So, A is singular . Thus, the given system of equations is either inconsistent or it is consistent with
\[\text{ infinitely many solutions because }\left( adj A \right)B \neq 0 \text{ or }\left( adj A \right) = 0 . \]
\[ {\text{ Let }C}_{ij} {\text{ be the co factors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = 6, C_{12} = - 9, C_{21} = - 4, C_{22} = 6\]
\[adj A = \begin{bmatrix}6 & - 9 \\ - 4 & 6\end{bmatrix}^T \]
\[ = \begin{bmatrix}6 & - 4 \\ - 9 & 6\end{bmatrix}\]
\[\left( adj A \right)B = \begin{bmatrix}6 & - 4 \\ - 9 & 6\end{bmatrix}\binom{2}{3}\]
\[ = \binom{12 - 12}{ - 18 + 18}\]
\[ = \binom{0}{0}\]
\[If\left| A \right|=0\text{ and }\left( adjA \right)B=0,\text{ then the system is consistent and has infinitely many solutions.}\]
\[\text{ Thus, }AX=\text{ Bhas infinitely many solutions.}\]
Substituting y=k in the eq. (1), we get
\[6x + 4k = 2\]
\[ \Rightarrow 6x = 2 - 4k\]
\[ \Rightarrow x = \frac{2 - 4k}{6}\]
\[ \Rightarrow x = \frac{1 - 2k}{3}\]
\[ \therefore x = \frac{1 - 2k}{3} and y = k\]
These values of x and y satisfy the third equation .
\[\text{ Thus, }x = \frac{1 - 2k}{3}\text{ and }y = k \left(\text{ where k is a real number} \right) \text{ satisfy the given system of equations }.\]
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