Question

Show that the following systems of linear equations is consistent and also find their solutions:
2x + 3y = 5
6x + 9y = 15

Answer 1

Here,
\[2x + 3y = 5 . . . (1) \]
\[6x + 9y = 15 . . . (2) \]
\[or , AX = B \]
where,
\[A = \begin{bmatrix}2 & 3 \\ 6 & 9\end{bmatrix}, X = \binom{x}{y}\text{ and }B = \binom{5}{15}\]
\[ \Rightarrow \begin{bmatrix}2 & 3 \\ 6 & 9\end{bmatrix}\binom{x}{y} = \binom{5}{15}\]
\[ \therefore \left| A \right| = \begin{vmatrix}2 & 3 \\ 6 & 9\end{vmatrix}\]
\[ = 18 - 18\]
\[ = 0\]
So, A is singular . Thus, the given system of equations is either inconsistent or it is consistent with 
\[\text{ infinitely many solutions because }\left( adj A \right)B \neq 0\text{ or }\left( adj A \right) = 0 . \]
\[ {\text{ Let }C}_{ij} {\text{ be the co-factors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = 9, C_{12} = - 6, C_{21} = - 3\text{ and }C_{22} = 2\]
\[ \therefore adj A = \begin{bmatrix}9 & - 6 \\ - 3 & 2\end{bmatrix}^T \]
\[ = \begin{bmatrix}2 & - 3 \\ - 6 & 9\end{bmatrix}\]
\[ \Rightarrow \left( adjA \right)B = \begin{bmatrix}9 & - 3 \\ - 6 & 2\end{bmatrix}\binom{5}{15}\]
\[ = \binom{45 - 45}{ - 30 + 30}\]
\[ = \binom{0}{0}\]
\[\text{ If}\left| A \right|=0\text{ and }\left( adjA \right)B=0,\text{ then the system is consistent and has infinitely many solutions.}\]
\[\text{ Thus, AX=B has infinitely many solutions.}\]
\[\text{ Substituting y=k in eq. (1), we get }\]
\[2x + 3k = 5\]
\[ \Rightarrow 2x = 5 - 3k\]
\[ \Rightarrow x = \frac{5 - 3k}{2}\text{ and }y = k\]
These values of x and y satisfy the third equation . 
\[\text{ Thus, }x = \frac{5 - 3k}{2}\text{ and }y = k \left( \text{ where k is a real number }\right) \text{ satisfy the given system of equations }.\]