Show that the following system of linear equations is consistent … - Vidyadip

Question

Show that the following system of linear equations is consistent and also find solution:
$6x + 4y = 2$
$9x + 6y =3$

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Answer

Here,
$6\text{x}+4\text{y}=2\ \dots(1)$
$9\text{x}+6\text{y}\ \dots(2)$
AX = B
Here,
$\text{A}=\begin{bmatrix}6&4\\ 9&6\end{bmatrix},\text{X}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\ 3\end{bmatrix}$
$\begin{bmatrix}6&4\\ 9&6\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}2\\ 3\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}6&4\\ 9&6\end{vmatrix}$
$= 36 -36$
$= 0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because (adj A)B ≠ 0 or (adj A) = 0.
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=6,\\ \text{C}_{12}=-9,\\ \text{C}_{21}=-4,\\ \text{C}_{22}=6$
$\text{adj A}=\begin{bmatrix}6&-9\\ -4&6\end{bmatrix}^\text{T}$
$=\begin{bmatrix}6&-4\\ -9&6\end{bmatrix}$
$\text{(adj A) B}=\begin{bmatrix}6&-4\\ -9&6\end{bmatrix}\begin{bmatrix}2\\ 3\end{bmatrix}$
$=\begin{bmatrix}12-12\\ -18+18\end{bmatrix}$
$=\begin{bmatrix}0\\ 0\end{bmatrix}$
If |A| = 0 and (adj A) B = 0, then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions.
Substituting y = k in the eq. (1), we get
$6\text{x}+4\text{k}=2$
$\Rightarrow6\text{x}=2-4\text{k}$
$\Rightarrow\text{x}=\frac{2-\text{4k}}{6}$
$\Rightarrow\text{x}=\frac{1-\text{2k}}{3}$
$\therefore \text{x}=\frac{1-2\text{k}}{3}\text{ and }\text{y}=\text{k}$
These values of x and y satisfy the third equation.
Thus, $\text{x}=\frac{1-2\text{k}}{3}$ and y = k ( where k is a real number ) satisfy the given system of equations.

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