Question
Show that the following system of linear equations is consistent and also find solution:
$2x + 3y = 5$
$6x + 9y = 15$
$2x + 3y = 5$
$6x + 9y = 15$
✓
Answer
$2\text{x}+3\text{y}=5\dots(1)$
$6\text{x}+9\text{y}=15\dots(2)$
Or , $AX = B$
Where,
$\text{A}=\begin{bmatrix}2&3\\ 6&9\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\begin{bmatrix}2&3\\ 6&9\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\therefore\ |\text{A}|=\begin{vmatrix}2&3\\6&9\end{vmatrix}$
$=18-18$
$=0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\text{adj A})\text{B}\neq0\text{ or }(\text{adj A})=0$.
$C_{11} = 9, C_{12} = -6, C_{21} = -3$ and $C_{22} = 2$
$\therefore\ \text{adj A}=\begin{bmatrix}9&-6\\-3&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}2&-3\\-6&9\end{bmatrix}$
$\Rightarrow(\text{adj A})\text{B}=\begin{bmatrix}9&-3\\-6&2\end{bmatrix}\begin{bmatrix}5\\15\end{bmatrix}$
$=\begin{bmatrix}45-45\\-30+30\end{bmatrix}$
$=\begin{bmatrix}0\\0\end{bmatrix}$
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 eq. (1), we get
$2\text{x} + 3\text{k}=5$
$\Rightarrow2\text{x}=5-3\text{k}$
$\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$
These values of x and y satisfy the third equation.
Thus, $\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$ (where k is a real number) satisfy the given system of equations.
$6\text{x}+9\text{y}=15\dots(2)$
Or , $AX = B$
Where,
$\text{A}=\begin{bmatrix}2&3\\ 6&9\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\begin{bmatrix}2&3\\ 6&9\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\therefore\ |\text{A}|=\begin{vmatrix}2&3\\6&9\end{vmatrix}$
$=18-18$
$=0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\text{adj A})\text{B}\neq0\text{ or }(\text{adj A})=0$.
$C_{11} = 9, C_{12} = -6, C_{21} = -3$ and $C_{22} = 2$
$\therefore\ \text{adj A}=\begin{bmatrix}9&-6\\-3&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}2&-3\\-6&9\end{bmatrix}$
$\Rightarrow(\text{adj A})\text{B}=\begin{bmatrix}9&-3\\-6&2\end{bmatrix}\begin{bmatrix}5\\15\end{bmatrix}$
$=\begin{bmatrix}45-45\\-30+30\end{bmatrix}$
$=\begin{bmatrix}0\\0\end{bmatrix}$
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 eq. (1), we get
$2\text{x} + 3\text{k}=5$
$\Rightarrow2\text{x}=5-3\text{k}$
$\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$
These values of x and y satisfy the third equation.
Thus, $\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$ (where k is a real number) satisfy the given system of equations.
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