MCQ
Solve system of linear equations, using matrix method. $2 x-y=-2$ ; $3 x+4 y=3$
- A$x=\frac{5}{11},\,y=\frac{12}{11}$
- B$x=\frac{-5}{11},\,y=\frac{-12}{11}$
- ✓$x=\frac{-5}{11},\,y=\frac{12}{11}$
- D$x=\frac{5}{11},\,y=\frac{-12}{11}$
✓
Answer
Correct option: C.
$x=\frac{-5}{11},\,y=\frac{12}{11}$
c
The given system of equation can be written in the form of $A X=B$, where
The given system of equation can be written in the form of $A X=B$, where
$A=\left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}-2 \\ 3\end{array}\right]$
Now,
Now $|A|=8+3=11 \neq 0$
Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$A^{-1}=\frac{1}{|A|}(a d J A)=\frac{1}{11}\left[\begin{array}{cc}4 & 1 \\ -3 & 2\end{array}\right]$
$\therefore X=A^{-1} B=\frac{1}{11}\left[\begin{array}{cc}4 & 1 \\ -3 & 2\end{array}\right]\left[\begin{array}{c}-2 \\ 3\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-8+3 \\ 6+6\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-5 \\ 12\end{array}\right]=\left[\begin{array}{c}-\frac{5}{11} \\ \frac{12}{11}\end{array}\right]$
Hence, $x=\frac{-5}{11}$ and $y=\frac{12}{11}$
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