MCQ
Solve system of linear equations, using matrix method. $4 x-3 y=3$ ; $3 x-5 y=7$
- A$x=\frac{6}{11},y=\frac{-19}{11}$
- B$x=\frac{-6}{11},y=\frac{19}{11}$
- C$x=\frac{6}{11},y=\frac{19}{11}$
- ✓$x=\frac{-6}{11},y=\frac{-19}{11}$
$A=\left[\begin{array}{ll}4 & -3 \\ 3 & -5\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 7\end{array}\right]$
Now,
$|A|=-20+9=-11 \neq 0$
Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$A^{-1}=\frac{1}{|A|}(a d j A)=-\frac{1}{11}\left[\begin{array}{ll}
-5 & 3 \\
-3 & 4
\end{array}\right]=\frac{1}{11}\left[\begin{array}{ll}
5 & -3 \\
3 & -4
\end{array}\right]$
$\therefore X=A^{-1} B=\frac{1}{11}\left[\begin{array}{ll}5 & -3 \\ 3 & -4\end{array}\right]\left[\begin{array}{l}3 \\ 7\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{ll}5 & -3 \\ 3 & -4\end{array}\right]\left[\begin{array}{l}3 \\ 7\end{array}\right]$
$=\frac{1}{11}\left[\begin{array}{c}15-21 \\ 9-28\end{array}\right]$
$=\frac{1}{11}\left[\begin{array}{l}-6 \\ -19\end{array}\right]$
$=\left[\begin{array}{r}-\frac{6}{11} \\ -\frac{19}{11}\end{array}\right]$
Hence, $x=\frac{-6}{11}$ and $y=\frac{-19}{11}$
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