2x-3y+6=0 and 6x+y+8=0 solve the following equations using matrix inversion method

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Solution

$G i v e n E q u a t i o n s : 2 x - 3 y + 6 = 0 a n d 6 x + y + 8 = 0 W e c a n t r a n s f o r m i t i n t o m a t r i x f o r m : (\begin{matrix} 2 & - 3 \\ 6 & 1 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 6 \\ - 8 \end{matrix}) H e r e, A = (\begin{matrix} 2 & - 3 \\ 6 & 1 \end{matrix}) a n d i t s i n v e r s e A^{- 1} = (\begin{matrix} \frac{1}{20} & \frac{6}{20} \\ \frac{- 3}{20} & \frac{2}{20} \end{matrix}) N o w, (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 6 \\ - 8 \end{matrix}) (\begin{matrix} \frac{1}{20} & \frac{6}{20} \\ \frac{- 3}{20} & \frac{2}{20} \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 6 \times \frac{1}{20} - 8 \times \frac{6}{20} \\ - 6 \times \frac{(- 3)}{20} - 8 \times \frac{2}{20} \end{matrix}) = (\begin{matrix} \frac{- 54}{20} \\ \frac{2}{20} \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} \frac{- 27}{10} \\ \frac{1}{10} \end{matrix}) T h e r e f o r e, x = - 27 / 10 a n d y = 1 / 10$ $G i v e n E q u a t i o n s : 2 x - 3 y + 6 = 0 a n d 6 x + y + 8 = 0 W e c a n t r a n s f o r m i t i n t o m a t r i x f o r m : (\begin{matrix} 2 & - 3 \\ 6 & 1 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 6 \\ - 8 \end{matrix}) H e r e, A = (\begin{matrix} 2 & - 3 \\ 6 & 1 \end{matrix}) a n d i t s i n v e r s e A^{- 1} = (\begin{matrix} \frac{1}{20} & \frac{6}{20} \\ \frac{- 3}{20} & \frac{2}{20} \end{matrix}) N o w, (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 6 \\ - 8 \end{matrix}) (\begin{matrix} \frac{1}{20} & \frac{6}{20} \\ \frac{- 3}{20} & \frac{2}{20} \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 6 \times \frac{1}{20} - 8 \times \frac{6}{20} \\ - 6 \times \frac{(- 3)}{20} - 8 \times \frac{2}{20} \end{matrix}) = (\begin{matrix} \frac{- 54}{20} \\ \frac{2}{20} \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} \frac{- 27}{10} \\ \frac{1}{10} \end{matrix}) T h e r e f o r e, x = - 27 / 10 a n d y = 1 / 10$

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