Question
Solve the system of linear equation, using matrix method 4x - 3y = 3; 3x - 5y = 7
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 4&{ - 3} \\ 3&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3 \\ 7 \end{array}} \right]$
Here $A = \left[ {\begin{array}{*{20}{c}} 4&{ - 3} \\ 3&{ - 5} \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 3 \\ 7 \end{array}} \right]$
$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 4&{ - 3} \\ 3&{ - 5} \end{array}} \right| $ = - 20 - (-9) = - 20 + 9 = - 11 $ \ne $ 0
Therefore, solution is unique and $X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}} { - 5}&3 \\ { - 3}&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3 \\ 7 \end{array}} \right]$
$= \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}} { - 15 + 21} \\ { - 9 + 28} \end{array}} \right] = \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}} 6 \\ 9 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {\frac{{ - 6}}{{11}}} \\ {\frac{{ - 19}}{{11}}} \end{array}} \right]$
Therefore, $x = \frac{{ - 6}}{{11}}$ and $y = \frac{{ - 19}}{{11}}$
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