Question
Solve the system of linear equation, using matrix method 5x + 2y = 4; 7x + 3y = 5
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 5&2 \\ 7&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4 \\ 5 \end{array}} \right]$
Here $A = \left[ {\begin{array}{*{20}{c}} 5&2 \\ 7&3 \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}} 4 \\ 5 \end{array}} \right]$
$\left| A \right| = \left| {\begin{array}{*{20}{c}} 5&2 \\ 7&3 \end{array}} \right| = 15 - 14 = 1 \ne 0$
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}{1}\left[ {\begin{array}{*{20}{c}} 3&{ - 2} \\ { - 7}&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4 \\ 5 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {12 - 10} \\ { - 28 + 25} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ { - 3} \end{array}} \right]$
Therefore, x = 2 and y = - 3
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