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Solve the system of linear equation, using matrix method 2x - y = - 2; 3x + 4y = 3
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## QuestionMathsClass 12

Solve the system of linear equation, using matrix method 2x - y = - 2; 3x + 4y = 3

Matrix form of given equations is AX = B
$$\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&{ - 1} \\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2} \\ 3 \end{array}} \right]$$
Here $$A = \left[ {\begin{array}{*{20}{c}} 2&{ - 1} \\ 3&4 \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]$$ and $$B = \left[ {\begin{array}{*{20}{c}} { - 2} \\ 3 \end{array}} \right]$$
$$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&{ - 1} \\ 3&4 \end{array}} \right|$$ = 8 - (-3) = 8 + 3 $$= 11 \ne 0$$
Therefore, solution is unique and $$X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B$$
$$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} 4&1 \\ { - 3}&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 2} \\ 3 \end{array}} \right]$$
$$= \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} { - 8 + 3} \\ {6 + 6} \end{array}} \right] = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} { - 5} \\ {12} \end{array}} \right]$$
$$= \left[ {\begin{array}{*{20}{c}} {\frac{{ - 5}}{{11}}} \\ {\frac{{12}}{{11}}} \end{array}} \right]$$
Therefore, $$x = \frac{{ - 5}}{{11}}$$ and $$y = \frac{{12}}{{11}}$$