Question
Solve the system of linear equation, using matrix method 2x - y = - 2; 3x + 4y = 3
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&{ - 1} \\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2} \\ 3 \end{array}} \right]$
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 $y = \frac{{12}}{{11}}$
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| Differential equation | Function |
| $\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$ | $\text{y}=\pm\sqrt{\text{a}^2-\text{x}^2}$ |