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Question-201282




Question Number 201282 by Supro last updated on 03/Dec/23
Answered by MM42 last updated on 03/Dec/23
x=( determinant (((a^2           b)),((ab      −a)))/ determinant (((a       b)),((b     −a))))=((−a^3 −ab^2 )/(−a^2 −b^2 )) =a  y=( determinant (((a        a^2 )),((b        ab)))/ determinant (((a       b)),((b     −a))))=((a^2 b−a^2 b)/(−a^2 −b^2 )) =0
$${x}=\frac{\begin{vmatrix}{{a}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:{b}}\\{{ab}\:\:\:\:\:\:−{a}}\end{vmatrix}}{\begin{vmatrix}{{a}\:\:\:\:\:\:\:{b}}\\{{b}\:\:\:\:\:−{a}}\end{vmatrix}}=\frac{−{a}^{\mathrm{3}} −{ab}^{\mathrm{2}} }{−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:={a} \\ $$$${y}=\frac{\begin{vmatrix}{{a}\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} }\\{{b}\:\:\:\:\:\:\:\:{ab}}\end{vmatrix}}{\begin{vmatrix}{{a}\:\:\:\:\:\:\:{b}}\\{{b}\:\:\:\:\:−{a}}\end{vmatrix}}=\frac{{a}^{\mathrm{2}} {b}−{a}^{\mathrm{2}} {b}}{−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:=\mathrm{0} \\ $$$$ \\ $$

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