Question Number 39289 by Rio Mike last updated on 04/Jul/18
$${Find}\:{the}\:{value}\:{of}\:{x}\:{if}\: \\ $$$${the}\:{matrix}\: \\ $$$$ \\ $$$$\begin{pmatrix}{\mathrm{3}{x}\:\:\:\:\:\:\:\:\mathrm{5}{x}}\\{{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\begin{pmatrix}{{x}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:{x}}\end{pmatrix}\: \\ $$$${has}\:{no}\:{inverse} \\ $$
Answered by Rio Mike last updated on 05/Jul/18
$$=\:\begin{pmatrix}{\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{15}{x}}&{\mathrm{3}{x}+\:\mathrm{5}{x}^{\mathrm{2}} }\\{{x}^{\mathrm{2}} \:\:\:\:+\:\mathrm{3}{x}\:\:\:\:\:}&{{x}\:\:\:\:+\:\mathrm{3}{x}}\end{pmatrix} \\ $$$$\left(\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{15}{x}\:\right)\left({x}+\mathrm{3}{x}\right)−\:\left(\mathrm{3}{x}+\mathrm{5}{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}\right)=\mathrm{0} \\ $$$${can}\:{some}\:{one}\:{solve}\:{for}\:{x}? \\ $$