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Question Number 205586 by necx122 last updated on 25/Mar/24
$${Givem}\:{that}\:{the}\:{matrix}\:{A}\:=\:\begin{pmatrix}{\mathrm{3}}&{\mathrm{1}}&{\mathrm{5}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{5}}\\{\mathrm{5}}&{\mathrm{1}}&{\mathrm{6}}\end{pmatrix}.\: \\ $$$${If}\:{Adj}.\:{A}\:=\:\begin{pmatrix}{\mathrm{13}}&{-\mathrm{1}}&{-\mathrm{10}}\\{\mathrm{13}}&{-\mathrm{7}}&{-\mathrm{5}}\\{-\mathrm{13}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix} \\ $$$$\left({i}\right)\:{find}\:{A}^{−\mathrm{1}} \\ $$$$\left({ii}\right)\:{Use}\:{the}\:{result}\:{in}\:\left({i}\right)\:{to}\:{find}\:{the} \\ $$$${values}\:{of}\:{x},\:{y}\:{and}\:{z}\:{that}\:{will}\:{satisfy}\:{the} \\ $$$${equations}: \\ $$$$\mathrm{3}{x}\:+\:{y}\:+\:\mathrm{5}{z}\:=\:\mathrm{8} \\ $$$$\mathrm{2}{x}\:+\mathrm{3}{y}\:+\:\mathrm{5}{z}\:=\:\mathrm{0} \\ $$$$\mathrm{5}{x}\:+\:{y}\:+\:\mathrm{6}{z}\:=\:\mathrm{13} \\ $$
Answered by som(math1967) last updated on 25/Mar/24
$$\:{A}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mid{A}\mid}×{Adj}.{A} \\ $$$$\mid{A}\mid=\mathrm{3}\left(\mathrm{18}−\mathrm{5}\right)−\mathrm{1}\left(\mathrm{12}−\mathrm{25}\right)+\mathrm{5}\left(\mathrm{2}−\mathrm{15}\right) \\ $$$$=\mathrm{39}+\mathrm{13}−\mathrm{65} \\ $$$$=\mathrm{13}\left(\mathrm{3}+\mathrm{1}−\mathrm{5}\right)=−\mathrm{13} \\ $$$$\therefore{A}^{−\mathrm{1}} =\begin{pmatrix}{−\mathrm{1}}&{\frac{\mathrm{1}}{\mathrm{13}}}&{\frac{\mathrm{10}}{\mathrm{13}}}\\{−\mathrm{1}}&{\frac{\mathrm{7}}{\mathrm{13}}}&{\frac{\mathrm{5}}{\mathrm{13}}}\\{\mathrm{1}}&{\frac{−\mathrm{2}}{\mathrm{13}}}&{\frac{−\mathrm{7}}{\mathrm{13}}}\end{pmatrix} \\ $$$$\:{equation} \\ $$$$\:\:{A}×{X}={B} \\ $$$${where}={A}=\begin{pmatrix}{\mathrm{3}}&{\mathrm{1}}&{\mathrm{5}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{5}}\\{\mathrm{5}}&{\mathrm{1}}&{\mathrm{6}}\end{pmatrix} \\ $$$${X}=\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:\:\:{B}=\begin{pmatrix}{\mathrm{8}}\\{\mathrm{0}}\\{\mathrm{13}}\end{pmatrix} \\ $$$${A}^{−\mathrm{1}} ×{A}×{X}={A}^{−\mathrm{1}} ×{B} \\ $$$$\:{I}×{X}=\begin{pmatrix}{−\mathrm{8}+\mathrm{0}+\mathrm{10}}\\{−\mathrm{8}+\mathrm{0}+\mathrm{5}}\\{\mathrm{8}+\mathrm{0}−\mathrm{7}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:\:=\begin{pmatrix}{\mathrm{2}}\\{−\mathrm{3}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\therefore{x}=\:\mathrm{2},\:{y}=−\mathrm{3}\:,{z}=\mathrm{1}\:{ans} \\ $$
Commented by necx122 last updated on 25/Mar/24
$${This}\:{is}\:{so}\:{clear}\:{and}\:{expressive}.\:{Thank}\:{you} \\ $$$${so}\:{much}\:{sir}. \\ $$