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Question Number 202371 by mnjuly1970 last updated on 25/Dec/23
$$ \\ $$$$\:\:\:{If}\:\:\:\:{A}\:\in\:{M}_{\mathrm{2}×\mathrm{2}} \:,\:{det}\left({A}\:\right)\neq\:\mathrm{0} \\ $$$$\:\:\:\:,\:\:{A}^{\:\mathrm{3}} \:=\:{A}^{\mathrm{2}} \:+{A}\:\Rightarrow\:{Find}\:{the}\: \\ $$$$\:\:\:\:{values}\:{of}\:\:\:{det}\:\left(\mathrm{2}{A}\:−{I}\:\right) \\ $$$$ \\ $$
Answered by witcher3 last updated on 25/Dec/23
$$\mathrm{det}\left(\mathrm{A}\right)\neq\mathrm{0}\Rightarrow\:\mathrm{A}\:\mathrm{is}\:\mathrm{inversible} \\ $$$$\Rightarrow\mathrm{A}^{−} .\mathrm{A}^{\mathrm{3}} =\mathrm{A}^{−} \left(\mathrm{A}^{\mathrm{2}} +\mathrm{A}\right) \\ $$$$\Rightarrow\mathrm{A}^{\mathrm{2}} =\mathrm{A}+\mathrm{I} \\ $$$$\mathrm{A}^{\mathrm{2}} −\mathrm{A}=\left(\mathrm{A}−\frac{\mathrm{I}}{\mathrm{2}}\right)\left(\mathrm{A}−\frac{\mathrm{I}}{\mathrm{2}}\right)−\frac{\mathrm{I}_{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\left(\mathrm{A}−\frac{\mathrm{I}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}}\mathrm{I}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{det}^{\mathrm{2}} \left(\mathrm{A}−\frac{\mathrm{I}}{\mathrm{2}}\right)=\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{det}\left(\mathrm{A}−\frac{\mathrm{I}}{\mathrm{2}}\right)=\underset{−} {+}\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{det}\left(\mathrm{2}\left(\mathrm{A}−\frac{\mathrm{I}}{\mathrm{2}}\right)\right)=\underset{−} {+}\mathrm{5} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 25/Dec/23
$$\:{please}\:{recheck}\:{sir}\:... \\ $$$$\:\:\:{det}\:\left(\mathrm{2}{A}−{I}\right)=\pm\mathrm{5}\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Commented by witcher3 last updated on 25/Dec/23
$$\mathrm{yest}\:\mathrm{det}\left(\mathrm{aI}_{\mathrm{n}} \right)=\mathrm{a}^{\mathrm{n}} .\mathrm{thanx} \\ $$
Commented by mnjuly1970 last updated on 25/Dec/23
$$\:\:\:\:{thanks}\:{alot}\:{sir}\:.{so}\:{nice}\:{solution} \\ $$
Answered by Rajpurohith last updated on 26/Dec/23
![So A is annihilated by the polynomial p(x)=x^3 −x^2 −x=x(x^2 −x−1) so x=0,α_1 ,α_2 where α_1 and α_2 are the roots of x^2 −x−1=0 so det(A)=0 and α_1 +α_2 =1 and α_1 α_2 =−1 the roots of annihilating polynomial of B=2A−I will be −1,2α_1 −1,2α_2 −1 so det(2A−I)=−(2α_1 −1)(2α_2 −1) =−[4α_1 α_2 −2(α_1 +α_2 )+1] =−[−4−2(1)+1]=5](Q202437.png)
$${So}\:{A}\:{is}\:{annihilated}\:{by}\:{the}\:{polynomial} \\ $$$${p}\left({x}\right)={x}^{\mathrm{3}} −{x}^{\mathrm{2}} −{x}={x}\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right) \\ $$$${so}\:{x}=\mathrm{0},\alpha_{\mathrm{1}} ,\alpha_{\mathrm{2}} \: \\ $$$${where}\:\alpha_{\mathrm{1}} {and}\:\alpha_{\mathrm{2}} \:{are}\:{the}\:{roots}\:{of}\:{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${so}\:{det}\left({A}\right)=\mathrm{0}\:{and}\:\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} =\mathrm{1}\:{and}\:\alpha_{\mathrm{1}} \alpha_{\mathrm{2}} =−\mathrm{1} \\ $$$${the}\:{roots}\:{of}\:{annihilating}\:{polynomial}\:{of}\: \\ $$$${B}=\mathrm{2}{A}−{I}\:{will}\:{be}\:−\mathrm{1},\mathrm{2}\alpha_{\mathrm{1}} −\mathrm{1},\mathrm{2}\alpha_{\mathrm{2}} −\mathrm{1} \\ $$$${so}\:{det}\left(\mathrm{2}{A}−{I}\right)=−\left(\mathrm{2}\alpha_{\mathrm{1}} −\mathrm{1}\right)\left(\mathrm{2}\alpha_{\mathrm{2}} −\mathrm{1}\right) \\ $$$$=−\left[\mathrm{4}\alpha_{\mathrm{1}} \alpha_{\mathrm{2}} −\mathrm{2}\left(\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} \right)+\mathrm{1}\right] \\ $$$$=−\left[−\mathrm{4}−\mathrm{2}\left(\mathrm{1}\right)+\mathrm{1}\right]=\mathrm{5} \\ $$$$ \\ $$