Question Number 54968 by gunawan last updated on 15/Feb/19
![Let A matrices order 2×2 the fill tr(A^2 )=[tr(A)]^2 a. Find det(A) b. If A can′t diagonalizing, find tr(A)](https://www.tinkutara.com/question/Q54968.png)
$$\mathrm{Let}\:{A}\:\mathrm{matrices}\:\mathrm{order}\:\mathrm{2}×\mathrm{2}\:\mathrm{the}\:\mathrm{fill} \\ $$$$\mathrm{tr}\left({A}^{\mathrm{2}} \right)=\left[{tr}\left({A}\right)\right]^{\mathrm{2}} \\ $$$$\mathrm{a}.\:\mathrm{Find}\:\mathrm{det}\left(\mathrm{A}\right) \\ $$$$\mathrm{b}.\:\mathrm{If}\:\mathrm{A}\:\mathrm{can}'\mathrm{t}\:\mathrm{diagonalizing},\:\mathrm{find}\:\mathrm{tr}\left({A}\right) \\ $$$$ \\ $$
Answered by kaivan.ahmadi last updated on 15/Feb/19
![let A= [((a b)),((c d)) ]⇒A^2 = [((a^2 +bc ab+bd)),((cd+ad bc+d^2 )) ] ⇒a^2 +bc+bc+d^2 =(a+d)^2 =a^2 +2ad+d^2 ⇒ 2bc=2ad⇒bc=ad a. detA=ad−bc=0](https://www.tinkutara.com/question/Q54981.png)
$${let}\:\:{A}=\begin{bmatrix}{{a}\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix}\Rightarrow{A}^{\mathrm{2}} =\begin{bmatrix}{{a}^{\mathrm{2}} +{bc}\:\:\:\:{ab}+{bd}}\\{{cd}+{ad}\:\:\:\:{bc}+{d}^{\mathrm{2}} }\end{bmatrix} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{bc}+{bc}+{d}^{\mathrm{2}} =\left({a}+{d}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{2}{ad}+{d}^{\mathrm{2}} \Rightarrow \\ $$$$\mathrm{2}{bc}=\mathrm{2}{ad}\Rightarrow{bc}={ad} \\ $$$${a}.\:\:{detA}={ad}−{bc}=\mathrm{0} \\ $$$$ \\ $$