Question and Answers Forum

All Questions      Topic List

Matrices and Determinants Questions

Previous in All Question      Next in All Question      

Previous in Matrices and Determinants      Next in Matrices and Determinants      

Question Number 100650 by bobhans last updated on 28/Jun/20

$$\mathrm{find}\mathrm{all}2\mathrm{x}2\mathrm{matrices}\mathrm{A}\mathrm{such}$$$$\mathrm{that}{\mathrm{A}}^3-{3\mathrm{A}}^2=\left(\begin{array}{c}-2-2\\ -2-2\end{array}\right)$$

Answered by bramlex last updated on 28/Jun/20

$$letA=\left(\begin{array}{c}ab\\ cd\end{array}\right)$$$$A^2(A-3I)=\left(\begin{array}{c}-2-2\\ -2-2\end{array}\right)$$$${\left(det\left(A\right)\right)}^{2}det(A-3I)=det\left(\begin{array}{c}-2-2\\ -2-2\end{array}\right)=0$$$$case1$$$$det\left(A\right)=0\Rightarrow p\left(\lambda \right)=det(\lambda I-A)=0$$$$\left|\begin{array}{c}\lambda -a-b\\ -c\lambda -d\end{array}\right|={\lambda }^2-(a+d)\lambda +ad-bc$$$${\lambda }^2-TR\left(A\right)\lambda +det\left(A\right)=0$$$$\Rightarrow {\lambda }^2-TR\left(A\right)\lambda =0;A^2=TR\left(A\right).A$$$$A^3=A\left(TR\left(A\right)\right)={\left(TR\left(A\right)\right)}^{2}A$$$$\Rightarrow A^3-3A=\left(\begin{array}{c}-2-2\\ -2-2\end{array}\right)$$$$TR{\left(A\right)}^3-3{\left(TR\left(A\right)\right)}^2+4=0$$$$TR\left(A\right)=2orTR\left(A\right)=-1$$$$case1a$$$$TR\left(A\right)=2\Rightarrow -2A=\left(\begin{array}{c}-2-2\\ -2-2\end{array}\right)$$$$A=\left(\begin{array}{c}11\\ 11\end{array}\right).TR\left(A\right)=-1$$$$4A=\left(\begin{array}{c}-2-2\\ -2-2\end{array}\right)\Rightarrow A=\left(\begin{array}{c}-\frac{1}{2}-\frac{1}{2}\\ -\frac{1}{2}-\frac{1}{2}\end{array}\right)$$$$case2$$$$det(A-3I)=0$$$$det{(A-2I)}^2(A+I)=0$$$$case2a$$$$det(A-2I)=0\Rightarrow p\left(\lambda \right)={\lambda }^2-5\lambda +6$$$$A^2=5A-6I\Rightarrow A^3-3A^2=(19A-30I)-3(5A-6I)$$$$=4A-12I=\left(\begin{array}{c}-2-2\\ -2-2\end{array}\right)$$$$4A=\left(\begin{array}{c}10-2\\ -210\end{array}\right)\Rightarrow A=\left(\begin{array}{c}\frac{5}{2}-\frac{1}{2}\\ -\frac{1}{2}\frac{5}{2}\end{array}\right)$$$$case2b$$$$det(A+I)=0\Rightarrow p\left(\lambda \right)={\lambda }^2-2\lambda -3$$$$A^2-2A-3I=0;A^2=2A+3I$$$$A^3-3A^2=7A+6I-3(2A+3I)=A-3I$$$$\Rightarrow A-3I=\left(\begin{array}{c}-2-2\\ -2-2\end{array}\right)$$$$A=\left(\begin{array}{c}1-2\\ -21\end{array}\right)$$$$\therefore A=\left(\begin{array}{c}11\\ 11\end{array}\right);\left(\begin{array}{c}-\frac{1}{2}-\frac{1}{2}\\ -\frac{1}{2}-\frac{1}{2}\end{array}\right)$$$$;\left(\begin{array}{c}\frac{5}{2}-\frac{1}{2}\\ -\frac{1}{2}\frac{5}{2}\end{array}\right);\left(\begin{array}{c}1-2\\ -21\end{array}\right)$$$$$$

Commented by bobhans last updated on 28/Jun/20

$$\mathrm{great}===$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com