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Question Number 32869 by Rio Mike last updated on 04/Apr/18

$$A=\left(\begin{array}{c}20\\ 12\end{array}\right);h\mathrm{and}k\mathrm{are}\mathrm{numbers}\mathrm{so}$$$$\mathrm{that}{\mathrm{A}}^2=hA+kI,\mathrm{where}\mathrm{I}=\left(\begin{array}{c}10\\ 01\end{array}\right).$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}h\mathrm{and}k.$$

Commented by abdo mathsup last updated on 09/Apr/18

$$thecsfactetisticpolynomialofAis$$$$p_A\left(X\right)=det(A-XI)=\left|\begin{array}{c}2-X0\\ 12-X\end{array}\right|$$$$={(2-X)}^2=4-4X+X^{2}andkayleyhsmilton$$$$theoremgive{p}_A\left(A\right)=0\Rightarrow 4I-4A+A^2=0\Rightarrow$$$$A^2=4A-4I\Rightarrow h=4andk=-4.$$

Answered by Joel578 last updated on 05/Apr/18

$$A^2=\left(\begin{array}{cc}2& 0\\ 1& 2\end{array}\right)\left(\begin{array}{cc}2& 0\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}4& 0\\ 4& 4\end{array}\right)$$$$$$$$\left(\begin{array}{cc}4& 0\\ 4& 4\end{array}\right)=\left(\begin{array}{cc}2h& 0\\ h& 2h\end{array}\right)+\left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right)=\left(\begin{array}{cc}2h+k& 0\\ h& 2h+k\end{array}\right)$$$$$$$$h=4$$$$2h+k=4\to k=-4$$

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