Question

If $A=\left[\begin{array}{cc}1& -3\\ 2& K\end{array}\right]$ and $A^2-4A+10I=A$, then $K=$

• A. $1$ or $4$
• B. $4$ and not $1$
• C. $-4$
• D. $0$

Answer 1

The correct option is A $1$ or $4$

$\begin{array}{c}A=\left(\begin{array}{cc}1& -3\\ 2& k\end{array}\right)\\ A^{2}4A+101=A\Rightarrow A^2-5A+101=0\\ Findk\\ A^2=A\cdot A\\ =\left(\begin{array}{cc}1& -3\\ 2& k\end{array}\right)\left(\begin{array}{cc}1& -3\\ 2& k\end{array}\right)\\ =\left(\begin{array}{cc}1-6& -3-3k\\ 2+2k& -6+k^2\end{array}\right)=\left(\begin{array}{cc}-5& -3\left(1+k\right)\\ 2\left(1+k\right)& -6+k^2\end{array}\right)\\ Now,\\ A=4A+I0\cdot I=A\\ \left(\begin{array}{cc}-5& -3-3k\\ 2+2k& -6+k^2\end{array}\right)-\left(\begin{array}{cc}4& -12\\ 8& 4k\end{array}\right)+\left(\begin{array}{cc}10& 0\\ 0& 10\end{array}\right)\cdot \left(\begin{array}{cc}1& -3\\ 2& k\end{array}\right)\\ \left(\begin{array}{cc}-5& -3-3k\\ 2+2k& -6+k^2\end{array}\right)-\left(\begin{array}{cc}4& -12\\ 8& 4k\end{array}\right)+\left(\begin{array}{cc}10& 0\\ 0& 10\end{array}\right)\cdot \left(\begin{array}{cc}1& -3\\ 2& k\end{array}\right)\\ \left(\begin{array}{cc}-5-4+10& -3-3k+12+0\\ 2+2k-8+0& -6+k^2-4k+10\end{array}\right)=\left(\begin{array}{cc}1& -3\\ 2& k\end{array}\right)\\ \left(\begin{array}{cc}1& 9-3k\\ 2k-6& k^2-4k+4\end{array}\right)=\left(\begin{array}{cc}1& -3\\ 2& k\end{array}\right)\\ 9-3k=-3\Rightarrow -3k=-3-9\\ -3k=+12\Rightarrow k=4\end{array}$