If the matrix $A = \begin{pmatrix} 0 & 2 \\ K & -1 \end{pmatrix}$ satisfies $A(A^3+3I)=2I$ then value of K is:
Answer: 1
Attempt: I thought of calculating polonium, and showing that the matrix is invertible and calculating the inverse knowing that $p(A) = 0$. Could you give me any suggestions?
The characteristic polynomial of that matrix is $z^2+z-2K$. This should be a divisor of $z(z^3+3)-2 = z^4+3z-2$. Now, $$ z^4+3z-2 = (z^2-z+2)(z^2+z-1). $$ Thus, $K = \tfrac 12$.