If A=pmatrix1&5&10pmatrix, A^-1= A+ I, and +=-2, then 4^2+^2+^2 is equal to:

Question

If $A=\begin{pmatrix}1&5\\lambda&10\end{pmatrix}$, $A^{-1}=\alpha A+\beta I$, and $\alpha+\beta=-2$, then $4\alpha^2+\beta^2+\lambda^2$ is equal to:

Accepted Answer

$|A-xI|=0\Rightarrow x^2-11x+(10-5\lambda)=0$, so by Cayley--Hamilton $A^2-11A+(10-5\lambda)I=0$, giving $(10-5\lambda)A^{-1}=11I-A$, i.e.
$$\alpha=\frac{-1}{10-5\lambda},\qquad \beta=\frac{11}{10-5\lambda}.$$

$\alpha+\beta=-2 \Rightarrow \dfrac{10}{10-5\lambda}=-2\Rightarrow10-5\lambda=-5\Rightarrow\lambda=3$.

So $\alpha=\dfrac15,\ \beta=-\dfrac{11}5$.

$4\alpha^2+\beta^2+\lambda^2=\dfrac4{25}+\dfrac{121}{25}+9=5+9=14$.

Solutions · 1