Let A=pmatrix1&2&74&-2&83&8&-7pmatrix, and let be a real number satisfying (A- I)=0.

Question

Let $A=\begin{pmatrix}1&2&7\4&-2&8\3&8&-7\end{pmatrix}$, and let $\alpha$ be a real number satisfying $\det(A-\alpha I)=0$. If the largest possible value of $\alpha$ is $p$, then the circle $(x-p)^2+(y-2p)^2=320$ intersects the coordinate axes at:

Accepted Answer

Expanding $\det(A-\alpha I)=0$ gives the cubic $-\alpha^3-8\alpha^2+88\alpha+320=0$, i.e. $\alpha^3+8\alpha^2-88\alpha-320=0$.

Testing $\alpha=8$: $512+512-704-320=0$ , so $(\alpha-8)$ is a factor: $\alpha^3+8\alpha^2-88\alpha-320=(\alpha-8)(\alpha^2+16\alpha+40)$.

Solving $\alpha^2+16\alpha+40=0$: $\alpha=\dfrac{-16\pm\sqrt{96}}{2}=-8\pm2\sqrt6$.

So the eigenvalues are $8,\ -8+2\sqrt6,\ -8-2\sqrt6$, and the largest is $p=8$.

The circle is $(x-8)^2+(y-16)^2=320$, centre $(8,16)$, radius $\sqrt{320}=8\sqrt5$.

On the $x$-axis ($y=0$): $(x-8)^2+256=320\Rightarrow(x-8)^2=64\Rightarrow x=0$ or $16$. Points: $(16,0),(0,0)$.

On the $y$-axis ($x=0$): $64+(y-16)^2=320\Rightarrow(y-16)^2=256\Rightarrow y=0$ or $32$. Points: $(0,32),(0,0)$.

Distinct points: $(16,0),(0,0),(0,32)$ --- 3 points.

Solutions · 1