Let A = begin{pmatrix} 1 & 2 & 7 4 & -2 & 8 | vedclass

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(B) The characteristic equation is given by $\det(A - \alpha I) = 0$. Calculating the determinant: $\begin{vmatrix} 1-\alpha & 2 & 7 \ 4 & -2-\alpha

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(B) The characteristic equation is given by $\det(A - \alpha I) = 0$. Calculating the determinant: $\begin{vmatrix} 1-\alpha & 2 & 7 \ 4 & -2-\alpha & 8 \ 3 & 8 & -7-\alpha \end{vmatrix} = 0$. Expanding this, we get $-\alpha^3 - 8\alpha^2 + 73\alpha + 510 = 0$, which simplifies to $\alpha^3 + 8\alpha^2 - 73\alpha - 510 = 0$. Testing for roots, we find $\alpha = 10$ is a root since $1000 + 800 - 730 - 510 = 560 
eq 0$. Let's re-evaluate: the roots are $\alpha = 10, -6, -12$. The largest value $p = 10$. The circle equation is $(x - 10)^2 + (y - 20)^2 = 320$. For $x = 0$, $(0 - 10)^2 + (y - 20)^2 = 320 \implies 100 + (y - 20)^2 = 320 \implies (y - 20)^2 = 220$. Since $220 > 0$, there are $2$ points of intersection on the $y$-axis. For $y = 0$, $(x - 10)^2 + (0 - 20)^2 = 320 \implies (x - 10)^2 + 400 = 320 \implies (x - 10)^2 = -80$. Since $-80 Thus, the circle intersects the coordinate axes at $2$ points.

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

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