Let $A = \begin{bmatrix} 1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7 \end{bmatrix}$ and $\det(A - aI) = 0$, where $a$ is a real number. If the largest possible value of $a$ is $p$, then the circle $(x - p)^2 + (y - 2p)^2 = 320$, intersects the co-ordinate axes at
- (1) 1 point
- (2) 2 points
- (3) 3 points
- (4) 4 points
Show answer and solution
Answer: Option (3)
Key idea: Eigenvalues and Circle-Axis Intersections
- Find eigenvalues by solving $\det(A - aI) = 0$, which gives $a^3 + 8a^2 - 88a - 320 = 0$
- Factor to get $(a-8)(a^2 + 16a + 40) = 0$, yielding eigenvalues $a = 8, -8 \pm 2\sqrt{6}$
- The largest eigenvalue is $p = 8$, so the circle is $(x-8)^2 + (y-16)^2 = 320$
- For x-axis intersection (y=0): $(x-8)^2 = 64 \Rightarrow x = 0, 16$
- For y-axis intersection (x=0): $(y-16)^2 = 256 \Rightarrow y = 0, 32$
- The distinct intersection points are $(0,0)$, $(16,0)$, and $(0,32)$ - total 3 points