Question

Let \(A = \begin{bmatrix} 1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7 \end{bmatrix}\) 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 co-ordinate axes at:

• A. 1 point
• B. 2 points
• C. 3 points
• D. 4 points

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The value \(\alpha\) represents the eigenvalues of matrix \(A\). We find the largest eigenvalue \(p\), then substitute it into the circle equation to find its intersections with the axes (\(x = 0\) and \(y = 0\)).

Step 2: Key Formula or Approach:
1. Characteristic Equation: \(\det(A - \lambda I) = 0\).
2. Circle intersections: solve for \(x\) with \(y=0\) and for \(y\) with \(x=0\).

Step 3: Detailed Explanation:
Characteristic equation expansion: \(\lambda^3 + 8\lambda^2 - 79\lambda - 320 = 0\).
Checking factors of 320, we find \(\lambda = 8\) is a root: \(512 + 512 - 632 - 320 = 72\) (incorrect).
Let's check \(\lambda = 8\) in \(\begin{vmatrix} 1-8 & 2 & 7 \\ 4 & -2-8 & 8 \\ 3 & 8 & -7-8 \end{vmatrix} = \begin{vmatrix} -7 & 2 & 7 \\ 4 & -10 & 8 \\ 3 & 8 & -15 \end{vmatrix}\).
Value \(= -7(150 - 64) - 2(-60 - 24) + 7(32 + 30) = -602 + 168 + 434 = 0\).
So \(p = 8\). Circle equation: \((x - 8)^2 + (y - 16)^2 = 320\).
x-intercept (\(y=0\)): \((x - 8)^2 + (-16)^2 = 320 \implies (x - 8)^2 = 320 - 256 = 64 \implies x = 16, 0\).
y-intercept (\(x=0\)): \((0 - 8)^2 + (y - 16)^2 = 320 \implies (y - 16)^2 = 320 - 64 = 256 \implies y = 32, 0\).
Distinct points: \((0,0), (16,0), (0,32)\). Total = 3 points.

Step 4: Final Answer:
The circle intersects the axes at 3 points.