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 equation \(\det(A - \alpha I) = 0\) is the characteristic equation that defines the eigenvalues of matrix \(A\). We must find the largest eigenvalue \(p\), substitute it into the given circle equation, and count its intersection points with the \(x\)-axis (\(y=0\)) and \(y\)-axis (\(x=0\)).
Step 2: Key Formula or Approach:
Characteristic polynomial for a \(3 \times 3\) matrix:
\[ \alpha^3 - \text{tr}(A)\alpha^2 + (M_{11} + M_{22} + M_{33})\alpha - \det(A) = 0 \] where \(\text{tr}(A)\) is the trace and \(M_{ii}\) are the principal minors.
Step 3: Detailed Explanation:
Trace: \(\text{tr}(A) = 1 - 2 - 7 = -8\).
Determinant:
\[ |A| = 1(14 - 64) - 2(-28 - 24) + 7(32 - (-6)) = -50 + 104 + 266 = 320 \] Sum of principal minors:
\(M_{11} = 14 - 64 = -50\)
\(M_{22} = -7 - 21 = -28\)
\(M_{33} = -2 - 8 = -10\)
Sum = \(-88\).
The characteristic equation:
\[ \alpha^3 + 8\alpha^2 - 88\alpha - 320 = 0 \] By trial, \(\alpha = 8\) is a root: \(8^3 + 8(8^2) - 88(8) - 320 = 512 + 512 - 704 - 320 = 0\).
Factor out \((\alpha - 8)\):
\[ (\alpha - 8)(\alpha^2 + 16\alpha + 40) = 0 \] Roots of quadratic: \(\alpha = \frac{-16 \pm \sqrt{256 - 160}}{2} = -8 \pm 2\sqrt{6}\).
Since \(2\sqrt{6} \approx 4.9\), the largest eigenvalue is \(p = 8\).
Substitute \(p = 8\) into the circle equation:
\[ (x - 8)^2 + (y - 16)^2 = 320 \] Intersect with \(x\)-axis (\(y = 0\)):
\[ (x - 8)^2 + 256 = 320 \implies (x - 8)^2 = 64 \implies x = 16 \text{ or } x = 0 \] Points: \((16, 0), (0, 0)\).
Intersect with \(y\)-axis (\(x = 0\)):
\[ 64 + (y - 16)^2 = 320 \implies (y - 16)^2 = 256 \implies y = 32 \text{ or } y = 0 \] Points: \((0, 32), (0, 0)\).
The unique intersection points are \((0, 0)\), \((16, 0)\), and \((0, 32)\).
Step 4: Final Answer:
The circle intersects the axes at 3 points.