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

Mathematics · Matrices

JEE Main 2026 — 4 April, Evening Shift — Question 30

Let $A = \left[ \begin{array}{ccc}1 & 2 & 7\\ 4 & -2 & 8\\ 3 & 8 & -7 \end{array} \right]$ 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

Option A:
1 point
Option B:
2 points
Option C:
3 points
Correct
Option D:
4 points

Answer: C

Step-by-step solution

$\left|\begin{array}{ccc}1-\alpha & 2 & 7\\ 4 & -2-\alpha & 8\\ 3 & 8 & -7-\alpha\end{array}\right|=0$ $\Rightarrow(1-\alpha)[(\alpha+2)(\alpha+7)-64]-2[-28-4 \alpha-24]+7[32+6+3 \alpha]=0$

$\Rightarrow \alpha^{3}+8 \alpha^{2}-88 \alpha-320=0$

(\alpha-8)\left(\alpha^{2}+16 \alpha+40\right)=0

\alpha=8, \alpha=-8 \pm 2 \sqrt{6}

So $\mathrm{p}=-8$ $\Rightarrow(\mathrm{x}-8)^{2}+(\mathrm{y}-16)^{2}=320=8^{2}+16^{2}$ put $\mathrm{y}=0 \Rightarrow \mathrm{x}=16,0$ put $x=0 \Rightarrow y=32,0$ $\Rightarrow(16,0),(0,32),(0,0)$ $⇒ 3$ points of intersection

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Matrices
Topic: Characteristic Equation & roots,application of cayley - hamilton theorem