JEE Main 2026 — Matrices Question with Solution

The characteristic equation is given by $\det (A-\alpha I)=0$.

$\det \left[\begin{array}{ccc}1-\alpha & 2& 7\\ 4& -2-\alpha & 8\\ 3& 8& -7-\alpha \end{array}\right]=0$

Expanding the determinant along the first row:

$(1-\alpha )[(-2-\alpha )(-7-\alpha )-64]-2[4(-7-\alpha )-24]+7[32-3(-2-\alpha )]=0$

$(1-\alpha )({\alpha }^2+9\alpha -50)-2(-4\alpha -52)+7(3\alpha +38)=0$

$-{\alpha }^3-8{\alpha }^2+59\alpha -50+8\alpha +104+21\alpha +266=0$

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

By inspection, $\alpha =8$ is a root since $8^3+8(8^2)-88(8)-320=512+512-704-320=0$.

Factoring out $(\alpha -8)$, we get:

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

The roots are $\alpha =8$ and $\alpha =\frac{-16\pm \sqrt{256-160}}{2}=-8\pm 2\sqrt{6}$.

The largest possible value of $\alpha$ is $p=8$.

Substituting $p=8$ into the given equation of the circle:

$(x-8{)}^2+(y-16{)}^2=320$

To find the intersection points with the x-axis, substitute $y=0$:

$(x-8{)}^2+(-16{)}^2=320$

$(x-8{)}^2+256=320\Rightarrow (x-8{)}^2=64\Rightarrow x-8=\pm 8$

$x=0$ or $x=16$

The circle intersects the x-axis at $(0,0)$ and $(16,0)$.

To find the intersection points with the y-axis, substitute $x=0$:

$(-8{)}^2+(y-16{)}^2=320$

$64+(y-16{)}^2=320\Rightarrow (y-16{)}^2=256\Rightarrow y-16=\pm 16$

$y=0$ or $y=32$

The circle intersects the y-axis at $(0,0)$ and $(0,32)$.

The distinct points of intersection with the coordinate axes are $(0,0)$, $(16,0)$, and $(0,32)$.

Thus, the circle intersects the coordinate axes at exactly $3$ points.

Answer: $3$ points