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Question #40908

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be a $3 \times 3$ matrix such that $A^2 = A$ . A point $P$ divides the line segment joining $M(2, 1, 5)$ and $N(6, 5, 9)$ in the ratio $k:1$ , where $k = \text{Tr}(A)$ . If $P$ lies on the plane $x + y - z = 0$ , let $d$ be the dimension of the null space of $A$ and $r$ be the rank of $A$ . Find the value of $\det(A + I) + d^2 + r^2$ .

Options

A.
4
B.
5
C.
7
D.
9

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Ratio of Division (k:1)|Inverse Existence Criterion|Matrix Trace Properties

Why This Question Is Tricky

Common Mistakes

This question integrates 3D coordinate geometry with matrix eigenvalue properties, specifically targeting the 'Inverse Existence Criterion' trap.

Students often implicitly assume matrices in such problems are invertible (which for $A^2=A$ would mean $A=I$ and $k=3$ ).

However, the geometric constraint (the plane equation) forces $k=1$ .

This mathematically forces the matrix into a singular state (rank 1, nullity 2), meaning $\det(A) = 0$ .

Recognizing that the trace dictates the exact distribution of eigenvalues ( $1, 0, 0$ ) is crucial to finding the rank, nullity, and the determinant of $A+I$ without ever knowing the exact elements of $A$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Ratio of Division (k:1)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free