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Question #41292

Difficulty: 8/10 | 🟢 Free

The Question

Let $\hat{u}$ be a unit column vector in $\mathbb{R}^3$ and let $A = I - \hat{u}\hat{u}^T$ , where $I$ is the $3 \times 3$ identity matrix. Let $B$ be a $3 \times 3$ matrix representing the coterminous edges of a parallelepiped $P$ with a volume of $5$ cubic units. A new parallelepiped $P'$ is formed whose coterminous edges are the column vectors of the matrix $C = (\text{adj}(A) + 2A)B$ . What is the volume of $P'$ ?

Options

A.
0
B.
20
C.
40
D.
45

Check Your Answer

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

Determinant as Geometric Area/Volume|Adjoint-Determinant Relation|Singular vs. Non-Singular Threshold

Why This Question Is Tricky

Common Mistakes

This question tests the 'Singular Threshold' trap regarding the Adjoint matrix.

Students frequently assume that if $|A|=0$ , then $\text{adj}(A)$ must be the zero matrix, or they incorrectly attempt to use the identity $\text{adj}(A) = |A|A^{-1}$ which is undefined here.

By using a projection matrix $A = I - \hat{u}\hat{u}^T$ , the determinant is zero, but the adjoint is a non-zero rank-1 matrix ( $\hat{u}\hat{u}^T$ ).

If a student falls for the trap and assumes $\text{adj}(A) = O$ , they will calculate the transformation matrix as $2A$ , which has a determinant of 0, leading them to incorrectly choose 0 as the collapsed volume.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Determinant as Geometric Area/Volume
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free