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Question #41205

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be a $3 \times 3$ matrix whose rows are given by the vectors $\vec{r}_1, \vec{r}_2, \vec{r}_3 \in \mathbb{R}^3$ . It is given that $A$ is an involutory matrix and $\vec{r}_1, \vec{r}_2, \vec{r}_3$ are unit vectors. If $\vec{r}_1 = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$ and $\vec{r}_2 = x \hat{i} + y \hat{j} + z \hat{k}$ , then the maximum possible value of $y$ is:

Options

A.
$\frac{3 - \sqrt{3}}{6}$
B.
$\frac{3 + \sqrt{3}}{6}$
C.
$\frac{\sqrt{3} - 1}{2}$
D.
$\frac{1}{\sqrt{3}}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Idempotent and Involutory Properties|Scalar Triple Product Volume Logic|Unit Vector Normalization

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of matrix transformations and vector volume logic.

The primary trap lies in the involutory condition ( $A^2 = I$ ), which dictates that the determinant is $\pm 1$ .

Students often fail to connect this with the fact that the rows are unit vectors; the Scalar Triple Product (volume of the parallelepiped) can only be $\pm 1$ for unit vectors if they are mutually orthogonal.

This forces $A$ to be an orthogonal matrix.

Combining orthogonality ( $A A^T = I$ ) with the involutory property ( $A^2 = I$ ) elegantly proves that $A$ must be symmetric ( $A = A^T$ ).

Without this deduction, students will have insufficient equations to solve for the components of the second row, falling into the trap of assuming arbitrary vector configurations or failing to normalize the basis vectors.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Matrices - Class 12
Concept	Idempotent and Involutory Properties
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free