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Question #41121

Difficulty: 8/10 | 🟢 Free

The Question

Let $A = \begin{pmatrix} 4-i & -2+2i \\ 1 & 1 \end{pmatrix}$ and $I$ be the $2 \times 2$ identity matrix. A linear transformation $T_x: \mathbb{C}^2 \to \mathbb{C}^2$ is defined as $T_x(\mathbf{v}) = (A - xI)\mathbf{v}$ , where $x \in \mathbb{R}$ . If $S$ is the set of all real values of $x$ for which $T_x$ is NOT an injective mapping, then $S$ is equal to:

Options

A.
\emptyset
B.
{3}
C.
{2, 3}
D.
{3, 2-i}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Injective Mapping (One-One)|System of Linear Equations (Cramer/Inverse)|Quadratic Equations with Complex Coefficients

Why This Question Is Tricky

Common Mistakes

This question tests the conceptual understanding of injective linear transformations and the properties of polynomials with complex coefficients.

The primary trap lies in the characteristic equation $x^2 - (5-i)x + 6 - 3i = 0$ .

Students often calculate the discriminant $D = 2i$ and, seeing it is not a non-negative real number, incorrectly conclude that the equation has no real roots (Option A).

They fail to recognize that the $D \ge 0$ condition for real roots only applies to quadratics with real coefficients.

Additionally, distractors are provided for students who only solve the real part (Option C) or ignore the domain constraint $x \in \mathbb{R}$ (Option D).

The correct approach decouples the real and imaginary parts, revealing the real eigenvalue $x=3$ that makes the determinant zero and the mapping non-injective.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Injective Mapping (One-One)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free