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Question #40287

Difficulty: 8/10 | 🟢 Free

The Question

Let $A = \begin{pmatrix} |x| & 2 \\ 2 & |x|-3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ . Let $S$ be the set of all real values of $x$ such that the system of linear equations $A \mathbf{y} = \mathbf{c}$ has a unique solution for every $2 \times 1$ column matrix $\mathbf{c}$ , and the matrices satisfy the equation: $(A+B)^2 = A^2 + B^2 + \begin{pmatrix} x^2-4|x|+8 & x^2-4|x|+9 \\ -x^2+4|x|-1 & x^2-4|x|+2 \end{pmatrix}$ Find the number of elements in $S$ .

Options

A.
0
B.
2
C.
3
D.
4

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

Well-definedness|System of Linear Equations (Cramer/Inverse)|Matrix Multiplication Non-Commutativity|Modulus Function

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to synthesize matrix algebra with system of equations constraints.

The primary trap lies in the non-commutativity of matrix multiplication; students who incorrectly assume $(A+B)^2 = A^2 + B^2 + 2AB$ will obtain a quadratic with a negative discriminant, leading them to erroneously conclude there are 0 solutions.

The secondary, more lethal trap is the 'well-definedness' constraint.

Students who correctly expand to $AB + BA$ will find four solutions ( $x = \pm 2, \pm 4$ ).

However, they must cross-reference these with the condition that $A\mathbf{y} = \mathbf{c}$ has a unique solution, which requires $\det(A) \neq 0$ .

Since $\det(A) = 0$ for $x = \pm 4$ , these must be rejected, leaving only 2 valid solutions.

Failing to check this domain restriction leads to the incorrect answer of 4.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sets - Class 11
Concept	Well-definedness
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free