Part BCSIR NET December 2024nondegenerate-does-not-mean-definite

Nondegenerate does not mean definite

Consider the bilinear form $B : \mathbb{R}^{4} \times \mathbb{R}^{4} \to \mathbb{R}$ defined by $B(x, y) = x_{1}y_{3} + x_{2}y_{4} - x_{3}y_{1} - x_{4}y_{2}$ , where $x = (x_{1}, x_{2}, x_{3}, x_{4})$ and $y = (y_{1}, y_{2}, y_{3}, y_{4})$ in $\mathbb{R}^{4}$ . Let A denote the matrix of B with respect to the standard ordered basis of $\mathbb{R}^{4}$ . Which of the following statements is true?

A.det A = 0
B.det A = −1
C.B(x, x) ≠ 0 for all nonzero $x \in \mathbb{R}^{4}$ .
D.If $x \in \mathbb{R}^{4}$ is nonzero, then there exists $y \in \mathbb{R}^{4}$ such that B(x, y) ≠ 0.✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests converse assumed.

See pricing

50 are analysed free — try those first.

The trap it tests

Converse assumed

The theorem runs one way. You used it in the other.

Drill statements like this

Nondegenerate does not mean definite

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