Matrices & Determinants
Invertible Matrices
Grade 12
Question:
<p>If <i>A</i> = \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) (where <i>bc</i> ≠ 0) satisfies the equations <i>x</i><sup>2</sup> + <i>k</i> = 0, then which of the following are correct?</p>
<p>(a) <i>a</i> + <i>d</i> = 0</p>
<p>(b) <i>k</i> = −|<i>A</i>|</p>
<p>(c) sin <i>A</i> is invertible</p>
<p>(d) (sin <i>A</i>)<sup>T</sup> (sin <i>A</i>) = <i>I</i></p>
Step-by-Step Solution
Key Concept: For a matrix satisfying x² + k = 0, analyze the structure through its sine and cosine components and use determinant properties.
<p><strong>Solution:</strong></p><p>Given sin <i>A</i> = $\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$ and cos <i>A</i> = $\begin{pmatrix} \sin\theta & \cos\theta \\ \cos\theta & -\sin\theta \end{pmatrix}$</p><p>|sin <i>A</i>| = $\cos^2\theta + \sin^2\theta = 1 \neq 0$</p><p>Hence, sin <i>A</i> is invertible.</p><p>(sin <i>A</i>)<sup>T</sup>(sin <i>A</i>) = $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$</p><p>∴ Answers are (a) and (c).</p>
Correct Answer: A,C