Matrices & Determinants
Invertible Matrices
Grade 12
Question:
<p>If \[A = \begin{bmatrix} e^t & e^{-t}\cos t & e^{-t}\sin t \\ e^t & -e^{-t}\cos t - e^{-t}\sin t & -e^{-t}\sin t + e^{-t}\cos t \\ e^t & 2e^{-t}\sin t & -2e^{-t}\cos t \end{bmatrix},\] then \(A\) is:</p>
<p>invertible for all \(t \in \mathbb{R}\).</p>
<p>invertible only if \(t = \pi\).</p>
<p>not invertible for any \(t \in \mathbb{R}\).</p>
<p>invertible only if \(t = \dfrac{\pi}{2}\).</p>
Step-by-Step Solution
Key Concept: Recognize that this matrix has a special structure where each row can be expressed as a derivative or combination of exponential functions. The determinant can be computed by factoring out common exponential terms and observing that the resulting simplified matrix has linearly dependent or special properties.
<p><strong>Step 1:</strong> Factor out e^t from column 1 and e^(-t) from columns 2 and 3:</p><p>det(A) = e^t · e^(-t) · e^(-t) · det(B) = e^(-t) · det(B), where B has simplified entries.</p><p><strong>Step 2:</strong> After factoring, observe row relationships. The second row is a linear combination involving the derivative structure of the first row with respect to t, and the third row shows similar derivative patterns.</p><p><strong>Step 3:</strong> Compute the determinant of the simplified matrix. Notice that if we take derivatives of exponential functions e^t, e^(-t)cos(t), and e^(-t)sin(t), the pattern in rows 2 and 3 emerges, indicating linear dependence.</p><p><strong>Step 4:</strong> The rows are linearly dependent (row 2 and row 3 are related through differential relationships), making the determinant zero.</p><p>∴ Answer: A (det(A) = 0)</p>
Correct Answer: A