<p>Let \[A = \begin{bmatrix} e^t & e^{-t}\cos t & e^{-t}\sin t \\ e^t & -e^{-t}\cos t - e^{-t}\sin t & -e^{-t}\sin t + e^{-t}\cos t \\ e^t & 2e^{-t}\sin t & -2e^{-t}\cos t \end{bmatrix}\] Then which of the following is true?</p>
<p>\(A\) is invertible for all \(t \in \mathbb{R}\)</p>
<p>\(A\) is not invertible for any \(t \in \mathbb{R}\)</p>
<p>\(A\) is invertible only for \(t = 0\)</p>
<p>\(|A| = 0\) for all \(t \in \mathbb{R}\)</p>
Step-by-Step Solution
Key Concept: A matrix is invertible if and only if its determinant is nonzero. We need to compute det(A) and verify it is never zero for any real t.
<p><strong>Step 1: Set up the determinant</strong> We need to find |A| by expanding along the first column (which has a common factor e^t):</p><p>|A| = e^t · |M₁| - e^t · |M₂| + e^t · |M₃|</p><p>where M₁, M₂, M₃ are 2×2 minors obtained by deleting row i and column 1.</p><p><strong>Step 2: Factor out e^t</strong> |A| = e^t[|M₁| - |M₂| + |M₃|]</p><p><strong>Step 3: Compute M₁</strong> (delete row 1, column 1):</p><p>|M₁| = |(-e^(-t)cos t - e^(-t)sin t, -e^(-t)sin t + e^(-t)cos t; 2e^(-t)sin t, -2e^(-t)cos t)|</p><p>= (-e^(-t)cos t - e^(-t)sin t)(-2e^(-t)cos t) - (-e^(-t)sin t + e^(-t)cos t)(2e^(-t)sin t)</p><p>= 2e^(-2t)cos t(cos t + sin t) - 2e^(-2t)sin t(cos t - sin t)</p><p>= 2e^(-2t)[cos²t + cos t sin t - sin t cos t + sin²t] = 2e^(-2t)</p><p><strong>Step 4: Compute M₂</strong> (delete row 2, column 1):</p><p>|M₂| = |(e^(-t)cos t, e^(-t)sin t; 2e^(-t)sin t, -2e^(-t)cos t)|</p><p>= e^(-t)cos t(-2e^(-t)cos t) - e^(-t)sin t(2e^(-t)sin t)</p><p>= -2e^(-2t)cos²t - 2e^(-2t)sin²t = -2e^(-2t)</p><p><strong>Step 5: Compute M₃</strong> (delete row 3, column 1):</p><p>|M₃| = |(e^(-t)cos t, e^(-t)sin t; -e^(-t)cos t - e^(-t)sin t, -e^(-t)sin t + e^(-t)cos t)|</p><p>= e^(-t)cos t(-e^(-t)sin t + e^(-t)cos t) - e^(-t)sin t(-e^(-t)cos t - e^(-t)sin t)</p><p>= e^(-2t)[cos t(cos t - sin t) + sin t(cos t + sin t)]</p><p>= e^(-2t)[cos²t - cos t sin t + sin t cos t + sin²t] = e^(-2t)</p><p><strong>Step 6: Combine results</strong></p><p>|A| = e^t[2e^(-2t) - (-2e^(-2t)) + e^(-2t)]</p><p>= e^t[2e^(-2t) + 2e^(-2t) + e^(-2t)]</p><p>= e^t · 5e^(-2t) = 5e^(-t)</p><p><strong>Step 7: Analyze invertibility</strong> Since |A| = 5e^(-t) and e^(-t) > 0 for all t ∈ ℝ, we have |A| ≠ 0 for all t ∈ ℝ. Therefore, A is invertible for all t ∈ ℝ.</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A