Matrices & Determinants
Invertible Matrices
Grade 12
Question:
<p>If <i>A</i> = <span>$$\begin{bmatrix} e^t & e^{-t}\cos t & e^{-t}\sin t \\ e^t & -e^{-t}\cos t - e^{-t}\sin t & -e^{-t}\sin t + e^{-t}\cos t \\ e^t & 2e^{-t}\sin t & -2e^{-t}\cos t \end{bmatrix}$$</span>, then <i>A</i> is</p>
<p>(a) invertible only when \(t = \pi\)</p>
<p>(b) invertible for every \(t \in \mathbb{R}\)</p>
<p>(c) not invertible for any \(t \in \mathbb{R}\)</p>
<p>(d) invertible only when \(t = \frac{\pi}{2}\)</p>
Step-by-Step Solution
Key Concept: A matrix is invertible if and only if its determinant is non-zero. We need to calculate det(A) and determine for which values of t it equals zero.
<p><strong>Step 1: Identify the invertibility condition</strong><br/>A matrix A is invertible if and only if det(A) ≠ 0. We must compute the determinant of the given 3×3 matrix.</p><p><strong>Step 2: Calculate det(A) using cofactor expansion along Column 1</strong><br/>$$\text{det}(A) = e^t \begin{vmatrix} -e^{-t}\cos t - e^{-t}\sin t & -e^{-t}\sin t + e^{-t}\cos t \\ 2e^{-t}\sin t & -2e^{-t}\cos t \end{vmatrix} - e^t \begin{vmatrix} e^{-t}\cos t & e^{-t}\sin t \\ 2e^{-t}\sin t & -2e^{-t}\cos t \end{vmatrix} + e^t \begin{vmatrix} e^{-t}\cos t & e^{-t}\sin t \\ -e^{-t}\cos t - e^{-t}\sin t & -e^{-t}\sin t + e^{-t}\cos t \end{vmatrix}$$</p><p><strong>Step 3: Compute the 2×2 determinants</strong><br/>First minor:<br/>$$= (-e^{-t}\cos t - e^{-t}\sin t)(-2e^{-t}\cos t) - (-e^{-t}\sin t + e^{-t}\cos t)(2e^{-t}\sin t)$$<br/>$$= 2e^{-2t}\cos t(\cos t + \sin t) - 2e^{-2t}\sin t(\cos t - \sin t)$$<br/>$$= 2e^{-2t}(\cos^2 t + \sin t\cos t - \sin t\cos t + \sin^2 t) = 2e^{-2t}$$<br/><br/>Second minor:<br/>$$= e^{-t}\cos t(-2e^{-t}\cos t) - e^{-t}\sin t(2e^{-t}\sin t)$$<br/>$$= -2e^{-2t}(\cos^2 t + \sin^2 t) = -2e^{-2t}$$<br/><br/>Third minor:<br/>$$= e^{-t}\cos t(-e^{-t}\sin t + e^{-t}\cos t) - e^{-t}\sin t(-e^{-t}\cos t - e^{-t}\sin t)$$<br/>$$= e^{-2t}(-\cos t\sin t + \cos^2 t + \sin t\cos t + \sin^2 t) = e^{-2t}$$</p><p><strong>Step 4: Combine results</strong><br/>$$\text{det}(A) = e^t(2e^{-2t}) - e^t(-2e^{-2t}) + e^t(e^{-2t})$$<br/>$$= 2e^{-t} + 2e^{-t} + e^{-t} = 5e^{-t}$$</p><p><strong>Step 5: Analyze invertibility</strong><br/>Since det(A) = 5e^{-t} and e^{-t} > 0 for all t ∈ ℝ, we have det(A) ≠ 0 for every t ∈ ℝ.</p><p><strong>∴ Answer:</strong> b</p>
Correct Answer: b