First slide

Applications of determinant

Question

$If A = [\begin{array}{ccc} e^{t} & e^{t} \cos t & e^{- t} \sin t \\ e^{t} & - e^{t} \cos t - e^{- t} \sin t & - e^{- t} \sin t + e^{- t} \cos t \\ e^{t} & 2 e^{- t} \sin t & - 2 e^{- t} \cos t \end{array}] then A is$

Moderate

A

$invertible only when t = π$
B

$invertible for every t \in R$
C

$not invertible for any t \in R$
D

$invertible only when t = \frac{π}{2}$

Solution

$| A | = |\begin{matrix} e^{t} & e^{- t} \cos^{t} & e^{- t} \sin t \\ e^{t} & - e^{t} \cos t - e^{- t} \sin t & - e^{- t} \sin t + e^{- t} \cos t \\ e^{t} & 2 e^{- t} \sin t & - 2 e^{- t} \cos t \end{matrix}|$
$= (e^{t}) (e^{- t}) (e^{- t}) |\begin{matrix} 1 & \cos t & \sin t \\ 1 & - \cos t - \sin t & - \sin t + \cos t \\ 1 & 2 \sin t & - 2 \cos t \end{matrix}|$
(taking common from each column)
$Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}, we got [∵ e^{t - t} = e^{0} = 1]$
$= e^{- t} |\begin{matrix} 1 & \cos t & \sin t \\ 0 & - 2 \cos t - \sin t & - 2 \sin t + \cos t \\ 0 & 2 \sin t - \cos t & - 2 \cos t - \sin t \end{matrix}|$
$= e^{- t} ({(2 \cos t + \sin t)}^{2} + {(2 \sin t - \cos t)}^{2})$
( expanding along column 1)
$\begin{aligned} = e^{- t} (5 \cos^{2} t + 5 \sin^{2} t) \\ = 5 e^{- t} (∵ \cos^{2} t + \sin^{2} t = 1) \\ \Rightarrow | A | = 5 e^{- t} \neq 0 for all t \in R \end{aligned}$
$∴ A is invertible for all t \in R$
$[∵ I f | A | \neq 0, then A is invertible]$

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