Solveeit Logo
Login

Question

Question: If matrix A can be given as \(\left[ \begin{matrix} {{e}^{t}} & {{e}^{-t}}\cos \theta & {{e}^{-...

If matrix A can be given as [ete−tcos⁡θe−tsin⁡θ et−e−tcos⁡θ−e−tsin⁡θ−e−tsin⁡θ+e−tcos⁡θ et2e−tsin⁡θ−2e−tcos⁡θ ]\left[ \begin{matrix} {{e}^{t}} & {{e}^{-t}}\cos \theta & {{e}^{-t}}\sin \theta \\\ {{e}^{t}} & -{{e}^{-t}}\cos \theta -{{e}^{-t}}\sin \theta & -{{e}^{-t}}\sin \theta +{{e}^{-t}}\cos \theta \\\ {{e}^{t}} & 2{{e}^{-t}}\sin \theta & -2{{e}^{-t}}\cos \theta \\\ \end{matrix} \right], then A is
(A) Invertible only if t=π2 (B) Not invertible for any t∈R (C) Invertible for all t∈R (D) Invertible only if t=π \begin{aligned} & \left( A \right)\text{ Invertible only if }t=\dfrac{\pi }{2} \\\ & \left( B \right)\text{ Not invertible for any }t\in R \\\ & \left( C \right)\text{ Invertible for all }t\in R \\\ & \left( D \right)\text{ Invertible only if }t=\pi \\\ \end{aligned}

Explanation

Solution

We solve this problem starting with taking the given matrix and taking the value et{{e}^{t}} and e−t{{e}^{-t}} common from the columns 1,2 and 3 respectively and then applying elementary row operations to simplify the matrix and then calculate the determinant of the matrix A. Then we find the condition for which the determinant is equal to zero. In all other cases than those, matrix A is invertible.

Complete step-by-step solution:
The matrix we were given is A=[ete−tcos⁡θe−tsin⁡θ et−e−tcos⁡θ−e−tsin⁡θ−e−tsin⁡θ+e−tcos⁡θ et2e−tsin⁡θ−2e−tcos⁡θ ]A=\left[ \begin{matrix} {{e}^{t}} & {{e}^{-t}}\cos \theta & {{e}^{-t}}\sin \theta \\\ {{e}^{t}} & -{{e}^{-t}}\cos \theta -{{e}^{-t}}\sin \theta & -{{e}^{-t}}\sin \theta +{{e}^{-t}}\cos \theta \\\ {{e}^{t}} & 2{{e}^{-t}}\sin \theta & -2{{e}^{-t}}\cos \theta \\\ \end{matrix} \right]
We need to find the condition for which the inverse of A exists.
Let us go through the concept of the inverse of a matrix.
A matrix A is said to have inverse if and only if its determinant is not equal to zero.
So, let us find the determinant of our given matrix A using elementary row operations.
⇒∣A∣=∣ete−tcos⁡θe−tsin⁡θ et−e−tcos⁡θ−e−tsin⁡θ−e−tsin⁡θ+e−tcos⁡θ et2e−tsin⁡θ−2e−tcos⁡θ ∣\Rightarrow \left| A \right|=\left| \begin{matrix} {{e}^{t}} & {{e}^{-t}}\cos \theta & {{e}^{-t}}\sin \theta \\\ {{e}^{t}} & -{{e}^{-t}}\cos \theta -{{e}^{-t}}\sin \theta & -{{e}^{-t}}\sin \theta +{{e}^{-t}}\cos \theta \\\ {{e}^{t}} & 2{{e}^{-t}}\sin \theta & -2{{e}^{-t}}\cos \theta \\\ \end{matrix} \right|
As we see at the matrix, we can take et{{e}^{t}} common from the first column, e−t{{e}^{-t}} common from the second and third column. Then we get,
⇒∣A∣=(et×e−t×e−t)∣1cos⁡θsin⁡θ 1−cos⁡θ−sin⁡θ−sin⁡θ+cos⁡θ 12sin⁡θ−2cos⁡θ ∣  ⇒∣A∣=(e−t)∣1cos⁡θsin⁡θ 1−cos⁡θ−sin⁡θ−sin⁡θ+cos⁡θ 12sin⁡θ−2cos⁡θ ∣ \begin{aligned} & \Rightarrow \left| A \right|=\left( {{e}^{t}}\times {{e}^{-t}}\times {{e}^{-t}} \right)\left| \begin{matrix} 1 & \cos \theta & \sin \theta \\\ 1 & -\cos \theta -\sin \theta & -\sin \theta +\cos \theta \\\ 1 & 2\sin \theta & -2\cos \theta \\\ \end{matrix} \right| \\\ & \\\ & \Rightarrow \left| A \right|=\left( {{e}^{-t}} \right)\left| \begin{matrix} 1 & \cos \theta & \sin \theta \\\ 1 & -\cos \theta -\sin \theta & -\sin \theta +\cos \theta \\\ 1 & 2\sin \theta & -2\cos \theta \\\ \end{matrix} \right| \\\ \end{aligned}
Now, let us apply the row operation, add Row 2 to Row 1
⇒∣A∣=(e−t)∣1+1cos⁡θ+(−cos⁡θ−sin⁡θ)sin⁡θ+(−sin⁡θ+cos⁡θ) 1−cos⁡θ−sin⁡θ−sin⁡θ+cos⁡θ 12sin⁡θ−2cos⁡θ ∣ ⇒∣A∣=(e−t)∣2−sin⁡θcos⁡θ 1−cos⁡θ−sin⁡θ−sin⁡θ+cos⁡θ 12sin⁡θ−2cos⁡θ ∣ \begin{aligned} & \Rightarrow \left| A \right|=\left( {{e}^{-t}} \right)\left| \begin{matrix} 1+1 & \cos \theta +\left( -\cos \theta -\sin \theta \right) & \sin \theta +\left( -\sin \theta +\cos \theta \right) \\\ 1 & -\cos \theta -\sin \theta & -\sin \theta +\cos \theta \\\ 1 & 2\sin \theta & -2\cos \theta \\\ \end{matrix} \right| \\\ & \Rightarrow \left| A \right|=\left( {{e}^{-t}} \right)\left| \begin{matrix} 2 & -\sin \theta & \cos \theta \\\ 1 & -\cos \theta -\sin \theta & -\sin \theta +\cos \theta \\\ 1 & 2\sin \theta & -2\cos \theta \\\ \end{matrix} \right| \\\ \end{aligned}
Now, let us divide Row 3 with 2 and then add it to Row 1.
⇒∣A∣=(e−t)∣2+12−sin⁡θ+2sin⁡θ2cos⁡θ+−2cos⁡θ2 1−cos⁡θ−sin⁡θ−sin⁡θ+cos⁡θ 12sin⁡θ−2cos⁡θ ∣  ⇒∣A∣=(e−t)∣52−sin⁡θ+sin⁡θcos⁡θ−cos⁡θ 1−cos⁡θ−sin⁡θ−sin⁡θ+cos⁡θ 12sin⁡θ−2cos⁡θ ∣  ⇒∣A∣=(e−t)∣5200 1−cos⁡θ−sin⁡θ−sin⁡θ+cos⁡θ 12sin⁡θ−2cos⁡θ ∣ \begin{aligned} & \Rightarrow \left| A \right|=\left( {{e}^{-t}} \right)\left| \begin{matrix} 2+\dfrac{1}{2} & -\sin \theta +\dfrac{2\sin \theta }{2} & \cos \theta +\dfrac{-2\cos \theta }{2} \\\ 1 & -\cos \theta -\sin \theta & -\sin \theta +\cos \theta \\\ 1 & 2\sin \theta & -2\cos \theta \\\ \end{matrix} \right| \\\ & \\\ & \Rightarrow \left| A \right|=\left( {{e}^{-t}} \right)\left| \begin{matrix} \dfrac{5}{2} & -\sin \theta +\sin \theta & \cos \theta -\cos \theta \\\ 1 & -\cos \theta -\sin \theta & -\sin \theta +\cos \theta \\\ 1 & 2\sin \theta & -2\cos \theta \\\ \end{matrix} \right| \\\ & \\\ & \Rightarrow \left| A \right|=\left( {{e}^{-t}} \right)\left| \begin{matrix} \dfrac{5}{2} & 0 & 0 \\\ 1 & -\cos \theta -\sin \theta & -\sin \theta +\cos \theta \\\ 1 & 2\sin \theta & -2\cos \theta \\\ \end{matrix} \right| \\\ \end{aligned}
Now let us find the value of determinant of above obtained matrix after transformation,
⇒∣5200 1−cos⁡θ−sin⁡θ−sin⁡θ+cos⁡θ 12sin⁡θ−2cos⁡θ ∣ ⇒52[(−2cos⁡θ)(−cos⁡θ−sin⁡θ)−(2sin⁡θ)(−sin⁡θ+cos⁡θ)] ⇒52[(2cos⁡2θ+2sin⁡θcos⁡θ)−(−2sin⁡2θ+2sin⁡θcos⁡θ)] ⇒52[2cos⁡2θ+2sin⁡θcos⁡θ+2sin⁡2θ−2sin⁡θcos⁡θ] ⇒52[2(cos⁡2θ+sin⁡2θ)] ⇒52(2) ⇒5 \begin{aligned} & \Rightarrow \left| \begin{matrix} \dfrac{5}{2} & 0 & 0 \\\ 1 & -\cos \theta -\sin \theta & -\sin \theta +\cos \theta \\\ 1 & 2\sin \theta & -2\cos \theta \\\ \end{matrix} \right| \\\ & \Rightarrow \dfrac{5}{2}\left[ \left( -2\cos \theta \right)\left( -\cos \theta -\sin \theta \right)-\left( 2\sin \theta \right)\left( -\sin \theta +\cos \theta \right) \right] \\\ & \Rightarrow \dfrac{5}{2}\left[ \left( 2{{\cos }^{2}}\theta +2\sin \theta \cos \theta \right)-\left( -2{{\sin }^{2}}\theta +2\sin \theta \cos \theta \right) \right] \\\ & \Rightarrow \dfrac{5}{2}\left[ 2{{\cos }^{2}}\theta +2\sin \theta \cos \theta +2{{\sin }^{2}}\theta -2\sin \theta \cos \theta \right] \\\ & \Rightarrow \dfrac{5}{2}\left[ 2\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right) \right] \\\ & \Rightarrow \dfrac{5}{2}\left( 2 \right) \\\ & \Rightarrow 5 \\\ \end{aligned}
Then determinant of A is
⇒∣A∣=(e−t)∣5200 1−cos⁡θ−sin⁡θ−sin⁡θ+cos⁡θ 12sin⁡θ−2cos⁡θ ∣ ⇒∣A∣=(e−t)(5) ⇒∣A∣=5e−t \begin{aligned} & \Rightarrow \left| A \right|=\left( {{e}^{-t}} \right)\left| \begin{matrix} \dfrac{5}{2} & 0 & 0 \\\ 1 & -\cos \theta -\sin \theta & -\sin \theta +\cos \theta \\\ 1 & 2\sin \theta & -2\cos \theta \\\ \end{matrix} \right| \\\ & \Rightarrow \left| A \right|=\left( {{e}^{-t}} \right)\left( 5 \right) \\\ & \Rightarrow \left| A \right|=5{{e}^{-t}} \\\ \end{aligned}
So, determinant of matrix A is 5e−t5{{e}^{-t}}.
But as we know e−t≠0{{e}^{-t}}\ne 0 for all t∈Rt\in R, ∣A∣≠0\left| A \right|\ne 0.
So, as the determinant is not equal to zero inverse of A exists for all t∈Rt\in R, that is A is invertible for all t∈Rt\in R.
Hence, answer is Option C.

Note: We can also solve this problem without applying the row operations and finding the determinant for matrix A as,
⇒∣A∣=∣ete−tcos⁡θe−tsin⁡θ et−e−tcos⁡θ−e−tsin⁡θ−e−tsin⁡θ+e−tcos⁡θ et2e−tsin⁡θ−2e−tcos⁡θ ∣ ⇒∣A∣=et∣−e−tcos⁡θ−e−tsin⁡θ−e−tsin⁡θ+e−tcos⁡θ 2e−tsin⁡θ−2e−tcos⁡θ ∣−e−tcos⁡θ∣et−e−tsin⁡θ+e−tcos⁡θ et−2e−tcos⁡θ ∣  +e−tsin⁡θ∣et−e−tcos⁡θ−e−tsin⁡θ et2e−tsin⁡θ ∣ \begin{aligned} & \Rightarrow \left| A \right|=\left| \begin{matrix} {{e}^{t}} & {{e}^{-t}}\cos \theta & {{e}^{-t}}\sin \theta \\\ {{e}^{t}} & -{{e}^{-t}}\cos \theta -{{e}^{-t}}\sin \theta & -{{e}^{-t}}\sin \theta +{{e}^{-t}}\cos \theta \\\ {{e}^{t}} & 2{{e}^{-t}}\sin \theta & -2{{e}^{-t}}\cos \theta \\\ \end{matrix} \right| \\\ & \Rightarrow \left| A \right|={{e}^{t}}\left| \begin{matrix} -{{e}^{-t}}\cos \theta -{{e}^{-t}}\sin \theta & -{{e}^{-t}}\sin \theta +{{e}^{-t}}\cos \theta \\\ 2{{e}^{-t}}\sin \theta & -2{{e}^{-t}}\cos \theta \\\ \end{matrix} \right|-{{e}^{-t}}\cos \theta \left| \begin{matrix} {{e}^{t}} & -{{e}^{-t}}\sin \theta +{{e}^{-t}}\cos \theta \\\ {{e}^{t}} & -2{{e}^{-t}}\cos \theta \\\ \end{matrix} \right| \\\ & \text{ }+{{e}^{-t}}\sin \theta \left| \begin{matrix} {{e}^{t}} & -{{e}^{-t}}\cos \theta -{{e}^{-t}}\sin \theta \\\ {{e}^{t}} & 2{{e}^{-t}}\sin \theta \\\ \end{matrix} \right| \\\ \end{aligned}
But, it a difficult way of solving and by using row operations we can simplify it easily.