7.[15] a) Consider the linear system X' =

7.[15] a) Consider the linear system $X' = \begin{pmatrix} 1 & 2 \ 4 & 3 \end{pmatrix} X$. a) Is $X_1 = \begin{pmatrix} -1 \ 1 \end{pmatrix} e^{5t}$ a solution vector for this system? Justify your answer. b) Is {$X_1 = \begin{pmatrix} -1 \ 1 \end{pmatrix} e^{-t}, X_2 = \begin{pmatrix} 2 \ -2 \end{pmatrix} e^{-t}$} a fundamental solution set for this system? Justify your answer.

Transcript

00:01 In the question we have given that the matrix x dash is equal to negative 1 1 over 9 1 times x where the matrix x is equal to negative 1 3 times e to the power negative 4 times 3 o 3.

00:21 To simplify it, we know that x dash is nothing but a derivative of x with respect to t so we will take the derivative of x.

00:32 Of matrix x that is negative 1 3 e to the power negative 4 times t over 3 so we can rewrite it as derivative of negative sorry negative e to the power negative 4 times 3 over 3 over 3 times e to the power negative 4 times 3 over 3 so simplifying it we have to take the derivative of each term which is equal to we know that derivative of e to the power t is e to the power t.

01:08 So e to the power negative 4 times t over 3.

01:11 Then derivative of 4 times t over 3 will be negative 4 over 3 and here we have negative negative sign.

01:20 Similarly we get 3 times e to the power negative 4 times 3 over 3 times negative 4 over 3.

01:28 That's simplifying gives us.

01:31 The required matrix x -tash is equals to, here we have negative and this negative will be positive.

01:39 So it will be 4 over 3 times e to the power negative 4 over 3 times t.

01:47 And here we cancel out 3 from numerator and denominator.

01:50 This is negative 4 times e to the power negative 4 times t over 3.

01:57 We will take it as equation number 1.

01:59 Now we have given that or we have to find the value of negative 1 1 over 9 1.

02:11 That is, negative 1.

02:14 1 over 9 1...

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