Hw10: Problem 6 (1 pt) Let M = < b m a t r i x >. Find...

Hw10: Problem 6 (1 pt) Let $M = \begin{bmatrix} 14 & -14 \\ 7 & -7 \end{bmatrix}$. Find a diagonal matrix $D$ and an invertible matrix $P$ such that $M = PDP^{-1}$. $D = \begin{bmatrix} \\\\ \\\\ \end{bmatrix}$ and $P = \begin{bmatrix} \\\\ \\\\ \end{bmatrix}$ Now, find formulas for the entries of $M^n$, where $n$ is a positive integer. $M^n = \begin{bmatrix} \\\\ \\\\ \end{bmatrix}$ Note: $D$ and $P$ will not be graded unless you enter something for both of them. Note: You can earn partial credit on this problem.

Transcript

00:01 Hello everyone from the question we are given that matrix a is equals to 2 minus 2 3 1 1 2 3 2 2 2 2 2 3 minus 2 2 2 3 2 2 3 and we have to verify that p inverse ap is a diagonal with the eigenvalues on the main diagonal so first we are going to find eigenvalues of matrix a so first we write characteristic equation of matrix a which is equals to determinant of a minus lambda is equal to 0 this implies determinant of 2 minus lambda minus 2 3, 0 3 minus lambda minus 2, 0 minus 1, 2 minus lambda is equals to 0.

00:38 By solving this we get 2 minus lambda multiply by 3 minus lambda into 2 minus lambda minus lambda minus 2 is equals to 0.

00:47 This implies 2 minus lambda multiply by 6 minus 3 lambda minus 2 lambda plus lambda square minus 2 is equals to 0.

00:57 This implies 2 minus lambda multiplied by lambda square minus 5 lambda plus four is equals to zero.

01:04 This implies 2 minus lambda multiply by lambda to 8 lambda minus lambda plus 4 is equals to 0.

01:12 This implies 2 minus lambda into lambda minus 4 into lambda minus 1 is equal to 0 this implies lambda equals to 1 .2 .4 so the eigen values of matrixa is 1 to 4.

01:25 Now we are going to find eigenvector for these eigenvalues.

01:29 So first we put lambda equals to 1.

01:31 So we can write a minus lambda i x is equals to 0, where x1 x2 x3 is an eigenvector for lambda equals to 1.

01:40 This implies a minus 1, a minus i into x is equals to 0.

01:47 This implies 1 minus 2 3, 0 2 minus 2, 0 minus 1 multiply by x 1 x1 x2 x3 is equal to 0.

01:55 Is equal to 0.

01:56 From here we can write x1 minus 2 x2 plus 3 x3 is equal to 0, 2x2 minus 2 x3 is equal to 0 minus x2 plus x3 is equal to 0.

02:10 By solving these equations we get x2 is equals to x3 and x1 is equal to minus x2 minus x2 x2 this is equals to x2 into minus 1 -1 1 1.

02:22 So the eigenvector for lambda equals to 1 is 1.

02:25 Minus 1 -1.

02:26 Now we are going to find eigenvector for lambda equals to 2.

02:30 So we can write a minus lambda i x is equal to 0 where x1 x2 x3 is an eigenvector for lambda equals to 2.

02:39 This implies a minus 2 i into x is equals to 0.

02:43 This implies 0 minus 2 3 01 minus 2 0 minus 1 0 minus 2 x3 is equals to 0...

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