I am stuck on this problem. I need to know how to solve it and part of that is knowing what I have done wrong in my current attempts to solve it. I have attached my work, which is incomplete, and a sample problem that was done by the homework website. I added notes to the websites work. I am kinda lost and don't know what I am doing. Please treat my like a 5 year old and over explain everything.

HW 5. 3 Diagonalize the following matrix, One eigenvalue is 1: 2 and one eigenvector is ( - 3, 3, 3 ). - 10 - 18 9 12 20 - 9 Av. 12 18 - 7 - 3 -10 - 18 9 3 12 20 -9 w A - 3 -7 2 = - 1 12 18 -7 3 - 3 A- XI : 0 - 10 - 18 9 -10 0 1 - 9 - 18 9 9-18 9 - 1 - 2 1 e 12 20 - 9 - 1 12 21 0- 10 - 12 21 - 9 12 21 - 9 -12 - 18 12 18 - 7 0 01 12 18 - 6 12 18 - 6 3 - 3 - 1 - 2 1 -12- 24 12 12 - 2 1 6 3 - 3 18 - 4 3 3 - 3 12 18 -6 O - 6 6 - 6 6 : - 6 then oat 1 2 - 1 2 - 1 1 O 1 + - 2 2 0 0 0 0 O 1 XitX3 = 0 X , = - X 3 X 3 X 3 X2 - * 3 = 0 X 2 = X3 X3 X b X 3 = X3 1 /- cDiagonalize the following matrix. One eigenvalue is A =7 and one eigenvector is ( - 1,2,3). I am working with, Lamda=2, -9 - 8 eigenvector (-3,3,3) 32 23 - 8 -10 -18 9 48 24 -5 12 20 -9 12 18 -7 An n n square matrix A is diagonalizable if A = PDP" for some invertible matrix P and some diagonal matrix D. In this case, the columns of P are linearly independent eigenvectors of A and the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P. First use the given eigenvector ( - 1,2,3) to determine the corresponding eigenvalue. Let v, = (- 1,2,3). If Av, is a scalar multiple of v, and v, is nonzero, then v, is an eigenvector of A. Note that v, is nonzero. To determine the corresponding eigenvalue of v1, begin by computing Av,- -9 -8 4 AV1 = 32 23 -8 W N 48 24 -5 10 Multiply and simplify. 15 Now determine the value of > that satisfies the equation Av, = Av, - The vector v, should be multiplied by - 5 to obtain Av, - I was able to get to Lamda on my problem. I took lamda as -1, not 3, because I thought the goal was to get something that matched the v1 we were provided. - 5/ Multiply and simplify. W N So * = - 5 is an eigenvalue of the matrix. To find a basis corresponding to this eigenvalue, substitute ) = - 5 into the matrix A - MI and simplify. -4 -8 32 28 -8 48 24 0 Now write the resulting matrix as an augmented matrix for (A + 51)x =0, as shown below. -4 -8 40 32 28 -8 0 48 24 0 0 Use row operations to reduce the augmented matrix for (A + 51)x =0 to reduced echelon form. 10 0 -4 -8 4 0 This is as far as I got working on my own. I got v1= which doesn't match v1 from before. Do i just multiply by -3 and call it a day? 32 28 - 8 0 ~ 2 Does this sequince of step provid any information or is it just a check? 0 1 O 48 24 0 0 0 0 0 0 Therefore, a basis for the eigenspace corresponding to a = - 5 is v, - Please go into step by step so I can follow along, and if reasonable, explain what is going on, and what we are looking for when doing the steps. Next find a basis corresponding to the given eigenvalue, A =7. Substitute A = 7 into the matrix A - XI and simplify. -16 - 8 4 32 16 -8 48 24 Now write the resulting matrix as an augmented matrix for (A -71)x =0, as shown below. -16 - 8 40 32 16 -8 0 18 24 - 12 0 Use row operations to reduce the augmented matrix for (A -71)x =0 to reduced echelon form. 16 -8 4 0 32 1 -8 0 48 24 - 12 0 The general solution of (A - 71)x =0 in terms of x2 and x3 is X2 2 4 x 3 0 Next construct P of the form P = [ v, v2 v3 ] Recall that vi - 2 v2 - o - and va -[ : ] Finally, construct the diagonal matrix D corresponding to the columns in P. The diagonal entries of D are eigenvalues of A. -50 0 D= 070 0 07 The matrices P and D can be verified by checking that AP = PD