anonymous
  • anonymous
Review: Diagonalization. Give me some time to put the question and my solution. I got a different answer, so I need help correcting my solution.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Serenity74
  • Serenity74
Oh, I was about to say, yo wheres da question at XD
anonymous
  • anonymous
Show that A is diagonalizable then diagonalize A (that is, find an invertible matrix P and diagonal matrix D such that \(\sf P^{-1}AP=D\)) |dw:1449372083955:dw| My solution: • Attachment 1: I found the eigenvalues, to know the algebraic multiplicity • Attachment 2: Next, eigenvectors, to know the geometric multiplicity So since algebraic multiplicity= geometric multiplicy, A is diagonalizable. • Attachment 3: I used the eigenvectors I got as my P matrix. I formed 3 x3 matrix, so it should be invertible. Then, I calculated for P inverse using \(\sf \frac{1}{det(P)} (cof(P))^T\) ( we are not allowed to use \(\sf P^{-1}P=I\) ) Lastly, I used \(\sf P^{-1}AP=D\) to diagonalize A... but I think I made a mistake somewhere since I got a different answer.
anonymous
  • anonymous
I think I made a mistake when I was looking for the determinant of P.. Btw, the right answer is: |dw:1449372592785:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Serenity74
  • Serenity74
You figured it out right? So do you still need help or nah?
anonymous
  • anonymous
I made a mistake somewhere
UnkleRhaukus
  • UnkleRhaukus
1.jgp looks fine
anonymous
  • anonymous
hmm I think it's on 3.jpg ... I recalculated the determinant of P and this time I got 3, not 1. Can you check it? I'll try try to diagonalize it again using that value
UnkleRhaukus
  • UnkleRhaukus
1 Attachment
anonymous
  • anonymous
oh right, it should be <0 0 2 > , but then if I used it, i'll get a determinant of 2 lol the way I got 3 is I used the 3rd row to solve |P| . I already calculated the cofactors, so it is not that hard. \(\sf |P| = 0C_{31} + 1C_{32}+1C_{33} = 3\) but still I got a different matrix D >.<
UnkleRhaukus
  • UnkleRhaukus
<0 -1 2>
anonymous
  • anonymous
omg okay. I'll try to do it again. I just hope I got it this time
UnkleRhaukus
  • UnkleRhaukus
\[\ddot\smile \]
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align} |P|&= \begin{vmatrix} -1&-1&1\\1&0&1\\0&1&1 \end{vmatrix}\\ R_2\to R_1+R_2\\ &=\begin{vmatrix} -1&-1&1\\0&-1&2\\0&1&1 \end{vmatrix}\\ R_2\leftrightarrow R_3\\ &=-\begin{vmatrix} -1&-1&1\\0&1&1\\0&-1&2 \end{vmatrix}\\ R_3\to R_2+R_3 &=-\begin{vmatrix} -1&-1&1\\0&1&1\\0&0&3 \end{vmatrix}\\ &=-1(-1(3-0)-0+0)\\&=3 \end{align}\]
anonymous
  • anonymous
yeah I got that part. sorry, I forgot to tell you.. I am solving it until I get the matrix D. I am almost done :)
anonymous
  • anonymous
Finally I got it ^_^ I realized I also made a mistake in \(\sf C_{13}\), it should be 1 Anyway, thank you so much for your help and patience! :D
UnkleRhaukus
  • UnkleRhaukus
Thank you for posting your working, 'twas clear and easy to follow.

Looking for something else?

Not the answer you are looking for? Search for more explanations.