00:03
So, the given matrix here is the given matrix a is equal to 3 1 1 1 0 2 and 1 2 0.
00:24
Now, x is equal to x 1 x 2 x 3 be the corresponding corresponding eigenvector eigenvector.
00:58
What we have is that a x is equal to lambda into x.
01:12
So, implies that a minus lambda into i into x is equal to 0.
01:28
Here let us consider first lambda is equal to 1.
01:33
So, what we are getting here then 3 minus lambda 1 1 1 0 minus lambda 2 1 2 and 0 minus lambda into x 1 x 2 x 3 is coming as 0 0 0.
02:06
So, now what we have is that to if we solve it for lambda is equal to 1 then it is coming as 2 1 1 1 minus 1 2 1 2 minus 1 into x 1 x 2 x 3 is coming as is equal to 0 0 0.
02:44
So, from there if we solve the equation then we can easily find that x 1 x 2 x 3 is coming as 3 minus 3 minus 3.
02:59
So, for lambda is equal to 1 lambda is equal to 1 the eigenvalue eigenvector x is equal to 3 minus 3 and minus 3.
03:17
Now, if we similarly find the eigenvector for lambda is equal to 4 then we will get the eigenvector as x is equal to 6 3 3 and for lambda is equal to minus 2 we will get the eigenvector x is equal to 0 minus 9 9.
03:56
So, now if we then calculate c or the orthogonal matrix then the c or orthogonal matrix will come as 3 by root over 27 minus 3 by root over 27 minus 3 by root over 27 6 by root over 54 3 by root over 54 3 by root over 54 0 by root over 162 minus 9 by root over 162 and 9 by root over 162.
04:52
So, this is our orthogonal matrix.
04:56
Now, is the answer for question a now if we go to question b then the d is given as 1 0 0 0 4 0 0 minus 2 we are asked to find c transpose a c is equal to d.
05:28
So, if we first find c transpose into a then that will be is equal to 3 by root over 27 6 by root over 54 0 and minus 3 by root over 27 12 by root over 54 18 by root over 162 and minus 3 by root over 27 12 by root over 54 and minus 18 by root over 162...