00:01
In this question we are given that a matrix a can be diagonalized if it exists in invertible matrix of p then diagonal matrix d such that a is equal to p d and p inverse.
00:22
So computing the aging values computing agent values here are determinant will be as to 0 .2 0 .0.
00:43
You have 0 .0 .0 .0.
00:47
0 .8.
00:49
0.
00:50
0.
00:50
0.
00:50
0.
00:50
0.
00:50
0.
00:51
0.
00:51
0.
00:52
0.
00:52
0.
00:52
This will be 0 .0.
01:05
This will be equal to 0.
01:07
And therefore when we calculate our value for lambda cube plus 1 .4 lambda squared minus 0 .4 lambda plus 0 .04 will be equal to 0.
01:22
And when we solve our value for lambda will be as 1 or 1 by 5 or 1 move into this 1 by 5.
01:33
You will get 3 roots because this is a cubic polynomial.
01:36
And now our value for d will be 1 .0, 0, 1 by 5, 0, 0 and 1 by 5.
01:52
So when lambda is equal to 1, we will get here 0 .0, 0 .2, 0, then 0 .2, 0, 0 .2, 0, 0 .8, 0 .8.
02:09
And 1 minus lambda so we will put here 1 on the place of lambda multiplied by 1 0 010 0 0 0 0 multiplied by b 1 v2 and b 3 is equal to 0 so then after we will get here our value will be as minus 4 by 5 0 0 0 then 0 minus 4 by 5 0 0 0 0 0 0 0 0 0 0 0 0 minus 4 by 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.
02:42
4 by 5, 4 by 5 and 0.
02:48
This will be similar to 1 .0 -0 -1 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -1.
02:58
And for the multiplied by 1 -0 -0 -0 -0 -1, multiplied by 1 -0 -0 -0 -1, multiplied by v1, v2, v3, and now this will be, to a matrix with just 0 and 0.
03:25
So then we can say here, v is equal to 0 and t is equal to 0 and t is equal to 0 0 of t and therefore our value when lambda is equal to 1 by a pile with the similar way when we solve our value for v will be equal to as minus s minus t and s so there is this is equal to minus 1 1 0 of t into s so here is this is equal to minus 1 1 0 1 into s so here diagnose when we diagnose this value where s is equal t is equal to 1 we will get lies our will be as 0 .2 0 0 0 .20 0 0 .80 0 .8 1 such that 0 001 minus 1 minus 1 0 0 0 1 multiplied by 0 1 0 0 1 .055 and 0 .055...