Question 3. Write the characteristic polynomial and find the...

Question 3. Write the characteristic polynomial and find the eigenvalues and eigenvectors of the matrix $A = \begin{bmatrix} A_1 & 0 \\ 0 & A_2 \end{bmatrix}$ where $A_1 = \begin{bmatrix} 1 & -3 & 3 \\ 3 & -5 & 3 \\ 6 & -6 & 4 \end{bmatrix}$, $A_2 = \begin{bmatrix} 4 & 0 \\ 2 & -1 \end{bmatrix}$

Transcript

00:01 In this question we have given a matrix a and we have to find out cractorcytic equation eigenvalues and eigenvector so for directorcytic equation a minus lambda i should be 0 the value of a is 6 minus 2 minus 3 1 and minus lambda into i is 0101 so it will be 6 minus lambda minus 2 minus 3 and 1 minus lambda this is a minus lambda matrix and its determinate should be 0 for characteristic equation so it will be 6 minus lambda minus 2 minus 3 1 minus lambda its determinant should be 0 if i solve this i will get 6 minus lambda 1 minus lambda plus 6 and outside it has a minus it is equal to 0 it can be solved as 6 minus 6 minus 7 lambda plus lambda square minus 6 is equal to 0 so it will be lambda square minus 7 lambda is equal to 0 or if i take lambda common it will be lambda minus 7 is equal to 0 this is its charactercetic equation and its eigenvalues are lambda is equal to 0 .7.

01:42 Now we have to find out eigenvector corresponding to lambda 1 is equal to 0 and lambda 2 is equal to 7.

01:52 We have matrix a is 6 minus 2 minus 3 and 1.

02:00 So for eigen vector we can write corresponding its eigenvector will be x1 and corresponding it will be x2 component of its will be xy and its will be u comma v so for characteristic equation we can write a x1 is equal to lambda 1 x1 and for we have to find out the value of x1 we can write a minus lambda 1 into i into matrix of x1 is equal to 0 the a minus lambda 1 because lambda 1 is 0 so it will be a so we can write 6 minus 2 minus 3 and 1 and the value of x 1 is x comma y it should be 0 if i solve this i will get 6 x minus 3 y is equal to 0 minus 2 x plus y is equal to 0 by equation 1 and 2 by equation 2 and 2 by equation 2 can say y is equal to 2x mean if x is equal to k y will be 2k so its eigenvector x1 is equal to k into the value of x is here 1 and it is 2k so it will be 2 here k k can't be 0 now the eigenvector corresponding to lambda 2 is equal to 7 we have lambda 2 is equal to 7 and a is equal to 6 minus 2 minus 3 and 1 and we have taken x2 is equal to u .m.

03:49 V for eigenvector we have to write a minus lambda 2 i into x2 is equal to 0 the value of a is 6 minus 2 minus 3 1 minus the value of lambda 2 is 7 the value of i is that and we we will multiply it x2.

04:15 The value of x2 is u...

Question

Question 3. Write the characteristic polynomial and find the eigenvalues and eigenvectors of the matrix $A = \begin{bmatrix} A_1 & 0 \\ 0 & A_2 \end{bmatrix}$ where $A_1 = \begin{bmatrix} 1 & -3 & 3 \\ 3 & -5 & 3 \\ 6 & -6 & 4 \end{bmatrix}$, $A_2 = \begin{bmatrix} 4 & 0 \\ 2 & -1 \end{bmatrix}$

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