Let A be the matrix A = [ 0 1 0; -4 4 0; -2 0 1 ] (a) Find...

Let A be the matrix A = [ 0 1 0; -4 4 0; -2 0 1 ] (a) Find the eigenvalues and eigenvectors of A. (b) Find the algebraic multiplicity a_?, and the geometric multiplicity, g_? of each eigenvalue. (c) For one of the eigenvalues you should have g_? < a_?. (If not, redo the preceding parts!) Find a generalized eigenvector for this eigenvalue. (d) Verify that the eigenvectors and generalized eigenvectors are all linearly independent. (e) Find a fundamental set of solutions for x' = Ax.

Transcript

00:01 Hello students, in this problem given the matrix a is equal to 0 1 0 minus 4 4 0 minus 2 0 1 we have to find eigenvalue and eigenvector.

00:16 First we have to find the characteristic equation of the given matrix.

00:21 Now the characteristic equation for 3 cross 3 matrix is lambda cube minus s1 lambda square plus s2 lambda minus s3 is equal to 0.

00:33 Here s1 represents sum of the diagonal elements which is equal to 0 plus 4 plus 1 which is equal to 5 and s2 is equal to sum of the minus of the elements which is equal to 4 0 0 1 plus 0 0 minus 2 1 plus 0 1 minus 4 4.

01:01 Simplifying we get s2 is equal to 8 and s3 is equal to its determinant value and which is equal to determinant 0 1 0 minus 4 4 0 minus 2 0 1 this is equal to 4.

01:22 Therefore the characteristic equation is lambda cube minus 5 lambda square plus 8 lambda minus 4 is equal to 0.

01:34 Now solving we get lambda is equal to 1 2 2.

01:41 Therefore the eigenvalues are lambda is equal to 1 2 2.

01:51 Next we have to find eigenvector.

01:55 Now eigenvector can be calculated by the formula a minus lambda i into x is equal to 0 which implies minus lambda x1 plus x2 is equal to 0.

02:07 Consider this as equation number 1 and minus 4 x1 plus 4 minus lambda x2 is equal to 0.

02:15 Consider this as equation number 2 and minus 2 x1 plus 1 minus lambda x3 is equal to 0.

02:23 Take this equation number 3...

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