Given matrix A = [1 0 0 -1 1 1 -1 -2 4] and matrix B = [1 0 0 0 2 0 0 -0 3]. Usi

Question

Given matrix A = [1 0 0 -1 1 1 -1 -2 4] and matrix B = [1 0 0 0 2 0 0 -0 3]. Using the fact that matrix A is similar to matrix B, determine the eigenvalues for matrix A. Given lambda_1 = 2, lambda_2 = -2, lambda_3 = 3 are the eigenvalues for matrix A where A = [1 -1 -1 1 3 1 -3 1 -1]. In addition, given the eigenvectors corresponding to the above eigenvalues respectively x_1 = (-1, 0, 1), x_1 = (1 -1, 4), x_3 = (-1, 1, 1), determine the matrix P such that P^-1AP = D where D = [2 0 0 0 -2 0 0 0 3] Matrix P: ___ Determine matrix A is diagonalizable given matrix A = [1 -2 1 0 0 1 0 0 -3]. Justify your answer. Matrix A is diagonalizable since ___ Matrix A is not diagonalizable since ___

Explanation / Answer

-1

1

-1

0

-1

1

1

4

1

3. The characteristic equation of A is det(A- I3) = 0 or, (-1)(+3) = 0. Thus, the eigenvalues of A are 1 =-3, 2 = 1 and 3 =0. Further , the eigenvectors of A corresponding to the eigenvalue are the solutions to the equation (A-I3)X = 0. The eigenvectors of A corresponding to the above eigenvalue are v = (-5,-4,12)T,v =(1,0,0)T and v = (2,1,0)T . Now, let B =

-5

1

2

-4

0

1

12

0

0

The RREF of B is I3 . Thus A has 3 distinct linearly independent eigenvectors. Therefore,A is diagonalizable.

-1

1

-1

0

-1

1

1

4

1

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