Homework SourseAnswer both questions ONLY Consider the following A 9 40 2
Answer both questions ONLY.
Consider the following. A = [9 -40 2 -9], P = [-4 -5 1 -1] (a) Verify that A is diagonalizable by computing P^-1 AP. P^-1 AP = [] (b) Use the result of part (a) and the theorem below to find the eigenvalues of A. Similar Matrices Have the Same Eigenvalues If A and B are similar n times n matrices, then they have the same eigenvalues. (lambda_1, lambda_2) = (1, -1) For the matrix A, find (if possible) a nonsingular matrix P such that P^-1 AP is diagonal. (If not possible, enter IMPOSSIBLE.) A = [12 -3 -4 1] P = [] Verify that P^-1 AP is a diagonal matrix with the eigenvalues on the main diagonal. P^-1 AP = []Solution
1.(a) P-1=
1
-5
-1
4
Therefore, P-1AP = D(say) =
-1
0
0
1
Since P-1AP = D is a diagonal matrix, and since P is an invertible matrix, hence A is diagonalizable.
(b) Since A and D are similar matrices, since similar matrices have the same eigenvalues, and since the eigenvalues of D are the entries on its leading diagonal, hence the eigenvalues of A are -1 and 1.
2. The characteristic equation of A is det(A- I2) = 0 or, 2-13 = 0 or, (-13) = 0. Thus, the eigenvalues of A are 1 =13 and 2 = 0. Further, the eigenvector of A correspondingto the eigenvalue 13 is the solution to the equation (A-13I2)X = 0. To solve this equation, we will reduce A-13I2 to its RREF as under:
Multiply the 1st row by -1
Add 4 times the 1st row to the 2nd row
Then the RREF of A-13I2 is
1
3
0
0
Thus,if X = (x,y)T, then the equation (A-13I2)X = 0 is equivalent to x+3y = 0 or, x = -3y. Then X = (-3y,y)T = y(-3,1)T. Hence, the eigenvector of A corresponding to the eigenvalue 13 is v1 = (-3,1)T.Similarly, the eigenvector of A correspondingto the eigenvalue 0 is v2 = (1,4)T. Since the two eigenvectors of A are distinct and linearly independent, hence A is diagonalizable. Now, we know that if P-1AP = D(say), a diagonal matrix, then the eigenvalues of A are the terms on the leading diagonal of D, and also, the eigenvectors of A are the columns of P in the same order. Then P =
-3
1
1
4
and P-1 =
-4/13
1/13
1/13
3/13
Also, P-1AP=
13
0
0
0
which is a diagonal matrix with the eigenvalues of A as the terms on its leading diagonal .
| 1 | -5 |
| -1 | 4 |
![Answer both questions ONLY. Consider the following. A = [9 -40 2 -9], P = [-4 -5 1 -1] (a) Verify that A is diagonalizable by computing P^-1 AP. P^-1 AP = [] (b](/WebImages/42/answer-both-questions-only-consider-the-following-a-9-40-2-1130198-1761603600-0.webp "Answer both questions ONLY. Consider the following. A = [9 -40 2 -9], P = [-4 -5 1 -1] (a) Verify that A is diagonalizable by computing P^-1 AP. P^-1 AP = [] (b")
![Answer both questions ONLY. Consider the following. A = [9 -40 2 -9], P = [-4 -5 1 -1] (a) Verify that A is diagonalizable by computing P^-1 AP. P^-1 AP = [] (b](/WebImages/42/answer-both-questions-only-consider-the-following-a-9-40-2-1130198-1761603600-1.webp "Answer both questions ONLY. Consider the following. A = [9 -40 2 -9], P = [-4 -5 1 -1] (a) Verify that A is diagonalizable by computing P^-1 AP. P^-1 AP = [] (b")
