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-2 -2 -2 A= -2 0-4 -2 -4 0 Start from forming a new matrix by subtracting from the diagonal entries of the given matrix: 1-2 -2 -2 -2 -1 -4 -2 -4 -1 Find the determinant of the obtained matrix: -2-2-2 -2 -2 - 2 -1 -4 -2 -4 -1 = 12 (-1-2) + 241 = This is a characteristic polynomial. Solve the equation 1(-1-2) + 241 = 0.

The roots are: l1 = 4 12 = -6 13 = 0 These are the eigenvalues. Next, find the eigenvectors. a.

L=4 -1 -2 -2 -2 -2 -1 -4 -2 -4 -1 6 -2 -2. -2 -4 -4 -2 -4 -4 Perform row operations to obtain the rref of the matrix:

-6 -2 -27 -2 -4 -4 -2 -4 -4 1 0 0 0 1 1 0 0 0 ar) Now, solve the matrix equation ſi 0 0] V1 0 0 1 1 V2 = 0 0 0 0 V3 0 If we take V3 = t, then vi = : 0, V2 , V3 = t. -t, = Therefore, 0 0 pº V = -t = -1 t t 1

b. =-6 X-2 -2 -2 -2 - -4 -2 -4 -2 -2 4 = -2 6 -4 -2 -4 6 Perform row operations to obtain the rref of the matrix: 4 -2 -2 -2 6 -4 6 -2 -4 1 0 -1 v 0 1 - 1 0 0 0

Now, solve the matrix equation [1 0 -1 V1 0 1 -1 V2 0 0 0 V3 - If we take v3 t, then vi =t, V2 = t, v3 = t. t 1 Therefore, v= t = 1t 1 C.

D=0 -X-2 -2 -2] -2 - -4 -2 -4 - -2 -2 -2 = -2 0 -4 -2 -4 0

Perform row operations to obtain the rref of the matrix: -2 -2 -2 -2 0 -4 --2 -4 0 1 0 2 0 1 - 1 으 0 0 o Now, solve the matrix equation 1 0 2 V1 0 0 1 - 1 = V2 0 0 0 0 V3 If we take V3 = t, then vi = :-2t, V2 =t, V3 = t. Therefore,

- 2t 2 -0-0 1 t 1 ANSWER 0 Eigenvalue: 4, eigenvector: -1 1 1 Eigenvalue: -6, eigenvector: 1

2 Eigenvalue: 0, eigenvector: 1 1

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