Question

(23). (15 Marks). Suppose A has eigenvalues 11 = 3, 12 = 1, 13 = 0) with corresponding eigenvectors 0 0 2 = [1] 12 8) 23 = (a) (5 points). How do you know that the third column of A contains all zeros? (b) (10 points). Find the matrix A.

Answer 1

Given :- A is a matrix and it's eigen values are A = 8. do 21 ts=0 with cooresponding eigen vectors are Ad; x = 0 0 1 0 x = O O 0 X32 . : + 1 0 O to find matrix - 1 Here is a formula A - XXX A -1 where x is a matrix whose are the eigene vectors of A columns Now, A = 1 0 0 3 0 L 1 0 0 0 - 1 0 s is diagonal matrix whose diagonal entries are the eigen values 1 1 0 0 -1 1 A. - O 0 3 O 3 0 1 0 X = o O 1 Ooo 3 1 0 - 1 0 1 O - 3 1 0 O -t Now X- Adj X А 3 0 0 1X1 1 0 2 2 1 0 Ans 1x1 = 1 (1-0) cofactor x Minor) of an element : МКИ Adj X = (cofactors of each element) 1 -1 o one O minor = it is the determinant of the square matric by deleting row and one colurm from some larger square matrix * Transpose - To switch rows with the column

☆ a 3rd eigen value of the matrix is zero henca ne the third column of A will contain call zelos. can say 3 O * b 2 2 0

Please feel free to ask doubts.