Consider the following.-1 10 1A =03 0P = 0 4 020 22(a) Verify that A is diagonalizable by computing P-lAP.|-113p-1AP =22(b) Use the result of part (a) and the theorem below to find the eigenvalues of A.Similar Matrices Have the Same EigenvaluesIf A and B are similar n x n matrices, then they have the same eigenvalues.(^1, d2, 13) =|2,3, – 1

A=-110030202P=01-3040122 Obtain the inverse of P As follows. Augment the matrix P with a 3x3 identity matrix and convert the augmented matrix to the identity matrix. P=01-304012201-3|100040|010122|001≡122|001010|014001-3|100 R1↔R3, R2→14R2≡122|001010|0140003|-1140 R3→R2-R3≡122|001010|0140001|-131120 R3→13R3≡100|23-231010|0140001|-131120 R1→R1-2R2-2R3 The inverse of P is P-1=23-2310140-131120.(a) Verify that A is diagonalizable as follows. P-1AP=23-2310140-131120-11003020201-3040122=43-432034013-112001-3040122=20003000-1(b) The matrix A and P-1AP are similar. By theorem similar matrices have the same eigenvalues. Both A and P-1AP have the same eigenvalues. The characteristic equation is as follows. 20003000-1-λ100010001=02-λ0003-λ000-1-λ=02-λ3-λ-1-λ=02-λ3-λλ+1=0λ=-1,2,3