Suppose the matrix A has eigenvalues lambda_1 = -1, lambda_2 = 1, lambda_3 = 2, with corresponding eigenvectors v_1 = [0 5 3]^T, v_2 = [2 0 1]^T, v_3 = [1 -1 0]^T. If you diagonalize A as A = PDP^-1 with P = [2 2 0; p_21 p_22 2; p_31 p_32 p_33], D = [2 0 0, 0 1 0; 0 0 -1], then p_32 = 1. Please explain how p_32 = 1 is found.

To find the value of p_32, we need to determine the eigenvectors and eigenvalues of the matrix A. Once we have the eigenvalues and eigenvectors, we can use them to form the diagonalization of A, which allows us to find the matrix P.

Given that A has eigenvalues lambda_1 = -1, lambda_2 = 1, lambda_3 = 2, and the corresponding eigenvectors v_1 = [0 5 3]^T, v_2 = [2 0 1]^T, v_3 = [1 -1 0]^T, we can represent these eigenvectors as the columns of a matrix V:

V = [v_1, v_2, v_3] = [0 2 1; 5 0 -1; 3 1 0]

To diagonalize A, we need to find the inverse of V. The inverse of V is denoted as V^(-1). We can compute the inverse using the formula:

V^(-1) = (1 / det(V)) * adj(V)

where adj(V) is the adjugate (or adjoint) of V. To calculate adj(V), we need to find the matrix of cofactors of V and then take its transpose.

Now, let's calculate the inverse of V step by step:

1. Calculate the determinant of V:
det(V) = 0*(0*0 - (-1*1)) - 2*(5*0 - (-1*3)) + 1*(5*1 - 3*0)

2. Calculate the matrix of cofactors of V:
cofactor(V) = [cofactor(V_11), cofactor(V_12), cofactor(V_13); cofactor(V_21), cofactor(V_22), cofactor(V_23); cofactor(V_31), cofactor(V_32), cofactor(V_33)]

where cofactor(V_ij) is (-1)^(i+j) times the determinant of the submatrix obtained by removing the ith row and jth column from V.

cofactor(V_11) = (-1)^(1+1) * det([0 -1; 1 0]) = 1 * (0*0 - (-1*1)) = 1
cofactor(V_12) = (-1)^(1+2) * det([5 -1; 3 0]) = -1 * (5*0 - (-1*3)) = 3
cofactor(V_13) = (-1)^(1+3) * det([5 0; 3 1]) = 1 * (5*1 - 3*0) = 5
cofactor(V_21) = (-1)^(2+1) * det([2 1; 3 0]) = -1 * (2*0 - 1*3) = -3
cofactor(V_22) = (-1)^(2+2) * det([0 1; 3 1]) = 1 * (0*1 - 3*1) = -3
cofactor(V_23) = (-1)^(2+3) * det([0 2; 3 5]) = -1 * (0*5 - 3*2) = -6
cofactor(V_31) = (-1)^(3+1) * det([2 1; 0 -1]) = 1 * (2*(-1) - 1*0) = -2
cofactor(V_32) = (-1)^(3+2) * det([0 1; 0 -1]) = -1 * (0*(-1) - 1*0) = 0
cofactor(V_33) = (-1)^(3+3) * det([0 2; 0 5]) = 1 * (0*5 - 2*0) = 0

Therefore, cofactor(V) = [1 3 5; -3 -3 -6; -2 0 0]

3. Transpose the matrix of cofactors to get adj(V):
adj(V) = [1 -3 -2; 3 -3 0; 5 -6 0]

4. Finally, calculate V^(-1):
V^(-1) = (1 / det(V)) * adj(V)
= (1 / det(V)) * [1 -3 -2; 3 -3 0; 5 -6 0]

Now that we have V and V^(-1), we can find matrix P, which is the matrix formed by the eigenvectors as columns:

P = V = [0 2 1; 5 0 -1; 3 1 0]

To find the value of p_32, we need to determine the element in the second row and third column of P:

p_32 = 1

Therefore, p_32 is indeed equal to 1.