According to the different values of the parameters, the rank of the first row 1 A - 1 2, the second row 2 - 1 a 5, and the third row 1 10 - 6 1 is obtained

According to the different values of the parameters, the rank of the first row 1 A - 1 2, the second row 2 - 1 a 5, and the third row 1 10 - 6 1 is obtained

So when a = 3, R (a) = 2, when a ≠ 3, R (a) = 3

Let a = the first row 3,2, - 2, the second row 0, - 1,0, the third row 4,2, - 3, find the invertible square matrix P so that P ^ - 1AP is a diagonal matrix
That's not right

|A - λ e | = 3 - λ 2 - 20 - 1 - λ 04 2 - 3 - λ = (- 1 - λ) [(3 - λ) (- 3 - λ) + 8] = - (λ - 1) (λ + 1) ^ 2. The eigenvalues of a are 1, - 1, - 1 (A-E) x = 0, the basic solution system is: A1 = (1,0,1) '(a + e) x = 0, the basic solution system is: A2 = (- 1,2,0)', A3 = (1,0,2) 'let P = (A1, a

Let the first row of matrix a be negative 4, negative 10,0, the second row 1,3,0, the third row 3,6,1. Find the invertible matrix P so that p-1ap can be diagonalized. Do me a favor

|A - λ e | = - 4 - λ - 10 01 3 - λ 03 6 1 - λ = (1 - λ) [(- 4 - λ) (3 - λ) + 10] = (1 - λ) (λ ^ 2 + λ - 2) = (1 - λ) (λ + 2) (λ - 1), so the eigenvalue of a is 1,1, the basic solution system of - 2 (A-E) x = 0 is A1 = (- 2,1,0) ^ t, A2 = (0,0,1) ^ t (a + 2e) x = 0, and the basic solution system is A3 = (- 5,1