Let d = | a - 580A + 1803a + 325 | = 0, and let a have three Let d = |a - 580A + 1803a + 325 | = 0 be the determinant of the third order, and a matrix of the third order has three eigenvalues 1 - 10 corresponding to three eigenvectors B1 = (12a - 1) TB2 = (a + 3A + 2) Tb3 = (A-2 - 1A + 1) t to determine the parameter a and find a

Let d = | a - 580A + 1803a + 325 | = 0, and let a have three Let d = |a - 580A + 1803a + 325 | = 0 be the determinant of the third order, and a matrix of the third order has three eigenvalues 1 - 10 corresponding to three eigenvectors B1 = (12a - 1) TB2 = (a + 3A + 2) Tb3 = (A-2 - 1A + 1) t to determine the parameter a and find a

The third order determinant d = | a - 580A + 1803a + 325 | = 0 can be divided into upper triangular determinant d = | a - 58,0a + 18,001 | = 0;
Then d = a * (a + 1) * 1 = 0, a = 0 or a = - 1 can be obtained;
When a = 0,
The third-order matrix a corresponds to three eigenvectors B1 = (10 - 1) t, B2 = (0 32) t, B3 = (- 2 - 1 1) T. in addition, it can be seen that the third-order matrix A has three eigenvalues 1 - 10. By diagonalizing the matrix (P ^ - 1) * a * P = V, we can make p = (B1, B2, B3), that is, P = (1 0 - 1,0 32, - 2 - 1), v = (1 00,0 - 10,0 00). Then we can obtain the inverse matrix (P ^ - 1) of P by elementary row change. Finally, we can get the inverse matrix (P ^ - 1) of P from the formula
(P ^ - 1) * a * P = V, a = P * V * (P ^ - 1) can be obtained
When a = - 1,
The third-order matrix a corresponds to three eigenvectors B1 = (1 - 2 - 1) t, B2 = (- 1, 21) t, B3 = (- 3, 10) T. in addition, it can be seen that the third-order matrix A has three eigenvalues 1 - 10, which can be diagonalized by the matrix (P ^ - 1) * a * P = V; P = (B1, B2, B3), that is, P = (1 - 2 - 1, - 1, 21, - 3 - 10), and V = (100,0 - 10,0 00); then the inverse matrix of P (P ^ - 1) can be obtained by elementary row change; finally, the formula
(P ^ - 1) * a * P = V, a = P * V * (P ^ - 1) can be obtained

Given that the adjoint matrix of a is a diagonal matrix, how can we find the matrix determinant of a,

AA*=|A|E
|AA * | = | a | ^ n, where n is the order of A
|A|^(n-1)=|A*|
Two convenient prescriptions will do

Is matrix · determinant = determinant · matrix equal?
Are the two formulas a + e (A-E) = (A-E) a + e equal? And tell me why! Is there a determinant that can be understood as a number?

A matrix is an ordered table composed of numbers, and the determinant is a number determined by certain operation rules. Your equation can be simply understood as C.A = A.C (C is a constant, a is a matrix)

Is there a determinant a of order n whose | a | = 0, but whose rank as a matrix is n
For example, the third-order example is OK

No
If the rank of the matrix is n
If and only if | a ≠ 0 is so contradictory