If the eigenvalue of the third-order square matrix A is known to be 1 (double), - 1, then the determinant of a & # 178; + 3A + 2E =?

If the eigenvalue of the third-order square matrix A is known to be 1 (double), - 1, then the determinant of a & # 178; + 3A + 2E =?

If the eigenvalues of the third order matrix A are 0,1,2, then the determinant of B = a ^ 2-3a + e | B | =? A 0 B 1 C - 1 D 2

Because the eigenvalues of a are 0,1,2
So the eigenvalues of B = a ^ 2-3a + e are 1, - 1, - 1
So | B | = 1 * (- 1) * (- 1) = 1
(B) Right

The eigenvalue of the third-order square matrix A is - 1,1,2 (1) a ^ 3, the eigenvalue of a ^ - 1 (2) f (a) = the eigenvalue of a ^ 2-A + e (3) calculate the determinant La ^ 2-A + E

The eigenvalues of a ^ 3 are: (- 1) ^ 3,1 ^ 3,2 ^ 3, that is - 1,1,8
The eigenvalues of a ^ (- 1) are: - 1,1,1 / 2 (the reciprocal of the eigenvalues of a)
The eigenvalues of F (a) are: F (- 1), f (1), f (2), namely 3,1,3
lA^2-A+E| = 3*1*3 = 9.