Let the eigenvalues of the third-order square matrix a be - 1, - 2, - 3, and find a *, a & # 178; + 3A + E

Let the eigenvalues of the third-order square matrix a be - 1, - 2, - 3, and find a *, a & # 178; + 3A + E

Is eigenvalue? A * eigenvalue = | a | / a eigenvalue, 6,2,3
The eigenvalues of a ^ 2 + 3A + e are - 1, - 1,1 brought in by the eigenvalues of A

Let the three eigenvalues of the third-order square matrix a be: λ 1 = 1, λ 2 = - 1, and λ 3 = 2, and find | a * + 3a-2i|

Let PAP ~ make a diagonal, i.e
PAP~=
1 0 0
0 -1 0
0 0 2
Then | a * + 3a-2 | = | P (a * + 3a-2) P ~ | = | PA * P ~ + PAP ~ - 2|
And PA * P ~ = P (| a | a ~) P ~ = | a | (PA ~ p ~) is the adjoint matrix of PAP ~
Calculate
Where a ~ represents the inverse matrix of a and P ~ is similar

Let a square matrix of order 3 have three eigenvalues λ 1, λ 2, λ 3. If | a | = 24, λ 1 = 2, λ 2 = 3, then λ 3=
Let a square matrix of order 3 have three eigenvalues λ 1, λ 2, λ 3. If | a | = 24, λ 1 = 2, λ 2 = 3, then λ 3=

|A| = λ1λ2λ3 = 2*3*λ3 = 24
So λ 3 = 4