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Given that the eigenvalues of matrix A of order 3 are 1,2,3, find | a * + A ^ 2 + 3E| Such as the title
AA*=|A|E
A*=|A|A^(-1)=6A^(-1)
therefore
A*+A^2+3E=6A^(-1)+A^2+3E
The eigenvalues are: 6 + 1 + 3 = 10; 6 △ 2 + 4 + 3 = 10; 6 △ 3 + 9 + 3 = 14
Namely
|A*+A^2+3E|=10×10×14=1400
Given that the eigenvalues of a matrix of order 3 are 1, 2 and 3, then | a * a-2a + 3E | =?
A * a in the title is a ^ 2
Let f (x) = x ^ 2-2 * x + 3
Then f (1) = 2, f (2) = 3, f (3) = 6
Because the eigenvalues of a are 1,2,3
So the eigenvalues of a ^ 2-2a + 3E are 2,3,6
So | a ^ 2-2a + 3E | = 2 * 3 * 6 = 36
Teacher, if a and B are n-order matrices, proving that AB and Ba have the same eigenvalues can be transformed into proving
|Ab - # e | = | ba - # e |?
If # denotes λ, that's OK
If the characteristic polynomials are the same, the eigenvalues are the same
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