Let a be a square matrix of order n and a be invertible. It is proved that det (adja) = (deta) is (n-1) power

Let a be a square matrix of order n and a be invertible. It is proved that det (adja) = (deta) is (n-1) power

There is an important relation: AA * = det (a) e, a * is the adjoint matrix of A
DET (a) det (a *) = det (a) ^ ndet (E) = det (a) ^ n, because det (a) is not equal to 0,
So there is det (a *) = (DET (a)) ^ (n-1)
By the way, it also holds when det (a) = 0

Let a be a square matrix of order 3 and | a | = 3, then | 3a-l|=________ Note: 3a-1 is actually the form of negative power of 3a, that is, invertible matrix

|The inverse of a | = 1 / | a|
|3 (inverse of a) | = 3 ^ 3 | = 27 / | = 27 / 3 = 9

Let the square matrix a satisfy the equation a ^ 2-3a-10e = 0, and prove that A-4E is invertible

A-4E is decomposed from a ^ 2-3a-10e, a ^ 2-3a-10e = (A-4E) (a + e) - 6e = 0, that is, (A-4E) (a + e) = 6e,
That is, (A-4E) (a + e) / 6 = E. according to the definition of the inverse of matrix, A-4E is invertible and its inverse is (a + e) / 6

A ^ 2-3a + 4E = 0, prove that a + e is invertible and find its inverse matrix
I'm in a hurry

Because a ^ 2-3a + 4E = (a + e) (A-4E) + 8e = 0
So (a + e) (A-4E) = - 8e
So (a + e) [(- 1 / 8) (A-4E)] = E
Because | a + e | A-4E | = | - 8e ≠ 0
So | a + e ≠ 0
So a + e is reversible, and (a + e) ^ (- 1) = (- 1 / 8) (A-4E)

A is a square matrix of order n, a ^ 2 + A-4E = O, prove that a and A-E are invertible matrices, and write a ^ - 1 and (A-E) ^ - 1

A^2+A-4E=O
A^2+A=4E
A(A+E)=4E
A(A+E)/4=E
Therefore, a is reversible and a ^ - 1 = (a + e) / 4
A^2+A-4E=O
A^2+A-2E＝2E
(A-E)(A+2E)=2E
(A-E)(A+2E)/2=E
Therefore, A-E is reversible and (A-E) ^ - 1 = (a + 2e)/

Let a be a square matrix of order n, a is not equal to 0, if the power of a2 - 3A = 0. It is proved that a-3e is irreversible

A ^ 2 - 3A = 0
A (a-3e) = 0
Because a ≠ 0,
So a-3e = 0,
0 matrix is irreversible, so a-3e is irreversible!

Let n-order square matrix a satisfy a2-3a-3e = 0, prove that A-E is invertible, and find (A-E) - 1

Certificate:
From a2-3a-3e = 0, we get that
(A-E)(A-2E)=5E
(A-E)[(A-2E)/5]=E
By definition, we get
(A-E) is reversible and (A-E) - 1 = (a-2e) / 5

If a is a third order square matrix, and | a + 2e | = 0, | 2A + e | = 0, | 3a-4e | = 0, then | a | =? Where e is the unit matrix

It shows that the three eigenvalues of a are - 2, - 1 / 2 and 4 / 3 respectively, so | a | = the multiplication of the three eigenvalues

Given that the square matrix A of order n satisfies a ^ 2-3a + e = 0, what is the inverse matrix of a?

A^2-3A+E=0
3A-A^2=E
(3E-A)A==E
A^(-1)=3E-A

A linear algebra problem, if a is a square matrix of third order, and | a + 2e | = 0, | 2A + e | = 0, | 3a-4e | = 0, then | a|=

Because | a + 2e | = 0, | 2A + e | = 0, | 3a-4e | = 0
So - 2, - 1 / 2, 4 / 3 are the eigenvalues of A
And a is a square matrix of order 3
So - 2, - 1 / 2, 4 / 3 are all eigenvalues of A
So | a | = (- 2) * (- 1 / 2) * (4 / 3) = 4 / 3