On "let square matrix a satisfy a ^ 2-a-2e = 0, prove that a and a + 2E are invertible, and find the inverse matrix of a and the inverse matrix of (a + 2e)" I can't see a big mistake when a teacher or a classmate does this, A ^ 2-e = a + e, left square difference formula, The results are as follows (A+E)(A-E)=A+E, The two sides are multiplied by the inverse of (a + e) to get a = 2E, So the inverse of (a + e) is equal to E / 3 There is another way From the known equation A(A-E) = 2E So a [(1 / 2) (A-E)] = e So a is reversible and a ^ - 1 = (1 / 2) (A-E)

On "let square matrix a satisfy a ^ 2-a-2e = 0, prove that a and a + 2E are invertible, and find the inverse matrix of a and the inverse matrix of (a + 2e)" I can't see a big mistake when a teacher or a classmate does this, A ^ 2-e = a + e, left square difference formula, The results are as follows (A+E)(A-E)=A+E, The two sides are multiplied by the inverse of (a + e) to get a = 2E, So the inverse of (a + e) is equal to E / 3 There is another way From the known equation A(A-E) = 2E So a [(1 / 2) (A-E)] = e So a is reversible and a ^ - 1 = (1 / 2) (A-E)

Knowledge point: if a and B are matrices of the same order, and ab = e, then a and B are all invertible, and a ^ - 1 = B, B ^ - 1 = A. since a [(1 / 2) (A-E)] = e, a is invertible, and a ^ - 1 = (1 / 2) (A-E)

Let a square matrix of order n satisfy a ^ 2-2a-3e = 0, then what is the inverse matrix of a_ #^^^^^^^
Let a satisfy a ^ 2-2a-3e = 0, then what is the inverse of a

Because a ^ 2-2a-3e = 0
So a (a-2e) = 3E
So a ^ - 1 = (1 / 3) (a-2e)

Given that the square matrix satisfies a ^ 2-2a + 2E = 0, it is proved that a and a-3e are invertible, and the inverse matrices of a and a-3e are obtained

Because a ^ 2-2a + 2E = 0, so a (a-2e) = - 2E, so a is reversible, and a ^ - 1 = - 1 / 2 (a-2e). Then by a ^ 2-2a + 2E = 0A (a-3e) + (a-3e) + 5E = 0, so (a + e) (a-3e) = - 5E, so a-3e is reversible, and (a-3e) ^ - 1 = - 1 / 5 (a + e)