Let a = {010 20-1 341}, I = {100 010 001}, find a + I ^ - 1

Let a = {010 20-1 341}, I = {100 010 001}, find a + I ^ - 1

Is the problem wrong? Is it (a + I) ^ - 1? Because the inverse of the identity matrix is itself
A+I^-1={110 21-1 342}
(A+I)^-1={110 21-1 342}^-1={-621 7-2-1 511}

Given that the matrix A = (200001,01x) is similar to B = (2, none, y, none, none-1), find the values of X and y, and find an invertible matrix P satisfying p ^ - 1AP = B

200
001
01X
And
2 none none
No y no
None none - 1
The determinants and traces of similar matrices are equal
So we have - 2 = - 2Y, 2 + x = 2 + Y-1
So y = 0, x = 1
So the eigenvalues of a are 2,1, - 1
To find P is to find the eigenvector

The first row 5-10 of matrix A is 3 * 3, the second row-23 1, the third row 2-16 of matrix B is 3 * 2, the first row 21, the second row 20, the third row 3 5 of matrix A is 3 * 3
Satisfy AX = B + 2x to find x
The inverse matrix of x = (a-2e) * B knows, but it's not right,

From AX = B + 2x, (a-2e) x = B (a-2e, e) = 3 - 10100-2110102 - 14001r1 + R2, R3 + R2 get 1011010-211010050011r2 + 2r1, R3 * (1 / 5) get 1011100