Given that a third-order matrix A has an eigenvalue of 2, then A2 + 2A + 3E must have an eigenvalue of 2

Given that a third-order matrix A has an eigenvalue of 2, then A2 + 2A + 3E must have an eigenvalue of 2

Is the first square? If so:
The square of 2 plus 2 times 2 plus 3 is 11
If AX = ax, a is the eigenvalue
Then a2x = a2x,
A-1x=1/ax,
A*x=|A|/ax

Given sin2a = - 24 / 25, a ∈ (0, π / 2), then cos (a + 2 π) =?
It's 24 / 25 without negative sign, and it's cos (a + π / 2) without semicolon. I'm wrong.

There's something wrong with the title!
A ∈ (0, π / 2), then the sin values in the interval 2A ∈ (0, π) are all positive! How can it be - 24 / 25
I now default that you are sin2a = 24 / 25
sin2a=2sinacosa=24/25
sinacosa=12/25
And Sina & # 178; + cosa & # 178; = 1
a∈(0,π/2) sina > 0 cosa > 0
(sina + cosa)²= sina²+cosa² + 2sinacosa=49/25
sina + cosa=7/5
Sina = 3 / 5 cosa = 4 / 5 or Sina = 4 / 5 cosa = 3 / 5
Cos (a + 2 π) = cosa = 3 / 5 or 4 / 5

If sin2a = 2 / 3, then cos & # 178; (a + 4 / π)=
emergency

A:
cos²(a+π/4)
=[ cos(2a+π/2)+1 ]/2
=-(1/2)sin(2a)+1/2
=-(1/2)*(2/3)+1/2
=1/6
If sin2a = 2 / 3, then cos & # 178; (a + 4 / π) = 1 / 6