Prove or give a counterexample:
if a is an n x n matrix with n distinct (real)
eigenvalues,then a is diagonalizable.

See proof below.

Step-by-step explanation:

True

For this case we need to use the following theorem "If are eigenvectors of an nxn matrix A and the associated eigenvalues are distinct, then are linearly independent". Now we can proof the statement like this:

Proof

Let A a nxn matrix and we can assume that A has n distinct real eingenvalues let's say

From definition of eigenvector for each one needs to have associated an eigenvector for

And using the theorem from before , the n eigenvectors are linearly independent since the are distinct so then we ensure that A is diagonalizable.

The answer to your question is: 30 in
Step-by-step explanation:
Data
1 piece
Width = 4 in
Length = 6 in
2 piece
Width = 6 inches
Proportions
4 : 6 :: 6 : x
x = (6 x 6) / 4
x = 9
Perimeter = 2(base) + 2(height)
= 2(9) + 2(6)
= 18 + 12
= 30 in