Consider the following 2 x 2 matrix -5 A = 3. For any matrix, we have a clear understanding of some exponents: A" refers to multiplying a matrix to itself n times and A-l refers to the inverse matrix. Combining this idea gives A- = (A-')", the nth power of the inverse. A natural question to ask is, can we take the nth root of a matrix? We're going to try to find A/3, i.e. a matrix that when we cube it gives A. (a) Calculate the eigenvalues and eigenvectors of A. (b) Find a matrix P and a diagonal matrix D such that A = PDP-1. (c) Find a matrix E such that E = D. Explain why this result is simple compared to finding the cube root of A.

Please solve max 20-25 minutes about chapter Linear Algebra thank u

Here matrix A 4 2- i 9. A= |2+i 0 3+i 3- i 9 7 Can you solve for A· A"(Transpose matrix)?

Show that if A is square matrix that satisfies the equation 2 A² - A - I = O, then its inverse is A-¹ = 2 A-I. The equation 24² - A - I = O implies that ---Select--- ✓. It follows that ---Select--- ✓. Notice the last equation means that ---Select--- multiplied with A is the identity, which is what we wanted to prove.