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91Ó°ÊÓ

For which values of constants a, b, and c are the matrices in Exercises 40 through 50 diagonalizable?

A=[11a1]

Short Answer

Expert verified

The matrix .a≠0 So, is diagonalizable.

Step by step solution

01

Characteristic equation 

Consider an n ×n matrix A and a scalar λ. Then λ is an eigenvalue5 of A.

det(A-λ±ô)=0

This is called the characteristic equation (or the secular equation) of matrix A.

02

Solution of the problem

We solve:

detA-λ±ô=01-λ1a1-λ=01-λ2-a=0λ2-2λ+1-a=0λ1,2=2±4-41-a2=1±a

For a=0, we only have one eigenvalue, so A is not diagonalizable. On the other hand, for a≠0, we have two distinct eigenvalues, so A is diagonalizable.

Here, the final result is a≠0.

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