Please Answer Fast

(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca

Yuri [45]

Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

$\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]$

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation $det(A-\lambdaI)=0$ where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

$\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6$

So the characteristic polynomial is $\lambda^2-5\lambda+6=0$ .

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

$\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0$

So $\lambda=3, \lambda=2$

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If $\lambda=3$ we get the following matrix

$\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right]$ .

Since both rows are equal, we have the equation

$x-2y=0$ . Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case $\lambda=2$ , using the same process, we get the vector (1,1).

d) By definition, to diagonalize a matrix A is to find a diagonal matrix D and a matrix P such that $A=PDP^{-1}$ . We can construct matrix D and P by choosing the eigenvalues as the diagonal of matrix D. So, if we pick the eigen value 3 in the first column of D, we must put the correspondent eigenvector (2,1) in the first column of P. In this case, the matrices that we get are

$P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]$

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

$P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]$

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

4 0

3 years ago

Please helpppppp fastttt

ValentinkaMS [17]

Answer:

I am pretty sure it is A. 85°

6 0

3 years ago

B= 8p+12 HELP ME FAST PLZ!!!!!!

MatroZZZ [7]

Simplify both sides of the equation and get the answer: p= -3/2+b/8

Hope this helps!
~Brooke❤️

8 0

3 years ago

A college administration has conducted a study of 190randomly selected students to determine the relationship between satisfacti

almond37 [142]

Answer:

The probability that a student selected at random is on academic probation and is not satisfied with advisement is 0.1789

Step-by-step explanation:

Consider the provided information.

Out of 68 students on academic probation, 34 are not satisfied with advisement; That means 68-34=34 are satisfied with advisement.

If 68 students are in academic then the non academic students probation will be: 190-68=122

Only 25of the students not on academic probation are dissatisfied with advisement.

That means 122-25=97 students are satisfied.

According to the provided information draw the table as shown.

Satisfied Not satisfied Total

Academic probation 34 34 68

Non academic probation 97 25 122

Total 131 59 190

Now we need to determine the probability that a student selected at random is on academic probation and is not satisfied with advisement.

On academic probation and is not satisfied with advisement = 34

The total number of cases is 200.

$Probability=\frac{\text{Favourable cases}}{\text{Total number of cases}}$

Substitute the respective values:

$Probability=\frac{34}{190}\approx 0.1789$

Hence, the probability that a student selected at random is on academic probation and is not satisfied with advisement is 0.1789

4 0

3 years ago

1/3 + 1/3= 5/10-2/10= = 4/5-2/5= 11+1= =

Mrac [35]

Answer:

Step-by-step explanation:

1/3 + 1/3= 2/3

5/10-2/10=3/10

4/5-2/5=2/5

11+1=12

5 0

2 years ago