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3 years ago

I need the answer of this with steps and solutions fast plz

Mathematics

1 answer:

Natali [406]3 years ago

6 0

Answer:

$1 \dfrac{13}{15}$

Step-by-step explanation:

$2 \dfrac{1}{5} - \dfrac{1}{3} =$

$= \dfrac{11}{5} - \dfrac{1}{3}$

$= \dfrac{33}{15} - \dfrac{5}{15}$

$= \dfrac{28}{15}$

$= 1 \dfrac{13}{15}$

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Step-by-step explanation:

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

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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,

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

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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.

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

Since both rows are equal, we have the equation

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

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

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

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3 years ago

I need the answer of this with steps and solutions fast plz​

I need the answer of this with steps and solutions fast plz