berljivx8

2022-01-18

The eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of the system.

$x}_{1}^{{}^{\prime}}=2{x}_{1}+{x}_{2}-{x}_{3$

$x}_{2}^{{}^{\prime}}=-4{x}_{1}-3{x}_{2}-{x}_{3$

$x}_{3}^{{}^{\prime}}=4{x}_{1}+4{x}_{2}+2{x}_{3$

How to find the eigenvalues by using the fact that $|A-\lambda I|=0$, how do it by inspection?

Imagine you have the matrix, $$A=(\begin{array}{ccc}2& -1& -1\\ -1& 2& -1\\ -1& -1& 2\end{array})$$

By noticing (or inspecting) that each row sums up to the same value, which is 0, we can easily see that [1, 1, 1] is an eigenvector with the associated eigenvalue of 0.

Elois Puryear

Beginner2022-01-19Added 30 answers

Let A - the matrix of coefficients, then $(A\cdot x)}_{k}=\lambda {x}_{k$.

Add the third and the second row.

So ${x}_{2}+{x}_{3}=\lambda ({x}_{2}+{x}_{3})$, thus $\lambda =1\text{}\text{or}\text{}{x}_{2}+{x}_{3}=0$. Since the trace of A equals 1, the sum of other two eigenvalues is 0.

For other eigenvalues, let $x}_{3}=-{x}_{2$, giving $2{x}_{1}+2{x}_{2}=\lambda {x}_{1}\text{}\text{and}\text{}-4{x}_{1}-2{x}_{2}=\lambda {x}_{2}$. The characteristic polynomial is easily found as ${\lambda}^{2}+4=0$.

Terry Ray

Beginner2022-01-20Added 50 answers

The first eigenvalue may be inferred from the fact that A-I has two equal rows and two equal columns.

The additional two eigenvalues are then provided to you by the trace and determinant.

alenahelenash

Expert2022-01-24Added 556 answers

The determinant of a matrix is the sum of the eigenvalues., so in this case ${\lambda}_{1}{\lambda}_{2}{\lambda}_{3}=4$. The sum of the diagonals is the sum of the eigenvalues so ${\lambda}_{1}+{\lambda}_{2}+{\lambda}_{3}=2-3+2=1$. $[-110{]}^{T}$ is an eigenvector with eigenvalue 1, then ${\lambda}_{1}=1\text{}\text{so}\text{}{\lambda}_{2}{\lambda}_{3}=4\text{}\text{and}\text{}{\lambda}_{2}+{\lambda}_{3}=0$. Thus ${\lambda}_{2},{\lambda}_{3}=\pm 2\sqrt{-1}$.In general there is no easy way to spot eigenvalues of $3\times 3$ matrices.With $2\times 2$ matrices you can use the determinant and trace. The fact that these are the only non-trivial coefficients of the characteristic polynomial is the primary cause of why this works (and thus determine the eigenvalues). Other characteristic polynomial coefficients aren't as simple to use or recall when working with larger matrices, which necessitates the employment of more invariants (i.e., coefficients of the characteristic polynomial).

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The Laplace transform of

is$u(t-2)$

(a)$\frac{1}{s}+2$

(b)$\frac{1}{s}-2$

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A) $\frac{9}{5}$ degree on fahrenheit scale

B) $\frac{5}{9}$ degree on fahrenheit scale

C) 1 degree on fahrenheit scale

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What is the Laplace transform of

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$\frac{-2s+1}{({s}^{2}+2s+5)}$

I Completed the square on the bottom but what do you do now?

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Find the inverse Laplace transform of $\frac{{s}^{2}-4s-4}{{s}^{4}+8{s}^{2}+16}$

inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$