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Point) Suppose that the matrix A has the following eigenvalues and eigenvectors: A1 = l with 31andAz = 4 with 8zWrite the solution to the linear system 7Av in the f...

Question

Point) Suppose that the matrix A has the following eigenvalues and eigenvectors: A1 = l with 31andAz = 4 with 8zWrite the solution to the linear system 7Av in the following forms.In eigenvalueleigenvector form:[wo] =01+ CAs two equations: (write "61" and "62" for Ciland Cy(t)Note: if you are feeling adventurous you could use other eigenvectors Iike 4 0132

point) Suppose that the matrix A has the following eigenvalues and eigenvectors: A1 = l with 31 and Az = 4 with 8z Write the solution to the linear system 7 Av in the following forms. In eigenvalueleigenvector form: [wo] =01 + C As two equations: (write "61" and "62" for Ciland C y(t) Note: if you are feeling adventurous you could use other eigenvectors Iike 4 01 32



Answers

For each of the following matrices, find all cigcnvalues and corresponding linearly independent eigenvectors: (a) $A=\left[\begin{array}{ll}2 & -3 \\ 2 & -5\end{array}\right]$ (b) $\quad B=\left[\begin{array}{rr}2 & 4 \\ -1 & 6\end{array}\right],(\mathrm{c}) \quad C=\left[\begin{array}{rr}1 & -4 \\ 3 & -7\end{array}\right]$

This problem gives them the tricks and asks us to solve for its Eigen values and Eigen vectors. We can do this by finding characteristic polynomial, which is found biting the determinant of the vector a minus Lambda Times The identity matrix I, which will come out to be seven minus, slammed up negative one for three minus slammed up. Taking that determine we get seven minus lambda times three minus lambda, minus four times negative one which will solve out to be a lander squared minus 10 Lambda plus 25 which is equivalent to Lambda minus five squared meaning that are Eigen. Value is going to be five with a knowledgeable algebraic multiplicity of To That means that this adding value can possibly have to Eigen vectors attached to it. So in order to solve for the Eiken vectors, we need to find a minus land I times X such that this is equal to zero. So for the five I'd in space will plug in Lambda equals five That will give us seven months. Five mega born for three months. I've Prince the Egg Director X is equal to 24 negative one negative, too Times a B is to 00 now. In this case, there's only one linearly independent vector that will be a blue solve this system and that is going to be our one Eigen value. And it is negative two one. So our Eigen values going to be five with multiplicity too with this single I director of negative 21


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