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Point) For which value of k does the matrixA = ~4 1|have one real eigenvalue of multiplicity 2?...

Question

Point) For which value of k does the matrixA = ~4 1|have one real eigenvalue of multiplicity 2?

point) For which value of k does the matrix A = ~4 1| have one real eigenvalue of multiplicity 2?



Answers

find the values of $k$ for which the matrix $A$ is invertible. $$A=\left[\begin{array}{cc} k-3 & -2 \\ -2 & k-2 \end{array}\right]$$

All right. We find, we want to find the values of K. For which the matrix A. Is convertible. Where A has rose K two and two. K. A is simply a two by two matrix given here in the upper right of the document. So in order to answer this question, we need to be able to take the determinant of K. Or rather disturbing enough Matrix A. And identify from that determinant where A is not convertible to do so we make a note of the fact that the determinant of a matrix A. B C. D. Is A. D minus C. B. And that matrix matrix A is inevitable if and only if the determinant does not equal zero. So we need to solve for the determinant and identify for what values of cable determinant does not equal zero. So to start off, the determinant of A is simply keychains K minus four or k squared minus four. We solve that for zero. Sochi experiments for equal zero, which reduces the case critical for or K equals plus or minus two. When the determinant zero, so monkey equals minus two, the determinant zero Ak is not inevitable, so that means that A is convertible whenever K does not equal plus or minus two.

Hello there. So for this exercise, we need to find the values off K such that a is not convertible or, in other words, a become singular. Okay, So what are the conditions for the matrix to become singular? Well, we need to check that The determinant. Actually, we need to ensure to the determinant is equal to zero. Okay, so the determine off this matrix is equals to four times K plus six, and this should be equals to zero on. We're going to solve this equation for K. So K is equal to minus six, divided four. But this is just gate. It's equals two minus three halfs. So if K is equal to three house, then the matrix is singular, and there is no inverse for these matrix.

We want to find the values. Okay, for which matrix A is convertible, matrix A is a three by three. Matrix throws 1 to 0 K. One K. M 0 to 1. As shown in the upper right. To solve this problem, we're gonna use the relationship between the determinant and matrix convertibility. Remember that to find the determined about three by three matrix with rows A B c, D e, f g h I. We use the formula here on the left. Then we can use the fact that A is inevitable if and only if it's determined does not equal zero. So we want to find where from the determinant formula. Certain values of K. Make are determined zero and go from there. So first applying the formula, the left, the determinant of A is one times one minus two K minus two. James, t minus 00 or two minus four K. So solving for zero, we have two minus four K equals zero or 14 equals two or k equals one half. That means that are determinant is zero and that is not inevitable when K equals one half. So we must have that Matrix A is convertible whenever K does not equal one half.

So we just multiply the big the vector with the corresponding matrix and see if this is equal to a multiple of this vector or not. I love matrixes this and our victor is. And if you multiply, we get minus three times one, which is minus three plus four and second entries minus three Danish one plus 32 it turns for and this is a quarter to one. And the second injury is 29 which is not equal toe multiple off the vector one for so 14 is not an Eigen Victor for this metrics.


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