Question 7 a) Find the eigenvalues of the given matrix: (i)...

Question 7 a) Find the eigenvalues of the given matrix: (i) A = egin{bmatrix} 1 & -2 \ 2 & -3 end{bmatrix} (ii) B = egin{bmatrix} 4 & 0 \ 5 & 1 end{bmatrix} (iii) C = egin{bmatrix} 0 & 1 & 3 \ -1 & 1 & 0 \ 1 & 1 & 2 end{bmatrix} (iv) C = egin{bmatrix} 3 & 1 & -3 \ 1 & 0 & 0 \ -2 & 1 & 2 end{bmatrix} b) Find the eigenvector corresponding to the given eigenvalue: (i) A = egin{bmatrix} 1 & 1 & 2 \ 0 & 2 & 0 \ 2 & 0 & -1 end{bmatrix}, (lambda = 2) (ii) B = egin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 end{bmatrix}, (lambda = 0) c) Find the eigenvalues for the given matrices. Hence, find the eigenvectors for each eigenvalue. (i) C = egin{bmatrix} 1 & 1 & -1 \ 0 & 2 & 1 \ 0 & 0 & 3 end{bmatrix} (iii) E = egin{bmatrix} 4 & 1 & 1 \ 1 & 4 & 1 \ 1 & 1 & 4 end{bmatrix} (ii) D = egin{bmatrix} 3 & 1 & -3 \ 1 & 0 & 0 \ -2 & 1 & 2 end{bmatrix}

Transcript

00:01 To find the eigenvalue of a matrix, what we need to do is replace diagonally these two for the following values.

00:07 We do 1 minus lambda, well this is still minus 2, 2 and then minus 3 minus lambda.

00:18 We're going to set this equal to 0 and we're going to find the determinant of the matrix.

00:24 The determinant of a 2x2 matrix is this times this.

00:28 So 1 minus lambda times minus 3 minus lambda and that's going to be added.

00:38 So plus the other one, which is 2 times minus 2.

00:46 2 times minus 2.

00:52 And actually this is not a positive, this is a negative.

00:55 So we do cross multiplication and we subtract them.

00:59 And this is equal to 0.

01:00 So let's distribute the parentheses here.

01:03 1 times minus 3 is minus 3.

01:05 1 times minus lambda is minus lambda.

01:09 Minus lambda times minus 3 lambda.

01:14 And minus lambda times minus lambda is positive lambda squared.

01:19 And then the two negatives cancel.

01:21 So positive 4 equals 0.

01:26 Let's combine like terms.

01:28 Minus 1 plus 3 is 2 giving us minus 3 plus 2 lambda plus lambda squared actually we can combine the plus 4 and the minus 3 as well leaving us with just 1 is equal to 0 and let's organize this lambda squared plus 2 lambda plus 1 equals 0 now we have a quadratic equation we need to solve for lambda.

02:01 We can either use factorization or the quadratic formula and if we do it we realize there is only one answer to this problem which is lambda is equal to minus one.

02:13 So the eigenvalue of this matrix is minus one.

02:20 And now let's find the eigenvalue of one of the three by three matrices.

02:24 We do the same procedure so let's put here minus lambda minus lambda.

02:30 Let's move it.

02:31 We don't have enough space there.

02:33 So here, minus lambda.

02:36 And this one as well, we have to do two minus lambda.

02:41 We set this equal to zero and we need to find the determinant of this matrix.

02:47 How do we find the determinant? well, let's start first, this one.

02:52 And we multiply across like this.

02:56 That gives us, this is just minus lambda.

03:01 Times.

03:03 So here across is 1 minus lambda times 2 minus lambda and we can close the parenthesis because it's a 0 times 1 which just turns to 0.

03:16 Then the next one we start right away with the negative.

03:20 We cross this and we go this times this and this times this.

03:25 So it's one times we have minus one times two minus lambda and this is just zero.

03:36 So we don't do it...

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