Question

Find the characteristic equation and the eigenvalues (and corresponding eigenvectors) of the matrix. 2 -2 7 0 3 -2 0 -1 2 (a) the characteristic equation (b) the eigenvalues (Enter your answers from smallest to largest.) (91, 12, 13) = 1, 2, 4 the corresponding eigenvectors X1 = x X2 = X3 =

Answer 1

$\text{Answer:} \\ \text{(a) The characteristic equation is } \fbox{\lambda^3-7\lambda^2+14\lambda-8=0} \\ \\ \text{(b) The eigenvalues are } \left(\lambda_1,\lambda_2,\lambda_3\right)=\fbox{(1,~2,~4)} \\ \\ \text{(c) The eigenvectors are: } x_1= \fbox{\begin{bmatrix} -5\\ 1\\ 1 \end{bmatrix}}~,~~x_2=\fbox{\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}}~,~x_3=\fbox{\begin{bmatrix} 11\\ -4\\ 2 \end{bmatrix}}$
2 - 2 7 let the matnin A = -3 -2 -1 2 of the matvin A (a) The characteristic equation ► det (A --I) = 0 2-) - 2 7 = 0 3-X -2 -1 2-2 =0 O → (2-»[ (3-1)(2-1)-(-2)(-1] (2-1) [6-52 +3 - 2 → (2-)) 22-52+4) = 0 22+722-147+8 = 0 > -(4-2) (0-4) (2-1) = 0 13-722 +142-8 = 0 >> (2-1)(2-2) (2-4)=0 Henee the characteristic equm is 3-72²+142-8 the characteristic equn eigenvalues of the matrix A, di=1,2₂=2, 13 = 4 eigen values 12., 72,40) = 1,2,4) are the (b) The roots of The roots are , Hence the are CS Scanned with CamScanner

for 1,-1, (A - 1)34 = 0 6 - 2 7 y = I 2 - 2 Gaussian Elimination 6 - 1 y o 5 - 2 7 10) Riley o - 0 2 -2 R2 + R₂R, R, 2 R₂ 7R, o O 0 0 0 A - 5 I3 -5c 7, +573 =0 92-92 = 0 let Yg=e, then Y = Yg=e, & y, a Y -5e 24 = 5 The eigen-recton for 2,=1,4 = for ^2=2, (A-21)x, = 0 71181-1 - 2 - -2 ร Y3 - 1 > -43-0 8,-2yz = 0, -29, + 7 93 = 0 y3 = 0, y y=0 → X2= - y, os The eigen vector for x=2,22 (9) for 7=4 (A-41) X3 = 0 2 - 2 71 -2 -29,-27+7=0 o => -92273 =0 -1 -1 -2 CS Scanned with CamScanner

let Yzze, then Ye=-24₂=-2e -29, -29, +783-0 >> -27,-26-2e)+7c = 0 y, le -27, = -11e + 2 11 10 the = é Xz JIN = Yz -4 -26 ya 2 e 2 11 a → The eigen-rector for A3=4, X3 -4 2 CS Scanned with CamScanner