Answer to Question #314303 in Linear Algebra for JKK

Question #314303

Find all Eigen values and corresponding Eigen vectors for the matrix A=

0 0 3

2 5 0

2 3 0



1
Expert's answer
2022-03-20T06:44:11-0400

"A=\\begin{pmatrix}\n 0 & 0&3 \\\\\n 2 & 5&0\\\\\n2&3&0\n\\end{pmatrix}\\\\\n|A-\\lambda I|=\\begin{vmatrix}\n 0-\\lambda & 0&3 \\\\\n 2 & 5-\\lambda&0\\\\\n2&3&0-\\lambda\n\\end{vmatrix}=\\\\\n=\\lambda^2(5-\\lambda)+0+18-6(5-\\lambda)-0-0=\\\\\n=-\\lambda^3+5\\lambda^2+6\\lambda-12=0\\\\\n\\lambda^3-5\\lambda^2-6\\lambda+12=0\\\\\n12:\\pm1, \\pm2,\\pm3,\\pm4,\\pm6,\\pm12\\\\\n\\lambda=\\pm1: (\\pm1)^3-5(\\pm1)^2-6(\\pm1)+12\\neq0\\\\\n\\lambda=\\pm2:(\\pm2)^3-5(\\pm2)^2-6(\\pm2)+12\\neq0\\\\\n\\lambda=\\pm3: (\\pm3)^3-5(\\pm3)^2-6(\\pm3)+12\\neq0\\\\\n\\lambda=\\pm4: (\\pm4)^3-5(\\pm4)^2-6(\\pm4)+12\\neq0\\\\\n\\lambda=\\pm6: (\\pm6)^3-5(\\pm6)^2-6(\\pm6)+12\\neq0\\\\\n\\lambda=\\pm12: (\\pm12)^3-5(\\pm12)^2-6(\\pm12)+12\\neq0"





"\\lambda_1\\approx -2\\\\\n\\lambda_2\\approx1\\\\\n\\lambda_3\\approx6"

All Eigen values

"(A-\\lambda I)\\vec{x}=\\vec{0}, \\vec{x}\\neq\\vec{0}\\\\\n1. \\lambda_1=-2\\\\\n\\vec{x_1}=(x_1,x_2,x_3)\\neq\\vec{0}\\\\\n(A-\\lambda_1I)\\vec{x_1}=\\vec{0}\\\\\n\\begin{pmatrix}\n 2 &0&3 \\\\\n 2 & 7&0\\\\\n2&3&2\n\\end{pmatrix}\\begin{pmatrix}\n x_1 \\\\\n x_2\\\\\nx_3\n\\end{pmatrix}=\\begin{pmatrix}\n 0\\\\0\\\\0\n\\end{pmatrix}\\\\\n2x_1+3x_3=0\\\\\n2x_1+7x_2=0\\\\\n2x_1+3x_2+2x_3=0\\\\\nx_3=-\\frac{2}{3}x_1\\\\\nx_2=-\\frac{2}{7}x_1\\\\\nx_1=1, x_2=-\\frac{2}{7}, x_3=-\\frac{2}{3}\\\\\n\\vec{x_1}=(1,-\\frac{2}{7}, -\\frac{2}{3})"

"2. \\lambda_2=1\\\\\n\\vec{x_2}=(x_1,x_2,x_3)\\neq\\vec{0}\\\\\n(A-\\lambda_2I)\\vec{x_2}=\\vec{0}\\\\\n\\begin{pmatrix}\n -1 &0&3 \\\\\n 2 & 4&0\\\\\n2&3&-1\n\\end{pmatrix}\\begin{pmatrix}\n x_1 \\\\\n x_2\\\\\nx_3\n\\end{pmatrix}=\\begin{pmatrix}\n 0\\\\0\\\\0\n\\end{pmatrix}\\\\\n-x_1+3x_3=0\\\\\n2x_1+4x_2=0\\\\\n2x_1+3x_2-x_3=0\\\\\nx_3=\\frac{1}{3}x_1\\\\\nx_2=-\\frac{1}{2}x_1\\\\\nx_1=1, x_2=-\\frac{1}{2}, x_3=\\frac{1}{3}\\\\\n\\vec{x_2}=(1,-\\frac{1}{2}, \\frac{1}{3})"

"3. \\lambda_3=6\\\\\n\\vec{x_3}=(x_1,x_2,x_3)\\neq\\vec{0}\\\\\n(A-\\lambda_3I)\\vec{x_3}=\\vec{0}\\\\\n\\begin{pmatrix}\n -6 &0&3 \\\\\n 2 & -1&0\\\\\n2&3&-6\n\\end{pmatrix}\\begin{pmatrix}\n x_1 \\\\\n x_2\\\\\nx_3\n\\end{pmatrix}=\\begin{pmatrix}\n 0\\\\0\\\\0\n\\end{pmatrix}\\\\\n-6x_1+3x_3=0\\\\\n2x_1-x_2=0\\\\\n2x_1+3x_2-6x_3=0\\\\\nx_3=2x_1\\\\\nx_2=2x_1\\\\\nx_1=1, x_2=2, x_3=2\\\\\n\\vec{x_3}=(1,2,2)"

"\\vec{x_1}=(1,-\\frac{2}{7}, -\\frac{2}{3}),\\vec{x_2}=(1,-\\frac{1}{2}, \\frac{1}{3}),\n\\vec{x_3}=(1,2,2)"

All corresponding Eigen vectors


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS