Answer to Question #350279 in Linear Algebra for Busi

Question #350279

Determine the inverse of A, and show that A


−1A = I.


A =


(2 1 0


2 −1 1


3 −2 4)





1
Expert's answer
2022-06-13T17:50:25-0400
"A=\\begin{pmatrix}\n 2 & 1 & 0\\\\\n 2 & -1 & 1\\\\ \n 3 & -2 & 4\\\\ \n\\end{pmatrix}"

Augment the matrix with the identity matrix:


"\\begin{pmatrix}\n 2 & 1 & 0 & & 1 & 0 & 0\\\\\n 2 & -1 & 1 & & 0 & 1 & 0\\\\ \n 3 & -2 & 4 & & 0 & 0 & 1\\\\ \n\\end{pmatrix}"

"R_1=R_1\/2"


"\\begin{pmatrix}\n 1 & 1\/2 & 0 & & 1\/2 & 0 & 0\\\\\n 2 & -1 & 1 & & 0 & 1 & 0\\\\ \n 3 & -2 & 4 & & 0 & 0 & 1\\\\ \n\\end{pmatrix}"

"R_2=R_2-2R_1"


"\\begin{pmatrix}\n 1 & 1\/2 & 0 & & 1\/2 & 0 & 0\\\\\n 0 & -2 & 1 & & -1 & 1 & 0\\\\ \n 3 & -2 & 4 & & 0 & 0 & 1\\\\ \n\\end{pmatrix}"

"R_3=R_3-3R_1"


"\\begin{pmatrix}\n 1 & 1\/2 & 0 & & 1\/2 & 0 & 0\\\\\n 0 & -2 & 1 & & -1 & 1 & 0\\\\ \n 0 & -7\/2 & 4 & & -3\/2 & 0 & 1\\\\ \n\\end{pmatrix}"

"R_2=-R_2\/2"


"\\begin{pmatrix}\n 1 & 1\/2 & 0 & & 1\/2 & 0 & 0\\\\\n 0 & 1 & -1\/2 & & 1\/2 & -1\/2 & 0\\\\ \n 0 & -7\/2 & 4 & & -3\/2 & 0 & 1\\\\ \n\\end{pmatrix}"

"R_1=R_1-R_2\/2"


"\\begin{pmatrix}\n 1 & 0 & 1\/4 & & 1\/4 & 1\/4 & 0\\\\\n 0 & 1 & -1\/2 & & 1\/2 & -1\/2 & 0\\\\ \n 0 & -7\/2 & 4 & & -3\/2 & 0 & 1\\\\ \n\\end{pmatrix}"

"R_3=R_3+7R_2\/2"


"\\begin{pmatrix}\n 1 & 0 & 1\/4 & & 1\/4 & 1\/4 & 0\\\\\n 0 & 1 & -1\/2 & & 1\/2 & -1\/2 & 0\\\\ \n 0 & 0 & 9\/4 & & 1\/4 & -7\/4 & 1\\\\ \n\\end{pmatrix}"

"R_3=4R_3\/9"


"\\begin{pmatrix}\n 1 & 0 & 1\/4 & & 1\/4 & 1\/4 & 0\\\\\n 0 & 1 & -1\/2 & & 1\/2 & -1\/2 & 0\\\\ \n 0 & 0 & 1 & & 1\/9 & -7\/9 & 4\/9\\\\ \n\\end{pmatrix}"

"R_1=R_1-R_3\/4"


"\\begin{pmatrix}\n 1 & 0 & 0 & & 2\/9 & 4\/9 & -1\/9\\\\\n 0 & 1 & -1\/2 & & 1\/2 & -1\/2 & 0\\\\ \n 0 & 0 & 1 & & 1\/9 & -7\/9 & 4\/9\\\\ \n\\end{pmatrix}"

"R_2=R_2+R_3\/2"


"\\begin{pmatrix}\n 1 & 0 & 0 & & 2\/9 & 4\/9 & -1\/9\\\\\n 0 & 1 & 0 & & 5\/9 & -8\/9 & 2\/9\\\\ \n 0 & 0 & 1 & & 1\/9 & -7\/9 & 4\/9\\\\ \n\\end{pmatrix}"

We are done. On the left is the identity matrix. On the right is the inverse matrix.


"A^{-1}=\\begin{pmatrix}\n 2\/9 & 4\/9 & -1\/9\\\\\n 5\/9 & -8\/9 & 2\/9\\\\ \n 1\/9 & -7\/9 & 4\/9\\\\ \n\\end{pmatrix}"

Check


"A^{-1}A=\\begin{pmatrix}\n 2\/9 & 4\/9 & -1\/9\\\\\n 5\/9 & -8\/9 & 2\/9\\\\ \n 1\/9 & -7\/9 & 4\/9\\\\ \n\\end{pmatrix}\\begin{pmatrix}\n 2 & 1 & 0\\\\\n 2 & -1 & 1\\\\ \n 3 & -2 & 4\\\\ \n\\end{pmatrix}"

"=\\begin{pmatrix}\n \\dfrac{4+8-3}{9} & \\dfrac{2-4+2}{9} & \\dfrac{0+4-4}{9}\\\\\n\\\\\n \\dfrac{10-16+6}{9} & \\dfrac{5+8-4}{9} & \\dfrac{0-8+8}{9}\\\\\n\\\\\n \\dfrac{2-14+12}{9} & \\dfrac{1+7-8}{9} & \\dfrac{0-7+16}{9}\\\\\n\\\\\n\\end{pmatrix}"

"=\\begin{pmatrix}\n 1 & 0 & 0\\\\\n 0 & 1 & 0\\\\ \n 0 & 0 & 1\\\\ \n\\end{pmatrix}=I"


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