Answer to Question #350968 in Linear Algebra for Nhm

Question #350968

Solve the system of linear equations using matrix inverse method.

x-y+z=1

2y-z=1

2x+3y=1


1
Expert's answer
2022-06-16T10:31:09-0400
"A=\\begin{pmatrix}\n 1 & -1 & 1 \\\\\n 0 & 2 & -1 \\\\ \n 2 & 3 & 0 \\\\\n\\end{pmatrix}, X=\\begin{pmatrix}\n x \\\\ \n y\\\\\nz\n\\end{pmatrix}, B=\\begin{pmatrix}\n 1\\\\ \n1\\\\\n1\n\\end{pmatrix}"

"AX=B"

"A^{-1}AX=A^{-1}B=>X=A^{-1}B"

augment the matrix with the identity matrix:


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

"R_3=R_3-2R_1"


"\\begin{pmatrix}\n 1 & -1 & 1 & & 1& 0 & 0 \\\\\n 0 & 2 & -1 & & 0& 1 & 0 \\\\ \n 0 & 5 & -2 & & -2 & 0 & 1 \\\\\n\\end{pmatrix}"

"R_2=R_2\/2"


"\\begin{pmatrix}\n 1 & -1 & 1 & & 1& 0 & 0 \\\\\n 0 & 1 & -1\/2 & & 0& 1\/2 & 0 \\\\ \n 0 & 5 & -2 & & -2 & 0 & 1 \\\\\n\\end{pmatrix}"

"R_1=R_1+R_2"


"\\begin{pmatrix}\n 1 & 0 & 1\/2 & & 1& 1\/2 & 0 \\\\\n 0 & 1 & -1\/2 & & 0& 1\/2 & 0 \\\\ \n 0 & 5 & -2 & & -2 & 0 & 1 \\\\\n\\end{pmatrix}"

"R_3=R_3-5R_2"


"\\begin{pmatrix}\n 1 & 0 & 1\/2 & & 1& 1\/2 & 0 \\\\\n 0 & 1 & -1\/2 & & 0& 1\/2 & 0 \\\\ \n 0 & 0 & 1\/2 & & -2 & -5\/2 & 1 \\\\\n\\end{pmatrix}"

"R_3=2R_3"


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

"R_1=R_1-R_3\/2"


"\\begin{pmatrix}\n 1 & 0 & 0 & & 3 & 3 & -1 \\\\\n 0 & 1 & -1\/2 & & 0& 1\/2 & 0 \\\\ \n 0 & 0 & 1 & & -4 & -5 & 2 \\\\\n\\end{pmatrix}"

"R_2=R_2+R_3\/2"


"\\begin{pmatrix}\n 1 & 0 & 0 & & 3 & 3 & -1 \\\\\n 0 & 1 & 0 & & -2 & -2 & 1 \\\\ \n 0 & 0 & 1 & & -4 & -5 & 2 \\\\\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 3 & 3 & -1 \\\\\n -2 & -2 & 1 \\\\ \n -4 & -5 & 2 \\\\\n\\end{pmatrix}"

"X=A^{-1}B"

"=\\begin{pmatrix}\n 3 & 3 & -1 \\\\\n -2 & -2 & 1 \\\\ \n -4 & -5 & 2 \\\\\n\\end{pmatrix}\\begin{pmatrix}\n 1\\\\ \n1\\\\\n1\n\\end{pmatrix}"

"=\\begin{pmatrix}\n 3+3-1\\\\ \n-2-2+1\\\\\n-4-5+2\n\\end{pmatrix}=\\begin{pmatrix}\n 5\\\\ \n-3\\\\\n-7\n\\end{pmatrix}"

"(5, -3, -7)"


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