Answer to Question #350512 in Linear Algebra for Busi

Question #350512

Use the matrix method (together with elementary row transformations) to solve the following:


2x −y +3z = 2


x +2y −z = 4


−4x +5y +z = 10


1
Expert's answer
2022-06-15T14:56:35-0400
"A=\\begin{pmatrix}\n 2 & -1 & 3 \\\\\n 1 & 2 & -1\\\\\n -4 & 5 & 1 \\\\\n\\end{pmatrix}, X=\\begin{pmatrix}\n x \\\\\n y \\\\\nz\n\\end{pmatrix}, B=\\begin{pmatrix}\n 2\\\\\n 4\\\\\n10\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 2 & -1 & 3 && 1& 0 & 0 \\\\\n 1 & 2 & -1 && 0 & 1 & 0\\\\\n -4 & 5 & 1 && 0 & 0 & 1 \\\\\n\\end{pmatrix}"

"R_1=R_1\/2"


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

"R_2=R_2-R_1"


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

"R_3=R_3+4R_1"


"\\begin{pmatrix}\n 1 & -1\/2 & 3\/2 && 1\/2& 0 & 0 \\\\\n 0 & 5\/2 & -5\/2 && -1\/2 & 1 & 0\\\\\n 0 & 3 & 7 && 2 & 0 & 1 \\\\\n\\end{pmatrix}"

"R_2=2R_2\/5"


"\\begin{pmatrix}\n 1 & -1\/2 & 3\/2 && 1\/2& 0 & 0 \\\\\n 0 & 1 & -1 && -1\/5 & 2\/5 & 0\\\\\n 0 & 3 & 7 && 2 & 0 & 1 \\\\\n\\end{pmatrix}"

"R_1=R_1+R_2\/2"


"\\begin{pmatrix}\n 1 & 0 & 1 && 2\/5 & 1\/5 & 0 \\\\\n 0 & 1 & -1 && -1\/5 & 2\/5 & 0\\\\\n 0 & 3 & 7 && 2 & 0 & 1 \\\\\n\\end{pmatrix}"

"R_3=R_3-3R_2"


"\\begin{pmatrix}\n 1 & 0 & 1 && 2\/5 & 1\/5 & 0 \\\\\n 0 & 1 & -1 && -1\/5 & 2\/5 & 0\\\\\n 0 & 0 & 10 && 13\/5 & -6\/5 & 1 \\\\\n\\end{pmatrix}"

"R_3=R_3\/10"


"\\begin{pmatrix}\n 1 & 0 & 1 && 2\/5 & 1\/5 & 0 \\\\\n 0 & 1 & -1 && -1\/5 & 2\/5 & 0\\\\\n 0 & 0 & 1 && 13\/50 & -6\/50 & 1\/10 \\\\\n\\end{pmatrix}"

"R_1=R_1-R_3"


"\\begin{pmatrix}\n 1 & 0 & 0 && 7\/50 & 8\/25 & -1\/10 \\\\\n 0 & 1 & -1 && -1\/5 & 2\/5 & 0\\\\\n 0 & 0 & 1 && 13\/50 & -6\/50 & 1\/10 \\\\\n\\end{pmatrix}"

"R_2=R_2+R_3"


"\\begin{pmatrix}\n 1 & 0 & 0 && 7\/50 & 8\/25 & -1\/10 \\\\\n 0 & 1 & 0 && 3\/50 & 7\/25 & 1\/10\\\\\n 0 & 0 & 1 && 13\/50 & -6\/50 & 1\/10 \\\\\n\\end{pmatrix}"

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

"A^{-1}B=\\begin{pmatrix}\n 0.14 & 0.32 & -0.1 \\\\\n 0.06 & 0.28 & 0.1 \\\\\n 0.26 & -0.12 & 0.1 \\\\\n\\end{pmatrix}\\begin{pmatrix}\n 2\\\\\n 4\\\\\n10\n\\end{pmatrix}"

"=\\begin{pmatrix}\n 0.28+1.28-1\\\\\n 0.12+1.12+1\\\\\n0.52-0.48+1\n\\end{pmatrix}=\\begin{pmatrix}\n 0.56\\\\\n 2.24\\\\\n1.04\n\\end{pmatrix}"

"(x,y,z)=(0.56, 2.24, 1.04)"



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