Answer to Question #186522 in Linear Algebra for Petra

Question #186522

Use the CRAMMER'S RULE to solve these systems of linear equations.

5x+6y+7z=40

2x+4y+2z=34

x+3y+5z=30


1
Expert's answer
2021-05-07T14:33:43-0400

Given:

{5x+6y+7z=402x+4y+2z=34x+3y+5z=30\left\{ \begin{array}{ c c c } 5x+6y+7z & = & 40 \\ 2x+4y+2z & = & 34\\ x+3y+5z & = & 30 \\ \end{array}\right.

Solution:

CRAMMER'S RULE 

deta11a12a13a21a22a23a31a32a33=a11a22a33+a12a23a31+a13a21a32a13a22a31a11a23a32a12a21a33\det\left| \begin{array}{ c c c } a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \\ \end{array}\right|=a_{11}*a_{22}*a_{33}+a_{12}*a_{23}*a_{31}+a_{13}*a_{21}*a_{32}-\\-a_{13}*a_{22}*a_{31}-a_{11}*a_{23}*a_{32}-a_{12}*a_{21}*a_{33}

x=DxD,y=DyD,z=DZDx=\frac{D_x}{D},y=\frac{D_y}{D},z=\frac{D_Z}{D}

D=det567242135=545+621+723741523625=36D=\det\left| \begin{array}{ c c c } 5 & 6 & 7 \\ 2 & 4 & 2\\ 1 & 3 & 5 \\ \end{array}\right|=5*4*5+6*2*1+7*2*3-7*4*1-\\ -5*2*3-6*2*5=36

Dx=det406734423035=4045+6230+7343743040236345=226D_x=\det\left| \begin{array}{ c c c } 40 & 6 & 7 \\ 34 & 4 & 2\\ 30 & 3 & 5 \\ \end{array}\right|=40*4*5+6*2*30+7*34*3-7*4*30-\\-40*2*3-6*34*5=-226

Dy=det540723421305=5345+4021+7230734152304025=412D_y=\det\left| \begin{array}{ c c c } 5 & 40 & 7 \\ 2 & 34 & 2\\ 1 & 30 & 5 \\ \end{array}\right|=5*34*5+40*2*1+7*2*30-7*34*1-\\-5*2*30-40*2*5=412

Dz=det564024341330=5430+6341+4023404153436230=14D_z=\det\left| \begin{array}{ c c c } 5 & 6 & 40 \\ 2 & 4 & 34\\ 1 & 3 & 30 \\ \end{array}\right|=5*4*30+6*34*1+40*2*3-40*4*1-\\-5*34*3-6*2*30=14

x=DxD=22636=11318=6518;y=DyD=41236=1039=1149;z=DzD=1436=718x=\frac{D_x}{D}=\frac{-226}{36}=-\frac{113}{18}=-6\frac{5}{18};y=\frac{D_y}{D}=\frac{412}{36}=\frac{103}{9}=11\frac{4}{9};z=\frac{D_z}{D}=\frac{14}{36}=\frac{7}{18}

Answer :

x=6518;y=1149;z=718x=-6\frac{5}{18};y=11\frac{4}{9};z=\frac{7}{18}


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