\[
\begin{cases}
3x+y-z=1\\
x-2y+3z=2\\
2x+y+z=3
\end{cases}
\]
wecanusethemethodofGaussianelimination.
First,wewritetheaugmentedmatrixforthesystem:
\[
\left[
\begin{array}{ccc|c}
3&1&-1&1\\
1&-2&3&2\\
2&1&1&3
\end{array}
\right]
\]
Step1:Makethefirstentryofthefirstrow1.
Wecanachievethisbyswappingthefirstandsecondrows:
\[
\left[
\begin{array}{ccc|c}
1&-2&3&2\\
3&1&-1&1\\
2&1&1&3
\end{array}
\right]
\]
Step2:Eliminatethex-termfromthesecondandthirdrows.
Toeliminate\(3x\)inthesecondrow,subtract3timesthefirstrowfromthesecondrow:
\[
3-3(1)=0,\quad1-3(-2)=1+6=7,\quad-1-3(3)=-1-9=-10,\quad1-3(2)=1-6=-5
\]
Toeliminate\(2x\)inthethirdrow,subtract2timesthefirstrowfromthethirdrow:
\[
2-2(1)=0,\quad1-2(-2)=1+4=5,\quad1-2(3)=1-6=-5,\quad3-2(2)=3-4=-1
\]
Thenewaugmentedmatrixis:
\[
\left[
\begin{array}{ccc|c}
1&-2&3&2\\
0&7&-10&-5\\
0&5&-5&-1
\end{array}
\right]
\]
Step3:Makethesecondentryofthesecondrow1.
Wecandividethesecondrowby7:
\[
\left[
\begin{array}{ccc|c}
1&-2&3&2\\
0&1&-\frac{10}{7}&-\frac{5}{7}\\
0&5&-5&-1
\end{array}
\right]
\]
Step4:Eliminatethey-termfromthethirdrow.
Toeliminate\(5y\)inthethirdrow,subtract5timesthesecondrowfromthethirdrow:
\[
0-5(0)=0,\quad5-5(1)=0,\quad-5-5\left(-\frac{10}{7}\right)=-5+\frac{50}{7}=\frac{-35+50}{7}=\frac{15}{7},\quad-1-5\left(-\frac{5}{7}\right)=-1+\frac{25}{7}=\frac{-7+25}{7}=\frac{18}{7}
\]
Thenewaugmentedmatrixis:
\[
\left[
\begin{array}{ccc|c}
1&-2&3&2\\
0&1&-\frac{10}{7}&-\frac{5}{7}\\
0&0&\frac{15}{7}&\frac{18}{7}
\end{array}
\right]
\]
Step5:Makethethirdentryofthethirdrow1.
Wecanmultiplythethirdrowby\(\frac{7}{15}\):
\[
\left[
\begin{array}{ccc|c}
1&-2&3&2\\
0&1&-\frac{10}{7}&-\frac{5}{7}\\
0&0&1&\frac{18}{15}\cdot\frac{7}{15}=\frac{6}{5}
\end{array}
\right]
\]
Step6:Backsubstitutiontosolvefor\(z\),\(y\),and\(x\).
\[
z=\frac{6}{5}
\]
Substitute\(z=\frac{6}{5}\)intothesecondrowtosolvefor\(y\):
\[
y-\frac{10}{7}\cdot\frac{6}{5}=-\frac{5}{7}
\]
\[
y-\frac{60}{35}=-\frac{5}{7}
\]
\[
y-\frac{12}{7}=-\frac{5}{7}
\]
\[
y=-\frac{5}{7}+\frac{12}{7}=\frac{7}{7}=1
\]
Substitute\(y=1\)and\(z=\frac{6}{5}\)intothefirstrowtosolvefor\(x\):
\[
x-2(1)+3\left(\frac{6}{5}\right)=2
\]
\[
x-2+\frac{18}{5}=2
\]
\[
x-2+3.6=2
\]
\[
x+1.6=2
\]
\[
x=2-1.6=0.4
\]
So,thesolutionis:
\[
x=0.4,\quady=1,\quadz=\frac{6}{5}
\]
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