Tosolvethesystemofequations

\begin{cases}

3x+y-z=1\\

x-2y+3z=2\\

2x+y+z=3

\end{cases}

wecanusethemethodofGaussianelimination.

First,wewritetheaugmentedmatrixforthesystem:

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\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}