Solve a system of equation using the Gaussian elimination Algebra I

The augmented matrix of this system is:

\begin{pmatrix} 2 & 3 & | & 2 \cr\cr 3 & 2 & | &4 \end{pmatrix}

Use the row operation to find the triangular form:

\begin{pmatrix} 2 & 3 & | & 2 \cr\cr 3 & 2 & | &4 \end{pmatrix} \ce{->[R_2: R_2-3/2R_1]}\begin{pmatrix} 2 & 3 & | & 2 \cr\cr 0 & -5/2 & | &1 \end{pmatrix}

The system is equivalent to:

\begin{cases} 2x+3y=2 \cr \cr -5/2y=1 \end{cases}

Solve the second equation:

-\dfrac{5}{2}y=1

y=-\dfrac{2}{5}

Now solve the first equation:

2x+3y=2 \Rightarrow 2x+3 \times\left(\dfrac{-2}{5}\right)= 2 \Rightarrow 2x-\dfrac{6}{5}=2 \Rightarrow 2x=\dfrac{16}{5} \Rightarrow x=\dfrac{8}{5}

The augmented matrix of this system is:

\begin{pmatrix} 1 & -5 & | & -5 \cr\cr 3 & 2 & | & 19 \end{pmatrix}

Use the row operation to find the triangular form:

\begin{pmatrix} 1 & -5 & | & -5 \cr\cr 3 & 2 & | & 19 \end{pmatrix} \ce{->[R_2: R_2-3R_1]}\begin{pmatrix} 1 & -5 & | & -5 \cr\cr 0 & 17 & | & 34 \end{pmatrix}

The system is equivalent to:

\begin{cases} x-5y=-5 \cr \cr 17y=34 \end{cases}

Solve the second equation:

17y=34

y=2

Now solve the first equation:

x-5y=-5 \Rightarrow x-5 \times\left(2\right)= -5 \Rightarrow x-10=-5 \Rightarrow x=5

The augmented matrix of this system is:

\begin{pmatrix} 1 & -3 & | & 5 \cr\cr -2& 6 & | &4 \end{pmatrix}

Use the row operation to find the triangular form:

\begin{pmatrix} 1 & -3 & | & 5 \cr\cr -2& 6 & | &4 \end{pmatrix}\ce{->[R_2: R_2+2R_1]}\begin{pmatrix} 1 & -3 & | & 5 \cr\cr 0& 0 & | &14 \end{pmatrix}

The system is equivalent to:

\begin{cases} x-3y=5 \cr \cr 0=14 \end{cases}

The augmented matrix of this system is:

\begin{pmatrix} 4 & -6 & | & 10 \cr\cr 2 & -3 & | &5 \end{pmatrix}

Use the row operation to find the triangular form:

\begin{pmatrix} 4 & -6 & | & 10 \cr\cr 2 & -3 & | &5 \end{pmatrix}\ce{->[R_2: R_2-1/2R_1]}\begin{pmatrix} 4 & -6 & | & 10 \cr\cr 0 & 0 & | &0 \end{pmatrix}

The system has infinitely many solutions. To find the general solution, assume that y=t.

Solve the first equation:

4x-6y=10 \Rightarrow 4x-6t=10 \Rightarrow x= \dfrac{10+6t}{4}

The augmented matrix of the equation is the following:

\begin{pmatrix} 1 & 1 &1& |&3 \cr\cr 2 &-1 &5 & |& 6 \cr\cr 4 & 3 & -8 & |& -1 \end{pmatrix}

To find a triangular system, the entries below the main diagonal must change to zero. We have:

\begin{pmatrix} 1 & 1 &1& |&3 \cr\cr 2 &-1 &5 & |& 6 \cr\cr 4 & 3 & -8 & |& -1 \end{pmatrix}

\ce{->[R_2:R_2-2R_1][R3:R_3-4R_1]}\begin{pmatrix} 1 & 1 &1& |&3 \cr\cr 0 &-3 &3 & |& 0 \cr\cr 0 & -1 & -12 & |& -13 \end{pmatrix}

\ce{->[R_3: R_3-1/3R_2]} \begin{pmatrix} 1 & 1 &1& |&3 \cr\cr 0 &-3 &3 & |& 0 \cr\cr 0 & 0& -13 & |& -13 \end{pmatrix}

Therefore, the system is equivalent to the following triangular system:

\begin{cases} x+y+z=3 \cr \cr -3y+3z=0 \cr \cr -13z=-13 \end{cases}

Solve the third equation:

-13z = -13

z=1

Now solve the second equation:

-3y+3z = 0 \Rightarrow -3y+ 3\left(1\right)= 0 \Rightarrow -3y=-3 \Rightarrow y=1

Finally, solve the first equation:

x+y+z=3 \Rightarrow x+ \left(1\right) + \left(1\right) = 3 \Rightarrow x+2 = 3 \Rightarrow x=1

The augmented matrix of the equation is the following:

\begin{pmatrix} 1 & 6 &-2& |&4 \cr\cr 4 &2 &-1 & |& 3 \cr\cr 2 & 4 & 1 & |& 0 \end{pmatrix}

To find a triangular system, the entries below the main diagonal must change to zero. We have:

\begin{pmatrix} 1 & 6 &-2& |&4 \cr\cr 4 &2 &-1 & |& 3 \cr\cr 2 & 4 & 1 & |& 0 \end{pmatrix}

\ce{->[R_2:R_2-2R_1][R3:R_3-4R_1]}\begin{pmatrix} 1 & 6 &-2& |&4 \cr\cr 0 &-22 &7 & |&-13 \cr\cr 0 & -8 & 5 & |& -8 \end{pmatrix}

\ce{->[R_3: R_3-4/11R_2]} \begin{pmatrix} 1 & 6 &-2& |&4 \cr\cr 0 &-22 &7 & |&-13 \cr\cr 0 & 0 & 5-\left(\dfrac{4}{11}\times7\right) & |& -8-\left(\dfrac{4}{11}\times-13\right) \end{pmatrix}

=\begin{pmatrix} 1 & 6 &-2& |&4 \cr\cr 0 &-22 &7 & |&-13 \cr\cr 0 & 0 & \dfrac{27}{11} & |& -\dfrac{36}{11} \end{pmatrix}

Therefore, the system is equivalent to the following triangular system:

\begin{cases} x+6y-2z=4 \cr \cr -22y+7z=-13 \cr \cr \dfrac{27}{11}z=-\dfrac{36}{11} \end{cases}

Solve the third equation:

\dfrac{27}{11}z=-\dfrac{36}{11}

z=-\dfrac{36}{27} =-\dfrac{4}{3}

Now solve the second equation:

-22y - 7z = -13 \Rightarrow -22y +7\left(-\dfrac{4}{3}\right)=-13 \Rightarrow -22y = -13+\dfrac{28}{3} = -\dfrac{11}{3}\\

y=\dfrac{1}{6}

Finally, solve the first equation:

x+6\left(\dfrac{1}{6}\right) -2\left(-\dfrac{4}{3}\right) = 4 \Rightarrow x=4-1-\dfrac{8}{3} = \dfrac{1}{3}

The augmented matrix of the equation is the following:

\begin{cases} 2x+2y-z=3 \cr \cr 4x+4y-2z=5 \cr \cr x+2y-z=1 \end{cases}

\begin{pmatrix} 2 & 2 &-1& |&3 \cr\cr 4 &4 &-2 & |& 5 \cr\cr 1 & 2 & -1 & |& 1 \end{pmatrix}

To find a triangular system, the entries below the main diagonal must change to zero. We have:

\begin{pmatrix} 2 & 2 &-1& |&3 \cr\cr 4 &4 &-2 & |& 5 \cr\cr 1 & 2 & -1 & |& 1 \end{pmatrix}

\ce{->[R_2:R_2-2R_1][R3:R_3-1/2R_1]}\begin{pmatrix} 2 & 2 &-1& |&3 \cr\cr 0&0 &0 & |& -1 \cr\cr 0 & 1 & -\dfrac{1}{2} & |& -\dfrac{1}{2} \end{pmatrix}

\ce{->[R_3: R_2]} \begin{pmatrix} 2 & 2 &-1& |&3 \cr\cr 0&0 &0 & |& -1 \cr\cr 0 & 1 & -\dfrac{1}{2} & |& -\dfrac{1}{2} \end{pmatrix}

\ce{->[R_2 \ce{ \lt = \gt } R_3]}\begin{pmatrix} 2 & 2 &-1& |&3\cr\cr 0 & 1 & -\dfrac{1}{2} & |& -\dfrac{1}{2} \cr\cr 0&0 &0 & |& -1 \end{pmatrix}

The system is equivalent to:

\begin{cases} 2x+2y-z=3 \cr \cr y -\dfrac{1}{2}z=-\dfrac{1}{2} \cr \cr 0=-1 \end{cases}