Find a triangular system that is equivalent to the following systems, using elementary row operations.
\begin{cases} x+3y-z=2 \cr \cr 3x+2y+3z=4 \cr \cr x+y-z=2 \end{cases}
The augmented matrix of the equation is the following:
\begin{pmatrix} 1 & 3 &-1& |&2 \cr\cr 3 &2 &3 & |& 4 \cr\cr 1 & 1 & -1 & |& 2 \end{pmatrix}
To find a triangular system, the entries below the main diagonal must change to zero. We have:
\begin{pmatrix} 1 & 3 &-1& |&2 \cr\cr 3 &2 &3 & |& 4 \cr\cr 1 & 1 & -1 & |& 2 \end{pmatrix}
\ce{->[R_2:R_2-3R_1][R3:R_3-R_1]}\begin{pmatrix} 1 & 3 &-1& |&2 \cr\cr 0 &-7 &6 & |& -2 \cr\cr 0 & -2 & 0 & |& 0 \end{pmatrix}
\ce{->[R_3: R_3-2/7R_2]} \begin{pmatrix} 1 & 3 &-1& |&2 \cr\cr 0 &-7 &6 & |& -2 \cr\cr 0 & 0 & -12/7 & |& 4/7 \end{pmatrix}
Therefore, the system is equivalent to the following triangular system:
\begin{cases} x+3y+z=2 \cr \cr -7y+6z=2 \cr \cr -12/7z=4/7 \end{cases}
\begin{cases} 2x-3y=2 \cr \cr x+2y=5 \end{cases}
The augmented matrix of the equation is the following:
\begin{pmatrix} 2 & -3& |&2 \cr\cr 1 & 2 & |& 5 \end{pmatrix}
To find a triangular system, the entry below the main diagonal must change to zero. We have:
\begin{pmatrix} 2 & -3& |&2 \cr\cr 1 & 2 & |& 5 \end{pmatrix}
\ce{->[R_2:R_2-1/2R_1]}\begin{pmatrix} 2 & -3& |&2 \cr\cr 0 &\dfrac{7}{2} & |& 4 \end{pmatrix}
Therefore, the system is equivalent to the following triangular system:
\begin{cases} 2x-3y=2 \cr \cr \dfrac{7}{2}y=4 \end{cases}
\begin{cases} x-4y=9 \cr \cr 3x+8y=7 \end{cases}
The augmented matrix of the equation is the following:
\begin{pmatrix} 1 & -4& |&9 \cr\cr 3 & 8 & |& 7 \end{pmatrix}
To find a triangular system, the entry below the main diagonal must change to zero. We have:
\ce{->[R_2:R_2-3R_1]}\begin{pmatrix} 1 & -4& |&9 \cr\cr 0 & 20 & |& -20 \end{pmatrix}
Therefore, the system is equivalent to the following triangular system:
\begin{cases} x-4y=9 \cr \cr 20y=-20 \end{cases}
\begin{cases} 2x+5y=3 \cr \cr 3x+4y=-1 \end{cases}
The augmented matrix of the equation is the following:
\begin{pmatrix} 2 & 5& |&3 \cr\cr 3 & 4 & |& -1 \end{pmatrix}
To find a triangular system, the entry below the main diagonal must change to zero. We have:
\ce{->[R_2:R_2-3/2R_1]}\begin{pmatrix} 2 & 5& |&3 \cr\cr 0 & -\dfrac{7}{2} & |& -\dfrac{11}{2} \end{pmatrix}
Therefore, the system is equivalent to the following triangular system:
\begin{cases} 2x+5y=3 \cr \cr -\dfrac{7}{2}y=-\dfrac{11}{2} \end{cases}
\begin{cases} x+y-z=2 \cr \cr 2x+2y+3z=7 \cr \cr 3x-y-2z=0 \end{cases}
The augmented matrix of the equation is the following:
\begin{pmatrix} 1 & 1 &-1& |&2 \cr\cr 2 &2 &3 & |& 7 \cr\cr 3 & -1 & -2 & |& 0 \end{pmatrix}
To find a triangular system, the entries below the main diagonal must change to zero. We have:
\ce{->[R_2:R_2-2R_1][R3:R_3-3R_1]} \begin{pmatrix} 1 & 1 &-1& |&2 \cr\cr 0 &0 &5 & |& 3 \cr\cr 0 & -4 & 1 & |& -6 \end{pmatrix}
\ce{->[R_3: R_3-2/7R_2]} \begin{pmatrix} 1 & 1 &-1& |&2 \cr\cr 0 &0 &5 & |& 3 \cr\cr 0 & -4 & 1 & |& -6 \end{pmatrix}
\ce{->[R_3\ce{ \lt = \gt } R_2]} \begin{pmatrix} 1 & 1 &-1& |&2 \cr\cr 0 &-4 &1 & |& -6 \cr\cr 0 & 0 & 5 & |& 3 \end{pmatrix}
Therefore, the system is equivalent to the following triangular system:
\begin{cases} x+y-z=2 \cr \cr -4y+z=-6 \cr \cr 5z=3 \end{cases}
\begin{cases} x-z=2 \cr \cr x+y=4 \cr \cr y-z=-2 \end{cases}
The augmented matrix of the equation is the following:
\begin{pmatrix} 1 & 0 &-1& |&2 \cr\cr 1 &1 &0 & |& 4 \cr\cr 0& 1 & -1 & |& -2 \end{pmatrix}
To find a triangular system, the entries below the main diagonal must change to zero. We have:
\begin{pmatrix} 1 & 0 &-1& |&2 \cr\cr 1 &1 &0 & |& 4 \cr\cr 0& 1 & -1 & |& -2 \end{pmatrix}
\ce{->[R_2: R_2-R_1]} \begin{pmatrix} 1 & 0 &-1& |&2 \cr\cr 0 &1 &1 & |& 2 \cr\cr 0& 1 & -1 & |& -2 \end{pmatrix}
\ce{->[R_3: R_3-R_2]} \begin{pmatrix} 1 & 0 &-1& |&2 \cr\cr 0 &1 &1 & |& 2 \cr\cr 0& 0 & -2 & |& -4 \end{pmatrix}
Therefore, the system is equivalent to the following triangular system:
\begin{cases} x-z=2 \cr \cr y+z=2 \cr \cr -2z=-4 \end{cases}
\begin{cases} x+y+z=5 \cr \cr 2x+2y+3z=10 \cr \cr 4x-5y+z=2 \end{cases}
The augmented matrix of the equation is the following:
\begin{pmatrix} 1 & 1 &1& |&5 \cr\cr 2 &2 &3 & |& 10 \cr\cr 4 & -5 & 1 & |& 2 \end{pmatrix}
To find a triangular system, the entries below the main diagonal must change to zero. We have:
\begin{pmatrix} 1 & 1 &1& |&5 \cr\cr 2 &2 &3 & |& 10 \cr\cr 4 & -5 & 1 & |& 2 \end{pmatrix}
\ce{->[R_2:R_2-2R_1][R3:R_3-4R_1]} \begin{pmatrix} 1 & 1 &1& |&5 \cr\cr 0 &0 &1 & |& 0 \cr\cr 0 & -9 & -3 & |& -18 \end{pmatrix}
\ce{->[R_2\ce{ \lt = \gt } R_3]} \begin{pmatrix} 1 & 1 &1& |&5 \cr\cr 0 & -9 & -3 & |& -18 \cr\cr 0 &0 &1 & |& 0 \end{pmatrix}
Therefore, the system is equivalent to the following triangular system:
\begin{cases} x+y+z=5 \cr \cr -9y-3z=-18 \cr \cr z=0 \end{cases}