Solve a system of linear equations using augmented matrix Algebra II

The system \begin{cases} 2x-y=4 \cr \cr 4x+2y=3 \end{cases} gives the below augmented matrix:

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

Adding twice the first row to the second gives:

\begin{pmatrix} 2 & -1 & 4 \cr\cr 8 & 0 &11 \end{pmatrix}

Dividing the second row by 8 gives:

\begin{pmatrix} 2 & -1 & 4 \cr\cr 1 & 0 &\dfrac{11}{8} \end{pmatrix}

Interchanging the two rows gives:

\begin{pmatrix} 1 & 0 &\dfrac{11}{8} \cr 2 & -1 & 4 \end{pmatrix}

Subtracting twice the first row from the second gives:

\begin{pmatrix} 1 & 0 &\dfrac{11}{8} \cr 0 & 1 & \dfrac{-5}{4} \end{pmatrix}

The solution appears in the matrix:

\begin{pmatrix} 1 & 0 &\textcolor{Red}{\dfrac{11}{8} }\cr 0 & 1 & \textcolor{Red}{\dfrac{-5}{4}} \end{pmatrix}

The system \begin{cases} 4x-2y=8 \cr \cr x+2y=6 \end{cases} provides the augmented matrix:

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

Adding the first row to the second gives:

\begin{pmatrix} 4 & -2 & 8 \cr\cr 5& 0 & 14 \end{pmatrix}

Interchanging the two rows gives:

\begin{pmatrix} 5 & 0 & 14 \cr \cr 4 & -2 & 8 \end{pmatrix}

Dividing the first row by 5 gives:

\begin{pmatrix} 1 & 0 & \dfrac{14}{5} \cr \cr 4 & -2 & 8 \end{pmatrix}

Subtracting four times the first row from the second gives:

\begin{pmatrix} 1 & 0 & \dfrac{14}{5} \cr \cr 0 & -2 & - \dfrac{16}{5} \end{pmatrix}

Dividing the second row by -2. The solution appears in the matrix:

\begin{pmatrix} 1 & 0 &\textcolor{Red}{\dfrac{14}{5} }\cr 0 & 1 & \textcolor{Red}{\dfrac{8}{5}} \end{pmatrix}

The system \begin{cases} 2x-6y=12 \cr \cr 4x+6y=24 \end{cases} provides the augmented matrix:

\begin{pmatrix} 2 & -6 & 12 \cr\cr 4 & 6 & 24 \end{pmatrix}

Adding the first row to the second gives:

\begin{pmatrix} 6 & 0 & 36 \cr\cr 4 & 6 & 24 \end{pmatrix}

Dividing the first row by 6 gives:

\begin{pmatrix} 1 & 0 & 6 \cr\cr 4 & 6 & 24 \end{pmatrix}

Subtracting four times the first row from the second gives:

\begin{pmatrix} 1 & 0 & 6 \cr\cr 0 & 6 & 0 \end{pmatrix}

Dividing the second row by 6 gives:

\begin{pmatrix} 1 & 0 &\textcolor{Red}{6 }\cr 0 & 1 & \textcolor{Red}{0} \end{pmatrix}

The system \begin{cases} x-y=3 \cr \cr x+y=7 \end{cases} provides the augmented matrix:

\begin{pmatrix} 1 & -1 & 3 \cr \cr 1 & 1 & 7 \end{pmatrix}

Adding the first row to the second gives:

\begin{pmatrix} 2 & 0 & 10 \cr \cr 1 & 1 & 7 \end{pmatrix}

Dividing the first row by 2 gives:

\begin{pmatrix} 1 & 0 & 5 \cr \cr 1 & 1 & 7 \end{pmatrix}

Subtracting the first row from the second gives:

\begin{pmatrix} 1 & 0 & 5 \cr \cr 0 & 1 & 2 \end{pmatrix}

The solution appears in the matrix:

\begin{pmatrix} 1 & 0 &\textcolor{Red}{5 }\cr 0 & 1 & \textcolor{Red}{2} \end{pmatrix}

The system \begin{cases} 5x-3y=12 \cr \cr -5x+y=10 \end{cases} provides the augmented matrix:

\begin{pmatrix} 5 & -3 & 12 \cr \cr -5 & 1 & 10 \end{pmatrix}

Adding the first row to the second gives:

\begin{pmatrix} 0& -2 & 22 \cr \cr -5 & 1 & 10 \end{pmatrix}

Dividing the first row by -2 gives:

\begin{pmatrix} 0& 1 & -11 \cr \cr -5 & 1 & 10 \end{pmatrix}

Subtracting the first row from the second gives:

\begin{pmatrix} 0& 1 & -11 \cr \cr -5 & 0 & 21 \end{pmatrix}

Interchanging the two rows gives:

\begin{pmatrix} -5 & 0 & 21 \cr \cr0& 1 & -11 \end{pmatrix}

Dividing the first row by -5 gives:

\begin{pmatrix} 1 & 0 & \dfrac{-21}{5} \cr \cr0& 1 & -11 \end{pmatrix}

The solution appears in the matrix:

\begin{pmatrix} 1 & 0 &\textcolor{Red}{\dfrac{-21}{5} }\cr 0 & 1 & \textcolor{Red}{-11} \end{pmatrix}

The system \begin{cases} x-2y=12 \cr \cr -x+y=8 \end{cases} provides the augmented matrix:

\begin{pmatrix} 1 & -2 & 12 \cr \cr -1 & 1 & 8 \end{pmatrix}

Adding the first row to the second gives:

\begin{pmatrix} 1 & -2 & 12 \cr \cr 0 & -1 & 20 \end{pmatrix}

Dividing the second row by -1 gives:

\begin{pmatrix} 1 & -2 & 12 \cr \cr 0 & 1 & -20 \end{pmatrix}

Adding twice the second row to the first gives:

\begin{pmatrix} 1 & 0 & -28 \cr \cr 0 & 1 & -20 \end{pmatrix}

The solution appears in the matrix:

\begin{pmatrix} 1 & 0 &\textcolor{Red}{-28 }\cr 0 & 1 & \textcolor{Red}{-20} \end{pmatrix}

The system \begin{cases} 3x-4y=8 \cr \cr -3x+y=10 \end{cases} provides the augmented matrix:

\begin{pmatrix} 3 & -4 & 8 \cr \cr -3 & 1 & 10 \end{pmatrix}

Adding the first row to the second gives:

\begin{pmatrix} 3 & -4 & 8 \cr \cr 0 & -3 & 18 \end{pmatrix}

Dividing the second row by -3 gives:

\begin{pmatrix} 3 & -4 & 8 \cr \cr 0 & 1 & -6 \end{pmatrix}

Adding four times the second row to the first gives:

\begin{pmatrix} 3 & 0 & -16 \cr \cr 0 & 1 & -6 \end{pmatrix}

Dividing the first row by 3 gives:

\begin{pmatrix} 1 & 0 & \dfrac{-16}{3} \cr \cr 0 & 1 & -6 \end{pmatrix}

The solution appears in the matrix:

\begin{pmatrix} 1 & 0 &\textcolor{Red}{\dfrac{-16}{3} }\cr 0 & 1 & \textcolor{Red}{-6} \end{pmatrix}