Step 1

To find the solution of system of equations reduce the matrix \(\displaystyle{\left[{A}{\mid}{B}\right]}\) where A is the matrix formed by the coefficients of LHS of the equations and B is the matrix formed by the RHS of the equations.

Step 2

Find the matrix \(\displaystyle{\left[{A}{\mid}{B}\right]}\) from the given system of equations and row reduce it.

\(\displaystyle{\left[\begin{array}{ccc|c} {1}&-{5}&{2}&{1}\\{3}&-{4}&{2}&-{1}\end{array}\right]}{R}_{{2}}\rightarrow{R}_{{2}}-{3}{R}_{{1}}\)

\(\displaystyle{\left[\begin{array}{ccc|c} {1}&-{5}&{2}&{1}\\{0}&{11}&-{4}&-{4}\end{array}\right]}{R}_{{2}}\rightarrow\frac{{R}_{{2}}}{{11}}\)

\(\displaystyle{\left[\begin{array}{ccc|c} {1}&-{5}&{2}&{1}\\{0}&{1}&-\frac{{4}}{{11}}&-\frac{{4}}{{11}}\end{array}\right]}{R}_{{1}}\rightarrow{R}_{{1}}+{5}{R}_{{2}}\)

\(\displaystyle{\left[\begin{array}{ccc|c} {1}&{0}&\frac{{2}}{{11}}&-\frac{{9}}{{11}}\\{0}&{1}&-\frac{{4}}{{11}}&-\frac{{4}}{{11}}\end{array}\right]}\)

Step 3

Form the equation from the above matrix.

\(\displaystyle{x}+\frac{{2}}{{11}}{z}=-\frac{{9}}{{11}}\)

\(\displaystyle{y}-\frac{{4}}{{11}}{z}=-\frac{{4}}{{11}}\)

Step 4

Let z=t. Find the value of x and y.

\(\displaystyle{x}+\frac{{2}}{{11}}{\left({t}\right)}=-\frac{{9}}{{11}}\)

\(\displaystyle{x}=-\frac{{9}}{{11}}-\frac{{{2}{t}}}{{11}}\)

\(\displaystyle{y}-\frac{{4}}{{11}}{\left({t}\right)}=-\frac{{4}}{{11}}\)

\(\displaystyle{y}=-\frac{{4}}{{11}}+\frac{{{4}{t}}}{{11}}\)

Step 5

Answer: For random z=t the solution set will be,

\(\displaystyle{\left\lbrace{\left(\frac{{-{9}}}{{11}}-\frac{{{2}{t}}}{{11}},\frac{{-{4}}}{{11}}+\frac{{{4}{t}}}{{11}}\right)},{t}\right.}\rbrace\)

To find the solution of system of equations reduce the matrix \(\displaystyle{\left[{A}{\mid}{B}\right]}\) where A is the matrix formed by the coefficients of LHS of the equations and B is the matrix formed by the RHS of the equations.

Step 2

Find the matrix \(\displaystyle{\left[{A}{\mid}{B}\right]}\) from the given system of equations and row reduce it.

\(\displaystyle{\left[\begin{array}{ccc|c} {1}&-{5}&{2}&{1}\\{3}&-{4}&{2}&-{1}\end{array}\right]}{R}_{{2}}\rightarrow{R}_{{2}}-{3}{R}_{{1}}\)

\(\displaystyle{\left[\begin{array}{ccc|c} {1}&-{5}&{2}&{1}\\{0}&{11}&-{4}&-{4}\end{array}\right]}{R}_{{2}}\rightarrow\frac{{R}_{{2}}}{{11}}\)

\(\displaystyle{\left[\begin{array}{ccc|c} {1}&-{5}&{2}&{1}\\{0}&{1}&-\frac{{4}}{{11}}&-\frac{{4}}{{11}}\end{array}\right]}{R}_{{1}}\rightarrow{R}_{{1}}+{5}{R}_{{2}}\)

\(\displaystyle{\left[\begin{array}{ccc|c} {1}&{0}&\frac{{2}}{{11}}&-\frac{{9}}{{11}}\\{0}&{1}&-\frac{{4}}{{11}}&-\frac{{4}}{{11}}\end{array}\right]}\)

Step 3

Form the equation from the above matrix.

\(\displaystyle{x}+\frac{{2}}{{11}}{z}=-\frac{{9}}{{11}}\)

\(\displaystyle{y}-\frac{{4}}{{11}}{z}=-\frac{{4}}{{11}}\)

Step 4

Let z=t. Find the value of x and y.

\(\displaystyle{x}+\frac{{2}}{{11}}{\left({t}\right)}=-\frac{{9}}{{11}}\)

\(\displaystyle{x}=-\frac{{9}}{{11}}-\frac{{{2}{t}}}{{11}}\)

\(\displaystyle{y}-\frac{{4}}{{11}}{\left({t}\right)}=-\frac{{4}}{{11}}\)

\(\displaystyle{y}=-\frac{{4}}{{11}}+\frac{{{4}{t}}}{{11}}\)

Step 5

Answer: For random z=t the solution set will be,

\(\displaystyle{\left\lbrace{\left(\frac{{-{9}}}{{11}}-\frac{{{2}{t}}}{{11}},\frac{{-{4}}}{{11}}+\frac{{{4}{t}}}{{11}}\right)},{t}\right.}\rbrace\)