Question

Solve the following system of equations by elementary row transformation (by coefficient matrix in echelon form) 2x-y+z=3, x+3y-2z=11 & 3x-2y+4z=1​

Answer 1

Elementary Row Transformation:

Step-by-step explanation:

In Elementary Row Transformation, only the rows of the matrices are transformed and NO changes are made in the columns. These row operations are executed according to a certain set of rules which make sure that the transformed matrix is equivalent to the original matrix. These rules are:

• Any two rows are interchangeable.
• All the elements of any row can be multiplied to any non-zero number.
• All the elements of a row can be added to corresponding elements of another row multiplied by any non-zero constant.

The given system of equations are:

                                     $2x-y+z=3\\x+3y-2z=11\\3x-2y+4z=1$

Then the coefficient matrix is

                                      $\left[\begin{array}{ccc}2&-1&1\\1&3&-2\\3&-2&4\end{array}\right]$

The augmented matrix will be

                                     $\left[\begin{array}{cccc}2&-1&1&3\\1&3&-2&11\\3&-2&4&1\end{array}\right]$

Now, we will do elementary row transformation.

                                     $\left[\begin{array}{cccc}1&-\frac{1}{2} &\frac{1}{2} &\frac{3}{2} \\1&3&-2&11\\3&-2&4&1\end{array}\right]$

                                     $\left[\begin{array}{cccc}1&-\frac{1}{2} &\frac{1}{2} &\frac{3}{2} \\0&\frac{7}{2} &-\frac{5}{2} &\frac{19}{2} \\3&-2&4&1\end{array}\right]$

                                     $\left[\begin{array}{cccc}1&-\frac{1}{2} &\frac{1}{2} &\frac{3}{2} \\0&\frac{7}{2} &-\frac{5}{2} &\frac{19}{2} \\0&-\frac{1}{2} &\frac{5}{2} &-\frac{7}{2} \end{array}\right]$

                                     $\left[\begin{array}{cccc}1&-\frac{1}{2} &\frac{1}{2} &\frac{3}{2} \\0&1 &-\frac{5}{7} &\frac{19}{7} \\0&-\frac{1}{2} &\frac{5}{2} &-\frac{7}{2} \end{array}\right]$

                                     $\left[\begin{array}{cccc}1&0&\frac{1}{7} &\frac{20}{7} \\0&1 &-\frac{5}{7} &\frac{19}{7} \\0&-\frac{1}{2} &\frac{5}{2} &-\frac{7}{2} \end{array}\right]$

                                     $\left[\begin{array}{cccc}1&0&\frac{1}{7} &\frac{20}{7} \\0&1 &-\frac{5}{7} &\frac{19}{7} \\0&0&\frac{30}{7} &-\frac{30}{7} \end{array}\right]$

                                     $\left[\begin{array}{cccc}1&0&\frac{1}{7} &\frac{20}{7} \\0&1 &-\frac{5}{7} &\frac{19}{7} \\0&0&1 &-1 \end{array}\right]$

                                     $\left[\begin{array}{cccc}1&0&0 &3 \\0&1 &-\frac{5}{7} &\frac{19}{7} \\0&0&1 &-1 \end{array}\right]$

                                     $\left[\begin{array}{cccc}1&0&0 &3 \\0&1 &0 &2\\0&0&1 &-1 \end{array}\right]$

Therefore, the solutions are $x=3,y=2,z=-1$