Chapter 7.4, Problem 62E

To determine

To write: the solution of the given system of linear equation.

Answer to Problem 62E

The system of linear equation has no solution

Explanation of Solution

Given information:

A system of linear equation is given as

  $\{\begin{array}{l}-x+y-2z=-4\\ 4x+2y-3z=8\\ 2x+4y-7z=1\end{array}$

Concept used:

Co-efficient matrix of a system of linear equation of the form $\{\begin{array}{l}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}$ is given by

  $\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]$

The augmented matrix of the system of linear equation is given by

  $\left[\begin{array}{cccc}{a}_1& b_1& ⋮& c_1\\ a_2& b_2& ⋮& c_2\end{array}\right]$

A matrix is in row-echelon form is

1. If any row containing all elements as zero is in the bottom of the matrix.
2. Each row that does not consist entirely of zeros has the first nonzero entry as 1.
3. For successive nonzero rows the leading 1 in the higher row is further than the leading in the lower row.

And a matrix is in reduced row echelon form if every column that has a leading 1 has zeros in every position above and below its leading 1.

To solve a system of linear equation, the corresponding augmented matrix need to reduce in row echelon form using Gaussian elimination method.

Consider the given system of equation.

  $\{\begin{array}{l}-x+y-2z=-4\\ 4x+2y-3z=8\\ 2x+4y-7z=1\end{array}$

Now, the augmented matrix is given by

  $\left[\begin{array}{ccccc}-1& 1& -2& ⋮& -4\\ 4& 2& -3& ⋮& 8\\ 2& 4& -7& ⋮& 1\end{array}\right]$

Now, to solve the system of equation the corresponding augmented matrix need to reduce in row echelon form using Gaussian elimination method.

Multiply row 1 by $\left(-1\right)$ .

Apply $R_1\to \left(-1\right)R_1,$

  $\left[\begin{array}{ccccc}1& -1& 2& ⋮& 4\\ 4& 2& -3& ⋮& 8\\ 2& 4& -7& ⋮& 1\end{array}\right]$

Multiply row 1 of the matrix by $\left(-4\right)$ and add with row 2.

Multiply row 1 of the matrix by $\left(-2\right)$ and add with row 3.

Apply $R_2\to \left(-4\right)R_1+R_2,R_3\to \left(-2\right)R_1+R_3$

  $\left[\begin{array}{ccccc}1& -1& 2& ⋮& 4\\ 0& 6& -11& ⋮& -8\\ 0& 6& -11& ⋮& -7\end{array}\right]$

Multiply row 2 by $\left(\frac{1}{6}\right)$ .

Apply $R_2\to \left(\frac{1}{6}\right)R_2,$

  $\left[\begin{array}{ccccc}1& -1& 2& ⋮& 4\\ 0& 1& -\frac{11}{6}& ⋮& -\frac{4}{3}\\ 0& 6& -11& ⋮& -7\end{array}\right]$

Multiply row 2 by $-6$ and add with row 3.

Apply $R_3\to \left(-6\right)R_2+R_3$

  $\left[\begin{array}{ccccc}1& -1& 2& ⋮& 4\\ 0& 1& -\frac{11}{6}& ⋮& -\frac{4}{3}\\ 0& 0& 0& ⋮& 1\end{array}\right]$

The third row of the augmented matrix gives us

  $0=1$

Therefore, the system of linear equation has no solution.