Chapter 7.3, Problem 43E

Solution

To identify: The system of equations by finding the reduced row echelon form for the augmented matrix.

  $\begin{array}{c}x+y-3z=1\\ x-z-w=2\\ 2x+y-4z-w=3\end{array}$

The value of $x,y,z,w$ is $\left(z+w+2,2z-w-1,z,w\right)$ .

Given information:

The given equation is.

  $\begin{array}{c}x+y-3z=1\\ x-z-w=2\\ 2x+y-4z-w=3\end{array}$

Explanation:

Consider the given information.

To obtain the augmented matrix of a system, write out the coefficients as entries, aligning those with the same variables, and placing 0 for variables that are not present. The last column corresponds to the constants.

  $\begin{array}{c}1x+1y-3z=1\\ 1x-1z-1w=2\\ 2x+1y-4z-1w=3\end{array}$

So, its corresponding augmented matrix is:

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

Apply elementary row operation to obtain a reduced row echelon from of the matrix.

  $\begin{array}{l}\left[\begin{array}{ccccc}1& 1& -3& 0& 1\\ 1& 0& -1& -1& 2\\ 2& 1& -4& -1& 3\end{array}\right]R_1-R_2\to \left[\begin{array}{ccccc}1& 1& -3& 0& 1\\ 0& 1& -2& 1& -1\\ 2& 1& -4& -1& 3\end{array}\right]\left(2\right)R_1-R_3\to \left[\begin{array}{ccccc}1& 1& -3& 0& 1\\ 0& 1& -2& 1& -1\\ 0& 1& -2& 1& -1\end{array}\right]\\ \left[\begin{array}{ccccc}1& 1& -3& 0& 1\\ 0& 1& -2& 1& -1\\ 0& 1& -2& 1& -1\end{array}\right]R_2-R_3\to \left[\begin{array}{ccccc}1& 1& -3& 0& 1\\ 0& 1& -2& 1& -1\\ 0& 0& 0& 0& 0\end{array}\right]\left(-1\right)R_2+R_1\to \left[\begin{array}{ccccc}1& 0& -1& -1& 2\\ 0& 1& -2& 1& -1\\ 0& 0& 0& 0& 0\end{array}\right]\end{array}$

This corresponds to the system of equation.

  $\begin{array}{c}x-z-w=2\\ y-2z+w=-1\end{array}$

The last equation is a true statement which tells us that there are infinitely which tells us that there are infinitely many solution Having four variables but only two equation tells us that there are infinitely many solution since can use any real value for $z\text{and}\text{w}$ to obtain its corresponding $x\text{and}y$ values solving for $x\text{and}y$ in term of $z\text{and}\text{w}$ the solution of the system is $\left(z+w+2,2z-w-1,z,w\right)$ .

So, the value of $x,y,z,w$ is $\left(z+w+2,2z-w-1,z,w\right)$ .