Chapter 7.3, Problem 72E

To determine

To find: The values of $x,y,\text{and }\lambda$ that satisfy the given system.

Answer to Problem 72E

The values of $x,y,\text{and }\lambda$ that satisfy the given system are $25,50,\text{and}-51$ , respectively.

Explanation of Solution

Given information:

The system of equations is

  $\{\begin{array}{c}2+2y+2\lambda =0\\ 2x+1+\lambda =0\\ 2x+y-100=0\end{array}$

Calculations:

The steps of solving the system of equations using Row-Echelon Form and Back substitutions are as follows:

Rewriting the system, we obtain

  $\{\begin{array}{c}2y+2\lambda =-2\\ 2x+\lambda =-1\\ 2x+y=100\end{array}$

Interchanging the first and second equations, we get

  $\{\begin{array}{c}2x+\lambda =-1\\ 2y+2\lambda =-2\\ 2x+y=100\end{array}$

Subtracting the third equation from the first equation to produce a new third equation, we get

  $\{\begin{array}{c}2x+\lambda =-1\\ 2y+2\lambda =-2\\ -y+\lambda =-101\end{array}$

Now, adding 2 times the third equation with the second equation to produce a new third equation, we get

  $\{\begin{array}{c}2x+\lambda =-1\\ 2y+2\lambda =-2\\ 4\lambda =-204\end{array}$

Solving $4\lambda =-204$ for $\lambda$ , we obtain $\lambda =-51$ .

Substituting this value in the second equation, we get

  $\begin{array}{l}2y+2(-51)=-2\\ \Rightarrow 2y-102=-2\\ \Rightarrow 2y=100\\ \Rightarrow y=50\end{array}$

Now, substituting the above finding values in the first equation, we get

  $\begin{array}{l}2x-51=-1\\ \Rightarrow 2x=50\\ \Rightarrow x=25\end{array}$

Conclusion:

Therefore, the values of $x,y,\text{and }\lambda$ that satisfy the given system are $25,50,\text{and}-51$ , respectively.