Chapter 16.6, Problem 15WE

To determine

To Find: Whether the given system has a unique solution. If yes, to find the solution.

  $\begin{array}{l}x+y=1\\ 2x+2y=5\end{array}$

Answer to Problem 15WE

The system does not have a unique solution.

Explanation of Solution

Given:

  $\begin{array}{l}x+y=1\\ 2x+2y=5\end{array}$

Formula Used: $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-cb$

If given system of equations are:

  $\begin{array}{l}{a}_{{}^1}x+b_{{}^1}y=c_1\\ a_{{}^2}x+b_{{}^2}y=c_2\end{array}$

The above system can be represented as:

  $\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]$

  $AX=C$

  $X=A^{-1}C$

The solution to the system of equations is the matrix $X$ .

There exists a unique solution to the system of equations if the matrix $A$ is invertible.

For checking whether a matrix is invertible it is needed to verify that the determinant of the matrix is non-zero or not.

If the determinant of the given matrix is non-zero then the matrix is invertible.

Finding the determinant:

  $\left|\begin{array}{cc}1& 1\\ 2& 2\end{array}\right|=(1)(2)-(2)(1)=0$

The determinant of the given matrix is zero so the matrix is not invertible.

Therefore, the system does not have a unique solution.