Chapter 16.6, Problem 13WE

To determine

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

  $\begin{array}{l}x+y=6\\ 2x=4\end{array}$

Answer to Problem 13WE

  $x=2,y=4$

Explanation of Solution

Given:

  $\begin{array}{l}x+y=6\\ 2x=4\end{array}$

Formula Used:

1. Determinant formula:
$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-cb$
2. Inverse formula:
${\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
3. Matrix Multiplication formula:

  $\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{cc}E& F\\ G& H\end{array}\right]=\left[\begin{array}{cc}AE+BG& AF+BH\\ CE+DG& CF+DH\end{array}\right]$

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:

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

The determinant of the given matrix is non zero so the matrix is invertible.

The solution to the given system of equations is:

  $\left[\begin{array}{c}x\\ y\end{array}\right]={\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]}^{-1}\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]$

  ${\left[\begin{array}{cc}1& 1\\ 2& 0\end{array}\right]}^{-1}=\frac{1}{-2}\left[\begin{array}{cc}0& -1\\ -2& 1\end{array}\right]$

  $\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-2}\left[\begin{array}{cc}0& -1\\ -2& 1\end{array}\right]\left[\begin{array}{c}6\\ 4\end{array}\right]=\left[\begin{array}{c}2\\ 4\end{array}\right]$

So, answer is:

  $x=2,y=4$