Chapter 7.6, Problem 43E

To determine

To solve: The given system of linear equations using an inverse matrix.

Answer to Problem 43E

The given system of linear equations has no solution.

Explanation of Solution

Given information :

The system of linear equations:

  $\begin{array}{l}-0.4x+0.8y=1.6\\ 2x-4y=5\end{array}$

Formula used :

For a $2\times 2$ matrix of the form:

  $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

The matrix $A$ is invertible if and only if $ad-bc\ne 0$ . Moreover, if $ad-bc\ne 0$ , the inverse is given by

  $A^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$ .

To solve a system of linear equations:

If $A$ is an invertible matrix, the system of linear equations represented by $AX=B$ has a unique solution given by $X=A^{-1}B$ .

Calculation :

As per the problem,

The given system of linear equations:

  $\begin{array}{l}-0.4x+0.8y=1.6\\ 2x-4y=5\end{array}$

Write the system in the matrix form $AX=B$

  $\left[\begin{array}{cc}-0.4& 0.8\\ 2& -4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}1.6\\ 5\end{array}\right]$

Where $A=\left[\begin{array}{cc}-0.4& 0.8\\ 2& -4\end{array}\right]$ , $X=\left[\begin{array}{c}x\\ y\end{array}\right]$ and $B=\left[\begin{array}{c}1.6\\ 5\end{array}\right]$

Find $A^{-1}$ using the above stated formula:

Since, $ad-bc=(-0.4)(-4)-(0.8)(2)=0$

The matrix $A$ is not invertible.

Hence, the given system of linear equations has no solution.