Chapter 7.6, Problem 40E

To determine

To solve: the system of linear equation-

  $\begin{array}{l}x-2y=1\\ 2x-3y=-2\end{array}$

Answer to Problem 40E

The solution of the given system of linear equation is:

  $\begin{array}{l}x=-7\\ y=-4\end{array}$

Explanation of Solution

Given information:

The following equations:

  $\begin{array}{l}x-2y=1\\ 2x-3y=-2\end{array}$

Formula used:

Inverse of a $2\times 2$ matrix is given by,

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

when $ad-bc\ne 0$ .

Calculation:

Begin by writing the system as $AX=B$

  $\left[\begin{array}{cc}1& -2\\ 2& -3\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}1\\ -2\end{array}\right]$

Then use the given formula to find $A^{-1}$ .

For matrix $A$ , apply the formula for inverse of a $2\times 2$ matrix to obtain:

  $\begin{array}{l}ad-bc=(1)(-3)-(-2)(2)\text{[Substitute]}\\ =-3+4\text{[Multiply]}\\ =1\text{[Simplify]}\end{array}$

Because this quantity is not zero, the inverse is formed by interchanging the entries on the main diagonal, changing the signs of the other two entries, and multiplying by scalar $1$ as follows,

  $\begin{array}{l}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right][\text{Formula]}\\ =\frac{1}{1}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\text{[Substitute]}\\ \text{=}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\left[\text{Multiply by scalar }1\right]\end{array}$

Finally, multiply $B$ by $A^{-1}$ on the left to obtain the solution.

  $\begin{array}{l}X=A^{-1}B\\ =\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\left[\begin{array}{c}1\\ -2\end{array}\right]\text{[Substitute]}\\ =\left[\begin{array}{c}-3-4\\ -2-2\end{array}\right][\text{Multiply]}\\ =\left[\begin{array}{c}-7\\ -4\end{array}\right]\text{[Simplify]}\end{array}$

Hence, the solution of the given system of linear equation is

  $\begin{array}{l}x=-7\\ y=-4\end{array}$