Q67. - Algebra 2

Question

Solve each system of equations by using inverse matrices.

$\begin{array}{l}x+4y=9\\ 3x+2y=-3\end{array}$

Step-by-Step Solution

Answer

The solution to the given system of equations is$(-3,3)$  .

Step 1. Given Information.

Given to solve the below system of equations by using inverse matrices

$\begin{array}{l}x+4y=9\\ 3x+2y=-3\end{array}$

Step 2. Explanation .

The matrix equation for the system of equations is:

$\left[\begin{array}{cc}1& 4\\ 3& 2\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}9\\ -3\end{array}\right]$

Where$A=\left[\begin{array}{cc}1& 4\\ 3& 2\end{array}\right]$  , $X=\left[\begin{array}{c}x\\ y\end{array}\right]$  and  $B=\left[\begin{array}{c}9\\ -3\end{array}\right]$

The inverse of coefficient matrix A is given by:$\begin{array}{l}A=\left[\begin{array}{cc}1& 4\\ 3& 2\end{array}\right]\\ A^{-1}=\frac{1}{\left(1\right)\cdot \left(2\right)-\left(4\right)\cdot \left(3\right)}\left[\begin{array}{cc}2& -4\\ -3& 1\end{array}\right]\\ A^{-1}=\frac{1}{2-12}\left[\begin{array}{cc}2& -4\\ -3& 1\end{array}\right]\\ A^{-1}=\frac{1}{-10}\left[\begin{array}{cc}2& -4\\ -3& 1\end{array}\right]\end{array}$

Multiplying each side of the matrix equation by the inverse matrix:

$\begin{array}{l}\frac{1}{-10}\left[\begin{array}{cc}2& -4\\ -3& 1\end{array}\right]\cdot \left[\begin{array}{cc}1& 4\\ 3& 2\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-10}\left[\begin{array}{cc}2& -4\\ -3& 1\end{array}\right]\cdot \left[\begin{array}{c}9\\ -3\end{array}\right]\\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-10}\left[\begin{array}{c}\left(2\right)\left(9\right)+(-4)(-3)\\ (-3)\left(9\right)+\left(1\right)(-3)\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-10}\left[\begin{array}{c}18+12\\ -27-3\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-10}\left[\begin{array}{c}30\\ -30\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-3\\ 3\end{array}\right]\end{array}$

Step 3. Conclusion .

Hence, the solution to the given system of equations is $(-3,3)$ .