Solving Matrix Equations Lesson

In the last lesson, we learned how to find the inverse of a square matrix. Now, we will go a step further and learn how to use the inverse of a matrix to solve a linear system. Let's look at an example.
Example #1: Solve each system. $$3x - 5y=4$$ $$3x + 5y=14$$ We will setup three matrices, A will be a matrix for the coefficients, B will be a matrix for the constants, and X will be a matrix for the variables. $$A=\left[ \begin{array}{cc}3&-5\\ 3&5\end{array}\right]$$ $$B=\left[ \begin{array}{cc}4 \\ 14\end{array}\right]$$ $$X=\left[ \begin{array}{cc}x \\ y\end{array}\right]$$ We can write our system as: $$AX=B$$ Now, we can multiply each side by the inverse of A to isolate X: $$A^{-1}AX=A^{-1}B$$ $$I_2X=A^{-1}B$$ $$X=A^{-1}B$$ In order to find X, our solution matrix, we must find the inverse of A and then multiply by B. $$A=\left[ \begin{array}{cc}3&-5\\ 3&5\end{array}\right]$$ $$A^{-1}=\left[ \begin{array}{cc}\frac{1}{6}&\frac{1}{6}\\ -\frac{1}{10}&\frac{1}{10}\end{array}\right]$$ $$B=\left[ \begin{array}{cc}4 \\ 14\end{array}\right]$$ $$A^{-1}B=X=\left[ \begin{array}{c}3 \\ 1\end{array}\right] $$ X gives us our solution to the system as (3, 1).

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