Using an Inverse to Solve a Matrix Equation

If a matrix $A$ has an inverse and the matrix equation $AX = B$ has a solution, then the solution is $X = A^{ 1}B$. To solve, you must multiply both sides of the equation by $A^{ 1}$ from the left side: $(A^{ 1})AX = (A^{ 1})B$, which simplifies to $X = A^{ 1}B$.

To solve $\begin{bmatrix} 5 & 7 \\ 2 & 3 \end{bmatrix} X = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, first find the inverse: $A^{ 1} = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$. Then, solve for $X$ by multiplying: $X = A^{ 1}B = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$. The system $2x+y=5$, $3x+2y=8$ becomes $AX=B$, where $A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$.

Solving $AX=B$ is like solving the simple equation $5x=10$. To get $x$, you multiply by the inverse of 5, which is $1/5$. Similarly, to get the matrix $X$, you multiply matrix $B$ by the inverse of matrix $A$. Just remember, matrix multiplication order matters, so you must multiply from the left side: $X = A^{ 1}B$.