The matrix inverse of a matrix $A$, denoted $A^{-1}$, is the matrix such that when multiplied by the matrix A the result is the identity matrix. (The identity matrix is the matrix with ones down the diagonal and zeroes everywhere else.)

For $2\times2$ matrices, if $$A=\left(\begin{array}{cc} a&b\\ c&d\\ \end{array} \right)$$ then $$A^{-1}={1\over\det(A)}\left(\begin{array}{cc} d&-b\\ -c&a\\ \end{array} \right).$$

For $3\times3$ matrices $B$, $B^{-1}$ is the transpose of the matrix made of all cofactors of $B$, divided by the determinant of $B$. This is easier said in symbols, so if $$B=\left(\begin{array}{ccc} a&b&c\\ d&e&f\\ g&h&i\\ \end{array} \right)$$ then $$B^{-1}={1\over\det(B)} \left(\begin{array}{ccc} \left|\begin{array}{cc} e&f\\ h&i\\ \end{array}\right| &-\left|\begin{array}{cc} b&c\\ h&i\\ \end{array}\right| &\left|\begin{array}{cc} b&c\\ e&f\\ \end{array}\right|\\ \\ -\left|\begin{array}{cc} d&f\\ g&i\\ \end{array}\right| &\left|\begin{array}{cc} a&c\\ g&i\\ \end{array}\right|& -\left|\begin{array}{cc} a&c\\ d&f\\ \end{array}\right|\\ \\ \left|\begin{array}{cc} d&e\\ g&h\\ \end{array}\right|& -\left|\begin{array}{cc} a&b\\ g&h\\ \end{array}\right| &\left|\begin{array}{cc} a&b\\ d&e\\ \end{array}\right|\\ \end{array}\right) $$

In both cases, the inverse only exists if the determinant is nonzero.

Try it for yourself for the matrix $$B=\left(\begin{array}{cc} 1&2\\ 3&4\\ \end{array} \right).$$ What is $B^{-1}$? Verify that $BB^{-1}$ is the identity matrix.