## Section3.10Inverses

The matrix inverse of a matrix $A\text{,}$ denoted $A^{-1}\text{,}$ 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

\begin{equation} A=\left(\begin{array}{cc} a\amp b\\ c\amp d \end{array} \right)\tag{3.10.1} \end{equation}

then

\begin{equation} A^{-1}={1\over\det(A)}\left(\begin{array}{cc} d\amp -b\\ -c\amp a \end{array} \right)\text{.}\tag{3.10.2} \end{equation}

For $3\times3$ matrices $B\text{,}$ $B^{-1}$ is the transpose of the matrix made of all cofactors of $B\text{,}$ divided by the determinant of $B\text{.}$ This is easier said in symbols, so if

\begin{equation} B=\left(\begin{array}{ccc} a\amp b\amp c\\ d\amp e\amp f\\ g\amp h\amp i \end{array} \right)\tag{3.10.3} \end{equation}

then

\begin{equation} B^{-1}={1\over\det(B)} \left(\begin{array}{ccc} \left|\begin{array}{cc} e\amp f\\ h\amp i \end{array} \right| \amp -\left|\begin{array}{cc} b\amp c\\ h\amp i \end{array} \right| \amp \left|\begin{array}{cc} b\amp c\\ e\amp f \end{array} \right|\\ \\ -\left|\begin{array}{cc} d\amp f\\ g\amp i \end{array} \right| \amp \left|\begin{array}{cc} a\amp c\\ g\amp i \end{array} \right|\amp -\left|\begin{array}{cc} a\amp c\\ d\amp f \end{array} \right|\\ \\ \left|\begin{array}{cc} d\amp e\\ g\amp h \end{array} \right|\amp -\left|\begin{array}{cc} a\amp b\\ g\amp h \end{array} \right| \amp \left|\begin{array}{cc} a\amp b\\ d\amp e \end{array} \right| \end{array} \right)\tag{3.10.4} \end{equation}

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

Try it for yourself for the matrix

\begin{equation} B=\left(\begin{array}{cc} 1\amp 2\\ 3\amp 4 \end{array} \right)\text{.}\tag{3.10.5} \end{equation}

What is $B^{-1}\text{?}$ Verify that $BB^{-1}$ is the identity matrix.