Section 3.10 Inverses
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.