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=\begin{pmatrix}
a\amp b\\
c\amp d
\end{pmatrix}\tag{3.10.1}
\end{equation}
then
\begin{equation}
A^{-1}={1\over\det(A)}\begin{pmatrix}
d\amp -b\\
-c\amp a
\end{pmatrix}\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=\begin{pmatrix}
a\amp b\amp c\\
d\amp e\amp f\\
g\amp h\amp i
\end{pmatrix}\tag{3.10.3}
\end{equation}
then
\begin{equation}
B^{-1}={1\over\det(B)}
\begin{pmatrix}
\begin{vmatrix}
e\amp f\\
h\amp i
\end{vmatrix}
\amp -\begin{vmatrix}
b\amp c\\
h\amp i
\end{vmatrix}
\amp \begin{vmatrix}
b\amp c\\
e\amp f
\end{vmatrix}\\
\\
-\begin{vmatrix}
d\amp f\\
g\amp i
\end{vmatrix}
\amp \begin{vmatrix}
a\amp c\\
g\amp i
\end{vmatrix}\amp
-\begin{vmatrix}
a\amp c\\
d\amp f
\end{vmatrix}\\
\\
\begin{vmatrix}
d\amp e\\
g\amp h
\end{vmatrix}\amp
-\begin{vmatrix}
a\amp b\\
g\amp h
\end{vmatrix}
\amp \begin{vmatrix}
a\amp b\\
d\amp e
\end{vmatrix}
\end{pmatrix}\tag{3.10.4}
\end{equation}
In both cases, the inverse only exists if the determinant is nonzero.
Checkpoint 3.7. Try it yourself: Inverse.
Suppose that
\begin{equation}
Q=\begin{pmatrix}
1\amp 2\\
3\amp 4
\end{pmatrix}\text{.}\tag{3.10.5}
\end{equation}
What is \(Q^{-1}\text{?}\) Verify that \(QQ^{-1}\) is the identity matrix.
Solution.
Using
(3.10.2), we have
\(\det(Q)=(1)(4)-(2)(3)=-2\text{,}\) so
\begin{equation}
Q^{-1}
= -\frac12 \begin{pmatrix}4\amp -2\\-3\amp 1\end{pmatrix}
= \begin{pmatrix}-2\amp 1\\\frac32\amp -\frac12\end{pmatrix}\tag{3.10.6}
\end{equation}
so that
\begin{align}
QQ^{-1}
\amp= \begin{pmatrix}1\amp 2\\3\amp 4\end{pmatrix}
\begin{pmatrix}-2\amp 1\\\frac32\amp -\frac12\end{pmatrix}\notag\\
\amp= \begin{pmatrix}
(1)(-2)+(2)(\frac32)\amp(1)(1)+(2)(-\frac12) \\
(3)(-2)+(4)(\frac32)\amp(3)(1)+(4)(-\frac12) \\
\end{pmatrix}
= \begin{pmatrix}1\amp0\\0\amp1\end{pmatrix}\tag{3.10.7}
\end{align}
