Solving the linear two–by–two system Eq. (5.1) for the components
x,
y of the
vector x,
is equivalent to the matrix equation
\begin{eqnarray}
Ax = \left (\array{
a&b\cr
c &d } \right )\left (\array{
x\cr
y } \right ) = \left (\array{
ax + by\cr
cx + dy } \right ) = v = \left (\array{
e\cr
f } \right ).& & %&(5.32)\\
\end{eqnarray}
We recognise that in order to explicitely solving this for
x, we have to invert
the operation A.
5.4.2 Definition and Theorem
The inverse {A}^{−1}of atwo–by–two matrix A is
defined as the matrix fulfilling {A}^{−1}A = A{A}^{−1} = I,\quad I = \left (\array{
1&0
\cr
0&1 } \right )
The determinant \mathop{det}(A) of
a two–by–two matrix A
is defined as \mathop{det}\left (\array{
a&b\cr
c &d } \right ) ≡\left |\array{
a&b\cr
c &d } \right | := ad−cb.
Theorem Consider the two–by–two matrix
\begin{eqnarray}
A = \left (\array{
a&b\cr
c &d } \right ).& & %&(5.33)\\
\end{eqnarray}
If the determinant of A
is non–zero, i.e. \mathop{det}(A) = ad − cb\mathrel{≠}0,
the inverse of A
exists and is given by
\begin{eqnarray}{
A}^{−1} = {1\over
ad − cb}\left (\array{
d & − b\cr
− c & a } \right ) ≡\left (\array{
{d\over
ad−cb}& {−b\over
ad−cb}
\cr
{−c\over
ad−cb}& {a\over
ad−cb} } \right ).& & %&(5.34)\\
\end{eqnarray}
For the proof of this, we just multiply A
with {A}^{−1}
and {A}^{−1}
with A:
\begin{eqnarray}
A{A}^{−1} = \left (\array{
a&b
\cr
c&d } \right ) {1\over
ad − cb}\left (\array{
d & − b\cr
− c & a } \right ) = {1\over
ad − cb}\left (\array{
ad − bc& − ab + ba\cr
cd − dc & − cb + da } \right ) = \left (\array{
1&0\cr
0 &1 } \right ).& & %&(5.35)\\
\end{eqnarray}