Inverse of a "Nondegenerate" 2×2 Matrix

Define the following $2\times 2$ matrix: $$A = \left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right]$$ We want to find a $2\times 2$ matrix $$A^{-1} = \left[ {\begin{array}{cc} w & x \\ y & z \\ \end{array} } \right]$$ such that $A^{-1}A=AA^{-1}=I$, where $$I = \left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]$$ $I$ is known as the identity matrix, because for any matrix $A$, $$AI=IA=A$$ $A^{-1}$ is called the inverse of $A$.

Let us evaluate the product $AA^{-1}$: $$AA^{-1} = \left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right] \left[ {\begin{array}{cc} w & x \\ y & z \\ \end{array} } \right]$$ $$AA^{-1} = \left[ {\begin{array}{cc} aw+by & ax+bz \\ cw+dy & cx+dz \\ \end{array} } \right]$$ We obtain the following system of equations: $$aw+by=1$$ $$ax+bz=0$$ $$cw+dy=0$$ $$cx+dz=1$$ It is quite tedious to solve this system of equations, but it is quite straightforward, since it is just $4$ variables in $4$ linear equations. Try to solve the system of equations yourself before moving on.

Here is the solution to the system of equations: $$w=\frac{d}{ad-bc}$$ $$x=\frac{-b}{ad-bc}$$ $$y=\frac{-c}{ad-bc}$$ $$z=\frac{a}{ad-bc}$$

Thus, we have solved for $A^{-1}$: $$A^{-1} = \frac{1}{ad-bc} \left[ {\begin{array}{cc} d & -b \\ -c & a \\ \end{array} } \right]$$ Notice that we only solved for $A^{-1}$ in the equation $AA^{-1}=I$. You are encouraged to work through the case $A^{-1}A=I$ and observe that you indeed get exactly the same answer for $A^{-1}$.

Therefore, if $ad-bc\neq 0$, then the matrix $A$ has an inverse, and it can be described as above. If $ad-bc=0$, then the matrix $A$ has no inverse.

The quantity $ad-bc$ is called the determinant of the matrix $A$, abbreviated $\det(A)$. The determinant can be expressed in a symbolic statement in one of the following ways: $$\det(A)=ad-bc$$ $$\left| {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right| = ad-bc$$ We say that a matrix $A$ is invertible if $\det(A)\neq 0$. If $\det(A)=0$, then we say that $A$ is singular or degenerate.