Consider a matrix of the form #A=((a,b),(c,d))# and let its inverse be of the form #A^{-1}=((\alpha, \beta),(\gamma, \delta))#.

Our job is to solve for #\alpha, \beta, \gamma # and #\delta# in terms of #a, b, c# and #d#.

Use the fact that #A\timesA^{-1}=I# to get the four equations to solve the four unknowns.

#((a,b),(c,d)).((\alpha,\beta),(\gamma,\delta))=((1,0),(0,1))#

#((a\alpha+b\gamma, a\beta+b\delta),(c\alpha+d\gamma,c\beta+d\delta))=((1,0),(0,1))#

Comparing component wise, we get four equations required to solve the four unknowns

#a\alpha + b\gamma = 1# ....... (**1**)#\quad\qquad # #a\beta + b\delta = 0# ........ (**2**)

#c\alpha + d\gamma = 0# ....... (**3**) #\qquad# #c\beta + d\delta = 1# ......... (**4**)

**Step 1**: Use (**3**) to eliminate #\gamma# in (**1**) and solve for #\alpha#.

From (**3**), #\gamma = - c/d\alpha#, substitute this in (**1**)

#\alpha(a-b c/d)=1 \qquad \rightarrow \alpha = d/(ad-bc)#

#\gamma = -c/d\alpha = -c/(ad-bc)#

**Step 2**: Use (**2**) to eliminate #\delta# in (**4**) and solve for #\beta#.

From (**2**), #\delta = - a/b\beta#, substitute this in (**4**)

# \beta(c-d a/b)=1 \qquad \rightarrow \beta = -b/(ad-bc)#

#\delta = -a/b\beta = a/(ad-bc)#

**Step 3**: Recognise that #(ad-bc)# is the determinant value of the matrix #A#

#|A| = |(a,b),(c,d)|=ad-bc#

#\alpha = d/|A|; \qquad \beta = -b/|A|; \qquad \gamma = -c/|A|; \qquad \delta = a/|A|#.

Therefore, #A^{-1}=1/|A|((d,-b),(-c,a))#.

You can see that this works only as long as #|A|\ne0#

If #|A|=0# then the matrix is said to be **singular** and non-invertible.