How do you find the inverse of #[(8,-3),(4,-5)]#?

Initial Matrix With Identity:
#[ ( 8.00000, -3.00000, 1.00000, 0.00000), ( 4.00000, -5.00000, 0.00000, 1.00000) ]#

Pivot Action #n#
pivot row = n; pivot column = n; pivot entry augmented matrix entry at (n,n)

1. convert pivot n row so pivot entry = 1
2. adjust non-pivot rows so entries in pivot column = 0

Pivot #color(black)(1)#
Pivot Row 1 reduced by dividing all entries by 8.00 so pivot entry = 1
#[ ( 1.00000, -0.37500, 0.12500, 0.00000), ( 4.00000, -5.00000, 0.00000, 1.00000) ]#

Non-pivot rows reduced for pivot column
by subtracting appropriate multiple of pivot row 1 from each non-pivot row

#[ ( 1.00000, -0.37500, 0.12500, 0.00000), ( 0.00000, -3.50000, -0.50000, 1.00000) ]#

Pivot #color(black)(2)#
Pivot Row 2 reduced by dividing all entries by -3.50 so pivot entry = 1
#[ ( 1.00000, -0.37500, 0.12500, 0.00000), ( 0.00000, 1.00000, 0.14286, -0.28571) ]#

Non-pivot rows reduced for pivot column
by subtracting appropriate multiple of pivot row 2 from each non-pivot row

#[ ( 1.00000, 0.00000, color(red)(0.17857), color(red)(-0.10714)), ( 0.00000, 1.00000, color(red)(0.14286), color(red)(-0.28571)) ]#

The last two columns are the required identity matrix.

With a bit of effort the decimal fractions can be converted to the fractions shown in the answer above.