Bobosharif S. answered 12/20/17
Tutor
4.4
(32)
PhD in Math, MS's in Calulus
Hard to type matrix here but anyway. First the idea: You have matrix A. Now you write this way A|I, I -unit matrix.
Apply row operations to this matrix until the left side is reduced to I.
[10 20 30]|[1 0 0]
[20 40 50]|[0 1 0]
[30 50 60]|]0 0 1]
Next step, apply row operations to the left side. Lets for now, multiply the first row by -2 and and add to the second and then multiply the it by -3 and add to the third row, etc...
At the end you have go get unit matrix in the left and
[1/10 -3/10, 1/5]
[-3/10 3/10, -1/10]
[1/5 -1/10 0]
This is the inverse of your matrix.
\!\(TagBox[RowBox[{"(", "", GridBox[{{FractionBox["1", "10"], RowBox[{"-", FractionBox["3", "10"]}], FractionBox["1", "5"]},{RowBox[{"-", FractionBox["3", "10"]}], FractionBox["3", "10"], RowBox[{"-", FractionBox["1", "10"]}]},{FractionBox["1", "5"], RowBox[{"-", FractionBox["1", "10"]}], "0"}},GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}},GridBoxSpacings->{"Columns" -> {Offset[0.27999999999999997`], {Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> {Offset[0.2], {Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "", ")"}],Function[BoxForm`e$, MatrixForm[BoxForm`e$]]]\)