Question 1133440


{{{A=matrix(2,2, 13 ,-4 ,3,1)}}}


Determinant is not zero, therefore inverse matrix exists:


Write the augmented matrix


{{{matrix(2,4,13,-4,1,	0,
3,	1,	0,	1)}}}


Make the pivot in the 1st column by dividing the 1st row by {{{13}}}



{{{matrix(2,4,1,-4/13,1/13,	0,
3,	1,	0,	1)}}}



Eliminate the 1st column


{{{matrix(2,4,1,	-4/13,	1/13,	0,
0,	25/13,	-3/13,	1)}}}



Make the pivot in the 2nd column by dividing the 2nd row by {{{25/13}}}


{{{matrix(2,4,1,	-4/13,	1/13,	0,
0,	1,	-3/25,	13/2)}}}




Eliminate the 2nd column


{{{matrix(2,4,1,	0,	1/25,	4/25,
0,	1,	-3/25,	13/25)}}}


There is the inverse matrix on the right


{{{matrix(2,4,1,	0,	1/25,	4/25,
0,	1,	-3/25,	13/25)}}}


Result:


{{{matrix(2,2,1/25,	4/25,
-3/25,	13/25)}}}


factor out {{{1/25}}}


{{{A^-1=(1/25)(matrix(2,2,1,4,-3,13))}}}