Question 174894
Find the inverse of A = {{{(matrix(2,2,-3,1,-5,5))}}} if it exists.
<pre><font size = 4 color = "indigo"><b>
Method 1:

Start with the matrix:
{{{ matrix(1,3, A, "=", (matrix(2,2,-3,1,-5,5))  ) }}}

Augment it on the right with the identity matrix, {{{(matrix(2,2,1,0,0,1)))}}},
like this

{{{(matrix(2,5,
-3,1,"|",1,0,
-5,5,"|",0,1))}}}
 
Now we do row operations to get the identity on the left:

To get a zero where the -5 is,

Add -5 times row 1 to 3 times row 2.  It's a good idea
to write the number you're multiplying row by, on
the far left:

{{{matrix(2,1,-5,3)}}}{{{(matrix(2,5,
-3,1,"|",1,0,
-5,5,"|",0,1))}}}

The result is:

{{{(matrix(2,5,
-3,1,"|",1,0,
0,10,"|",-5,3))}}}

Next we get a zero where the 1 is in row 1, column 2
by adding 1 times row 2 to -10 times row 1.

{{{matrix(2,1,-10,1)}}}{{{(matrix(2,5,
-3,1,"|",1,0,
0,10,"|",-5,3))}}}

The result is:

{{{(matrix(2,5,
30,0,"|",-15,3,
0,10,"|",-5,3))}}}
 
Get a 1 where the 30 is by multiplying row 1
by {{{1/30}}}

Get a 1 where the 10 is by multiplying row 2
by {{{1/10}}}

{{{matrix(2,1,1/30,1/10)}}}{{{(matrix(2,5,
30,0,"|",-15,3,
0,10,"|",-5,3))}}}

The result is:

{{{(matrix(2,5,
1,0,"|",-1/2,1/10,
0,1,"|",-1/2,3/10))}}}
 
So the inverse of {{{A}}} or {{{A^(-1)}}} is the part on the right:

{{{ matrix(1,3, A^(-1), "=", (matrix(2,2,-1/2,1/10,-1/2,3/10))  ) }}}

That method works for all matrices.

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Here's a method that works only for 2x2 matrices:

Method 2:

Start with the matrix:

{{{(matrix(2,2,-3,1,-5,5))}}}

Find the determinant of the matrix:

{{{matrix(1,7,abs( matrix(2,2,-3,1,-5,5) ), "=", (-3)(5)-(1)(-5), "=", -15+5, "=", -10 )}}}

Swap the upper left and lower right elements:

{{{(matrix(2,2,5,1,-5,-3))}}}

Change the signs of the upper right and lower left elements:

{{{(matrix(2,2,5,-1,5,-3))}}}

Divide every element by the -10 which was the determinant of the
original matrix:

{{{ (matrix(2,2,5/(-10),(-1)/(-10),5/(-10),(-3)/(-10)))}}}

Simplify the fractions:

{{{ matrix(1,3, A^(-1), "=", (matrix(2,2,-1/2,1/10,-1/2,3/10))  ) }}}

Edwin</pre>