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Solving Systems Using Matrices

I know you don't believe me, so watch this:

Let's find the inverse of A:

A = [ row 1: 2 , -5 row 2: 3 , 4 ]
 

[ row 1: 2 , -5 row 2: 3 , 4 | row 1: 1 , 0 row 2: 0 , 1 ]

 

STEP 1:

-3 times Row 1 ... 2 times Row 2 ... [ row 1: -6 , 15 row 2: 6 , 8 | row 1: -3 , 0 row 2: 0 , 2 ]


 

Row 1 back
to normal

Row 1 +
Row 2

[ row 1: 2 , -5 row 2: 0 , 23 | row 1: 1 , 0 row 2: -3 , 2 ]

 

STEP 2:

23 times Row 1 ... 5 times Row 2 ... [ row 1: 46 , -115 row 2: 0 , 115 | row 1: 23 , 0 row 2: -15 , 10 ]


 

Row 2 +
Row 1

Row 2 back
to normal

[ row 1: 46 , 0 row 2: 0 , 23 | row 1: 8 , 10 row 2: -3 , 2 ]

 

STEP 3:

( 1 / 46 ) times Row 1 ... ( 1 / 23 ) times Row 2 ... [ row 1: 1, 0 row 2: 0 , 1 | row 1: ( 4 / 23 ) , ( 5 / 23 ) row 2: -( 3 / 23 ) , ( 2 / 23 ) ]


A^( -1 ) = [ row 1: ( 4 / 23 ) , ( 5 / 23 ) row 2: -( 3 / 23 ) , ( 2 / 23 ) ]

 

I'll let you check it (rememberA^( -1 ) times A = I).