+0

# matrix

0
357
1 Jan 6, 2016

#1
+10

Note that we have this matrix equation :

Ax  = b        where A =   [5  -2             and     b = [ 2

2  -1]                             -4]

If I multiply both sides of the equation [on the left] by the inverse of A, we have

(A-1) (A )x  =( A-1 ) b

But    (A-1) (A)  is just the Identity matrix,  I

So.....the left side just simplifies to  x .......and we have

x = A-1 b

So.....we just   need to find the inverse of A  = A-1

Note, for a 2 x 2 matrix  in the form of      [ a     b

c     d ]

The inverse is given by :    [1/ (ad - bc)] *  [  d   - b    =

- c    a]

[1 / (-5 +4)] * [ -1  2     =     [-1] [ -1  2

-2  5]                   -2  5]

So.....the inverse of A   =  [1    - 2

2    -5]

So......

x = A-1 * b     =     [ 1    - 2   *    [2             =

2    -5 ]       -4]

[1*2  +    -2*-4      =    [ 2 +8        =  [10

2*2  +  - 5* -4 ]           4 +20]           24]

So....the last matrix is the answer.......

And that's it !!!!   Jan 6, 2016
edited by CPhill  Jan 6, 2016
edited by CPhill  Jan 6, 2016
edited by CPhill  Jan 6, 2016
edited by CPhill  Jan 6, 2016

#1
+10

Note that we have this matrix equation :

Ax  = b        where A =   [5  -2             and     b = [ 2

2  -1]                             -4]

If I multiply both sides of the equation [on the left] by the inverse of A, we have

(A-1) (A )x  =( A-1 ) b

But    (A-1) (A)  is just the Identity matrix,  I

So.....the left side just simplifies to  x .......and we have

x = A-1 b

So.....we just   need to find the inverse of A  = A-1

Note, for a 2 x 2 matrix  in the form of      [ a     b

c     d ]

The inverse is given by :    [1/ (ad - bc)] *  [  d   - b    =

- c    a]

[1 / (-5 +4)] * [ -1  2     =     [-1] [ -1  2

-2  5]                   -2  5]

So.....the inverse of A   =  [1    - 2

2    -5]

So......

x = A-1 * b     =     [ 1    - 2   *    [2             =

2    -5 ]       -4]

[1*2  +    -2*-4      =    [ 2 +8        =  [10

2*2  +  - 5* -4 ]           4 +20]           24]

So....the last matrix is the answer.......

And that's it !!!!   CPhill Jan 6, 2016
edited by CPhill  Jan 6, 2016
edited by CPhill  Jan 6, 2016
edited by CPhill  Jan 6, 2016
edited by CPhill  Jan 6, 2016