SOLUTION: Use the fact that if {{{A=(matrix(2,2,a,b,c,d))}}} then {{{A^(-1) = (1/(ad-bc))(matrix(2,2,d,-b,-c,a))}}} to find the inverse of each matrix, if possible.

Algebra ->  Matrices-and-determiminant -> SOLUTION: Use the fact that if {{{A=(matrix(2,2,a,b,c,d))}}} then {{{A^(-1) = (1/(ad-bc))(matrix(2,2,d,-b,-c,a))}}} to find the inverse of each matrix, if possible.      Log On


   



Question 274093: Use the fact that if A=%28matrix%282%2C2%2Ca%2Cb%2Cc%2Cd%29%29
then
A%5E%28-1%29+=+%281%2F%28ad-bc%29%29%28matrix%282%2C2%2Cd%2C-b%2C-c%2Ca%29%29 to find the inverse of each matrix, if possible.

Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Use the fact that if A=%28matrix%282%2C2%2Ca%2Cb%2Cc%2Cd%29%29
then
A%5E%28-1%29+=+%281%2F%28ad-bc%29%29%28matrix%282%2C2%2Cd%2C-b%2C-c%2Ca%29%29 to find the inverse of each matrix, if possible.

-------------------------------------

A=+%28matrix%282%2C2%2C0%2C3%2C4%2C-2%29%29

You just need to learn how to substitute numbers for letters,
which is very easy to do.  All you do is:

1. Put 0 or (0) everywhere there is an "a",
2. Put 3 or (3) everywhere there is a "b".
3. Put 4 or (4) everywhere there is a "c".
4. Put -2 or (-2) everywhere there is a "d".

So comparing the matrix

A=%28matrix%282%2C2%2Ca%2Cb%2Cc%2Cd%29%29

to your matrix,

A=+%28matrix%282%2C2%2C0%2C3%2C4%2C-2%29%29

then do the substituting in

A%5E%28-1%29+=+%281%2F%28ad-bc%29%29%28matrix%282%2C2%2Cd%2C-b%2C-c%2Ca%29%29

and you will have:

  

which is the inverse of that matrix, unsimplified.

Now you only have to simplify everything:

A%5E%28-1%29+=+%281%2F%280-12%29%29%28matrix%282%2C2%2C-2%2C-3%2C-4%2C0%29%29

A%5E%28-1%29+=+%28-1%2F12%29%28matrix%282%2C2%2C-2%2C-3%2C-4%2C0%29%29

Now distribute by multiplying the -1%2F12 by every 
element in the matrix.




Then simplify

A%5E%28-1%29+=+%28matrix%282%2C2%2C1%2F6%2C1%2F4%2C1%2F3%2C0%29%29

Edwin