Inverse Matrix Method

[Guest]

Hey y'all. Okay so I am just a little bit lost on how to come up with an answer in corrdinate notation using the Inverse Matrix Method.

5x-7y=54
2x-47=30

That is the system that I have to solve using Matracies so I put that problem into a set of matracties:

|5 -7| |X|    |54|
|2 -4| |Y| = |30|

Then i find the inverse by switching the places of the 5 and -4 and changing the sign on the -7 and 2 and multiplying that matrix by -1/6 which gives me:

| 2/3 -7/6|
| 1/3 -5/6|

Then I take this matrix and multiply it by the original matrix and the inverse of that matrix so it would look like this:

| 2/3 -7/6| |5 -7|X= |2/3 -7/6| |-4 7|
| 1/3 -5/6| |2 -4|X= |1/3 -5/6| |-2 5|

and it equals:

|1 0| = |-1/3 -7/6|
|0 1| = |1/3 -11/6|

Now from this point on, I am totally confused as to how I am supposed to end up with an (x,y) answer which is supposed to come out to be (1,7). Thanks for the help that anyone can give!! <3

 

[Guest]

oops correction on the system that i have to solve. Its supposed to be

5x - 7y = 54
2x - 4y = 30

 

[stapel]

The identity matrix is not equal to the matrix to which you have set it equal. I'm not sure what you're trying to accomplish with that "equation"...?

To solve matrix equations by using inverses, you start with the equation Ax = y, where x and y are column vectors and A is a square matrix, find the inverse of A, being A^-1, and multiply (on the left) on both sides:

. . . . .Ax = y

. . . . .A^-1Ax = A^-1y

. . . . .x = A^-1y

Multiply out the right-hand side, and read off the solutions for x contained in A^-1y.

Eliz.