Solve system of equations with inverse matrix

username12345

1. The problem statement, all variables and given/known data

Find the inverse matrix of A, then use this inverse to solve system of equation.

A is a given 3 x 3 matrix and the system of equations is 3 equations in 3 unknowns.

2. Relevant equations



3. The attempt at a solution

I have found the inverse of A using an augmented identity matrix. Checked this inverse by taking product of A and inverse A and got identity matrix.

The system of equations is something like this (these are not the actual values):

6x - 2y = 10
14x - 4y + 4z = 3
6x - 2y + 2z = 14

Now when I take the product of inverse A and [10 3 14] to solve for x, y and z, putting these values into the original equations does not hold.

I was told to try re-arrange the equations and compare to the inverse matrix, but can't seem to find a solution.

Note that by swapping 2 rows in my inverse matrix the co-effiecients of the unkowns in the system of equations are the same as the entries in the matrix.

Gee I hope that made sense.

Mark44

Without knowing what the system of equations is, it's not possible to know whether the system is inconsistent or you have made a mistake. If you have matrix equation A[x y z]^T = [c₁ c₂ c₃]^T, and A has an inverse, then the solution to the matrix equation is [x y z]^T = A^-1[c₁ c₂ c₃]^T.

Cyosis

It would be helpful if you showed us what you took as A and whether or not you inverted it correctly.

To repeat what Mark said in a slightly less abstract manner, this is the system you're trying to solve:
[tex]
\left( \begin{matrix} 6 & -2 & 0 \\ 14 & -4 & 4 \\ 6 & -2 & 2 \end{matrix}\right) \left( \begin{matrix} x \\ y \\ z \end{matrix}\right) =\left( \begin{matrix} 10 \\ 3 \\14\end{matrix}\right)
[/tex]

Did you take the matrix on the left hand side as A?

username12345

The system is consistent, I can solve it using row operations, however we are explicitly asked to use the inverse matrix of A to solve this particular system.

I am given this matrix A:
[tex]
A = \left( \begin{matrix} 1 & 0 & -2 \\ 3 & 1 & -6 \\ 0 & 1 & 1 \end{matrix}\right)
[/tex]

And produce this inverse:
[tex]A^{-1} = \left( \begin{matrix} 7 & -2 & 2 \\ -3 & 1 & 0 \\ 3 & -1 & 1 \end{matrix}\right) [/tex]

now I must use this inverse to solve the system. This is a 2 part question, each part with a system to solve. Part a works with this inverse. Part b (the system given in this thread) does not. The actual given system is:

3x - y = 5
7x - 2y + 2z = 3
3x - y + z = 7

Using row operations I get to row-echelon form and use back substitution to get z = 2, y = -38 and x = -11 and the original equations hold.

Using the inverse matrix gives z = 19, y = -12 and x = 43 and using these the original equations do not hold. I do this:

[tex]\left( \begin{matrix} x & y & z \end{matrix}\right) = \left( \begin{matrix} 7 & -2 & 2 \\ -3 & 1 & 0 \\ 3 & -1 & 1 \end{matrix}\right) \left( \begin{matrix} 5 & 3 & 7 \\ \end{matrix}\right)[/tex]

The co-efficients in the system are almost the same as the entries in the inverse matrix.

rock.freak667

3x - y = 5
7x - 2y + 2z = 3
3x - y + z = 7

Re-write this as

7x - 2y + 2z = 3
-3x + y = -5
3x-y+z=7

Now write down this in the matrix for AX=B.

Take a look at the matrix A and the inverse you calculated before.

username12345

The entries are the same, but then? Perhaps I am missing some idea or principle.

If AX = B, then A is the co-efficients, B is the numbers after the equals, X is the vector (x y z) but where does the inverse come into the story?

username12345

rock.freak667 had the solution.

3x - y = 5
7x - 2y + 2z = 3
3x - y + z = 7

Re-write this as

7x - 2y + 2z = 3
-3x + y = -5
3x-y+z=7

By letting B = the co-efficients of the system then B = inverse A, so A = inverse B.

then, Bx = b, where b = [3 -5 7]
so, x = inverse B b
thus x = A b

solving this for x, y and z and checking original system holds. YES!!!

Thanks to everyone for the help.