## Solve system of equations with inverse matrix

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.
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 Mentor 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.
 Recognitions: Homework Help 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: $$\left( \begin{matrix} 6 & -2 & 0 \\ 14 & -4 & 4 \\ 6 & -2 & 2 \end{matrix}\right) \left( \begin{matrix} x \\ y \\ z \end{matrix}\right) =\left( \begin{matrix} 10 \\ 3 \\14\end{matrix}\right)$$ Did you take the matrix on the left hand side as A?

## Solve system of equations with inverse matrix

 Quote by 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.
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.

 Quote by Cyosis It would be helpful if you showed us what you took as A and whether or not you inverted it correctly.
I am given this matrix A:
$$A = \left( \begin{matrix} 1 & 0 & -2 \\ 3 & 1 & -6 \\ 0 & 1 & 1 \end{matrix}\right)$$

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

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:

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

The co-efficients in the system are almost the same as the entries in the inverse matrix.
 Recognitions: Homework Help 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.

 Quote by rock.freak667 Take a look at the matrix A and the inverse you calculated before.
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?
 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.

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