# Solve system of equations with inverse matrix

In summary, I found the inverse matrix of A using an augmented identity matrix. I 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 = 1014x - 4y + 4z = 36x - 2y + 2z = 14However, 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

## Homework Statement

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.

## 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.

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.

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?

Last edited:
Mark44 said:
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.

Cyosis said:
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.

Last edited:
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.

rock.freak667 said:
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?

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.

## What is a system of equations?

A system of equations is a set of two or more equations with multiple variables that need to be solved simultaneously.

## What is an inverse matrix?

An inverse matrix is a matrix that, when multiplied by the original matrix, results in the identity matrix. This allows for solving systems of equations by using matrix multiplication.

## Why is using an inverse matrix helpful in solving systems of equations?

Using an inverse matrix eliminates the need for traditional algebraic methods, making it a more efficient and accurate way to solve systems of equations.

## What is the process for solving a system of equations with an inverse matrix?

The process involves converting the system of equations into matrix form, finding the inverse of the coefficient matrix, multiplying the inverse matrix by the constant matrix, and then using the resulting values to solve for the variables.

## In what situations is using an inverse matrix not possible?

An inverse matrix is not possible if the coefficient matrix is singular (non-invertible) or if the system of equations has infinitely many solutions.

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