# Application of Inverse Matrix in Linear Equation Systems

In this article we will
produce more about linear equations and invertibility matrix. So we will find
a new method for solving unknown unknowns.

Theorem 1.

If A is a matrix of n x n
which can be reversed, then for each B matrix measuring n x 1, the system
equation AX = B has exactly one solution, namely X = A

^{-1}B.
Proof:

Because A (A

^{-1}B) = B, then X = A^{-1}B is a solution AX = B. To show that this is one solution, we will assume that X*is any solution, and then show that X*_{0}*must be a solution of A*_{0}^{-1}B..
If X

*is a solution, then AX*_{0}*= B. By multiplying the two sections A*_{0}^{-1}, we get X*= A*_{0}^{-1}B.
Example 1

Review the system of linear equations below,

X1 + 2X2 + 3X3 = 3

2X1 + 5X2 + 3X3 = 3

X1 + 8X3 = 17

In the form of a matrix, this system can be written as AX = B, where

Because the A-1 matrix is known in Example 4 of the article "Elementary Line Operation Methods and Determinants for Finding the Inverse Matrix", in this article we only display the results of the A-1 matrix, namely,

According to the theorem 1 in this article, the completion of this system is

or x

*= 1, x*_{1}*= -1, x*_{2}*= 2.*_{3}
Let A be a reversible
matrix, then the theorem 1 solves this problem completely by specifying that
for each B matrix measuring m x 1, the equation AX = B has a unique solution X
= A

^{-1}B. If A is not a quadratic matrix, or if A is a quadratic matrix but cannot be reversed, then this 1 theorem cannot be applied. In these cases we want to determine what conditions, if there is something that must be met by the B Matrix so that AX = B is consistent. The following example illustrates how Gauss-Jordan elimination can be used to determine such conditions.
Example 2.

What conditions must be
met by b

*, b*_{1}*, b*_{2}*, so that the systems of the following equations are consistent?*_{3}
x

*+ x*_{1}*+ 2x*_{2}*= b*_{3}_{1}
x

*+ 2x*_{1 }*= b*_{3}_{2}
2x

*+ x*_{2}*+ 3x*_{2}*= b*_{3}_{3}

_{}
Solution:

The enlarged matrix for
the system is

Which can be reduced to the form of row echelon as follows,

It is clear now from the
third row in the matrix that the system has a solution if and only if b

*, b*_{1}*, and b*_{2}*meet the conditions.*_{3}
b

*- b*_{3 }*- b*_{2}*= 0 or*_{1}*b**= b*_{3 }*+ b*_{1}_{2}

_{}
to state this condition in
another way, then AX = B is consistent if and only if B is a shaped matrix,

Where b1 and b2 are any.

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