This method can be applied only when the coefficient matrix is a square matrix and non-singular.

**Matrix Inversion Method**

This method can be
applied only when the coefficient matrix is a square matrix and non-singular.

Consider the matrix equation

* AX *= *B *, â€¦ (1)

where *A *is a square matrix and non-singular. Since *A *is
non-singular, *A*^{âˆ’}^{1} exists and *A*^{âˆ’}^{1} *A *= *AA*^{âˆ’}^{1} = *I*. Pre-multiplying both sides of (1) by *A*^{âˆ’}^{1}, we get *A*^{âˆ’}^{1} ( *AX *) = *A*^{âˆ’}^{1}*B*. That is, ( *A*^{âˆ’}^{1} *A*) *X *= *A*^{âˆ’}^{1}*B*. Hence, we get *X *= *A*^{âˆ’}^{1}*B*.

Solve the following system of linear equations, using matrix
inversion method:

5*x *+
2 *y *= 3, 3*x *+ 2 *y *= 5 .

The matrix form of the system is AX = B , where

We find |A| = =
10 - 6= 4 â‰ 0. So, A^{âˆ’1} exists
and A^{âˆ’1} =

Then, applying the formula X = A^{âˆ’1}B , we get

So the solution is (*x* =
âˆ’1, *y* = 4).

Solve the following
system of equations, using matrix inversion method:

2*x*_{1} + 3*x*_{2}
+ 3*x*_{3} = 5,

*x*_{1} â€“ 2*x*_{2} + *x*_{3} = -4,

3*x*_{1} â€“ x_{2} â€“ 2*x*_{3} = 3

The matrix form of the system is AX = B,where

So,
the solution is ( *x*_{1} = 1, *x*_{2} = 2, *x*_{3} = âˆ’1) .

**Example 1.24**

If , find the products AB and BA and hence solve the system of equations x âˆ’ y + z = 4, x â€“ 2y â€“ 2z = 9, 2x + y +3z =1.

**Solution**

Writing
the given system of equations in matrix form, we get

Hence,
the solution is (*x* = 3, *y* = - 2,
*z* = âˆ’1).

Tags : Definition, Formulas, Solved Example Problems , 12th Mathematics : UNIT 1 : Applications of Matrices and Determinants

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12th Mathematics : UNIT 1 : Applications of Matrices and Determinants : Matrix Inversion Method | Definition, Formulas, Solved Example Problems

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