Let A be an n*n matrix. An n*n matrix A^{-1} such that

AA^{-1}=A^{-1}A=I_{n}

is the **inverse** of A. A matrix A is **nonsingular** if A^{-1} exists (i.e., if A has an inverse). If a matrix does not have an inverse, then it is **singular**.

Examples Explanation

Given a square matrix M, we know the size of its inverse (the same size as M) and the product of M and its inverse. Using this information, the inverse of M can be calculated by assigning variables to the elements of M ^{-1} and representing the product MM ^{-1} as a system of n equations in n unknowns, where M has dimension n*n

**Examples**

**Explanation**

This system can be expressed as an augmented matrix. Recall that an augmented matrix includes the constant coefficients of a system of equations.

The solution to this system will provide the elements of M ^{-1}.

Systems of linear equations can be solved using the matrix inverse. The system must be represented by the equation

AX=B

where A is the coefficient matrix of the system, X is the column matrix of variables, and B is the column matrix of constant coefficients. The solution to the system will then be A ^{-1}B because

**Examples**

**Explanation**

The solution is (x,y)=(3,4)

**Solve**

Show that

Show that the inverse of M is

Show that

Find the inverse of N using a system of four equations in four unknowns.- Is the identity matrix singular or nonsingular?
- Solve the system

using elementary row operations and matrix inverses. - Solve the system

using elementary row orations and matrix inverses. - Solve the system

using elementary row operations and matrix inverses. - Is it possible to solve a system composed of less unknowns than equations using elementary row operations and the matrix inverse?

By substitution,

No elementary row operations are required to transform I_{n}into In, so I_{n}-1=I_{n}. Since In has an inverse, it is nonsingular.

- Yes, but the extraneous equations must be eliminated so that the coefficient matrix is square (having the same number of equations as unknowns).

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