In Exercises 37 - 42,a. Write each linear system as a matrix equation in the form AX = B.b. Solve the system using the inverse that is given for the coefficient matrix.2x + 6y + 6z = 82x + 7y + 6z = 102x + 7y + 7z = 9The inverse of is

A linear system can be expressed in matrix form as AX = B, where A is the coefficient matrix, X is the column matrix of variables, and B is the column matrix of constants. This representation simplifies the process of solving systems of equations, allowing for the use of matrix operations to find solutions.

The inverse of a matrix A, denoted as A⁻¹, is a matrix that, when multiplied by A, yields the identity matrix. For a system of equations represented as AX = B, if A is invertible, the solution can be found using X = A⁻¹B. The existence of an inverse is crucial for solving linear systems efficiently.

To solve a linear system using the inverse of the coefficient matrix, one first writes the system in matrix form. Then, by multiplying both sides of the equation AX = B by A⁻¹, the solution for X can be obtained. This method is particularly useful when dealing with larger systems where traditional substitution or elimination methods may be cumbersome.