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

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1. Write the system of equations in matrix form: AX = B, where A is the coefficient matrix, X is the column matrix of variables, and B is the column matrix of constants.
2. Identify the coefficient matrix A as \( \begin{bmatrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \end{bmatrix} \), the variable matrix X as \( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), and the constant matrix B as \( \begin{bmatrix} 8 \\ 10 \\ 9 \end{bmatrix} \).
3. Express the matrix equation as \( \begin{bmatrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 8 \\ 10 \\ 9 \end{bmatrix} \).
4. Use the given inverse matrix \( \begin{bmatrix} \frac{7}{2} & 0 & -3 \\ -1 & 0 & 0 \\ 0 & -1 & 1 \end{bmatrix} \) to solve for X by multiplying both sides of the equation by the inverse of A.
5. Calculate \( X = A^{-1}B \) by performing the matrix multiplication \( \begin{bmatrix} \frac{7}{2} & 0 & -3 \\ -1 & 0 & 0 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 8 \\ 10 \\ 9 \end{bmatrix} \) to find the values of x, y, and z.

Key Concepts

Matrix Representation of Linear Systems

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.

Matrix Inverse

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.

Solving Linear Systems

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.