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 = 8 2x + 7y + 6z = 10 2x + 7y + 7z = 9 The inverse of is

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1. Write the system of equations in matrix form as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the column matrix of variables, and \(B\) is the constants matrix. Specifically, \(A = \begin{bmatrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), and \(B = \begin{bmatrix} 8 \\ 10 \\ 9 \end{bmatrix}\).
2. Use the given inverse matrix \(A^{-1} = \begin{bmatrix} \frac{7}{2} & 0 & -3 \\ 1 & 0 & 0 \\ 0 & -1 & 1 \end{bmatrix}\) to solve for \(X\) by multiplying both sides of the matrix equation by \(A^{-1}\), resulting in \(X = A^{-1}B\).
3. Set up the multiplication \(X = \begin{bmatrix} \frac{7}{2} & 0 & -3 \\ 1 & 0 & 0 \\ 0 & -1 & 1 \end{bmatrix} \times \begin{bmatrix} 8 \\ 10 \\ 9 \end{bmatrix}\).
4. Perform the matrix multiplication by calculating each element of \(X\) as the dot product of the corresponding row of \(A^{-1}\) with the column matrix \(B\). For example, the first element of \(X\) is \(\left( \frac{7}{2} \times 8 \right) + (0 \times 10) + (-3 \times 9)\).
5. Write the resulting expressions for \(x\), \(y\), and \(z\) from the multiplication and simplify each to find the solution to the system.

Key Concepts

Matrix Representation of Linear Systems

A system of linear equations can be expressed as a matrix equation AX = B, where A is the coefficient matrix, X is the column matrix of variables, and B is the constants matrix. This form simplifies solving and analyzing the system using matrix operations.

Matrix Inverse and Its Role in Solving Systems

The inverse of a square matrix A, denoted A⁻¹, is a matrix that when multiplied by A yields the identity matrix. If A is invertible, the solution to AX = B can be found by X = A⁻¹B, providing a direct method to solve linear systems.

Matrix Multiplication for Solution Computation

To find the solution vector X, multiply the inverse matrix A⁻¹ by the constants matrix B. This operation combines the inverse coefficients with the constants, yielding the values of variables that satisfy all equations simultaneously.