In Exercises 1 - 12, find the products AB and BA to determine whether B is the multiplicative inverse of A. 1 2 3 7/2 - 3 1/2 A = 1 3 4 B = - 1/2 0 1/2 1 4 3 - 1/2 1 - 1/2
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7. Systems of Equations & Matrices
Determinants and Cramer's Rule
Problem 37
Textbook Question
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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Step 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 constants matrix. Here, A = \( \begin{bmatrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \end{bmatrix} \), X = \( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), and B = \( \begin{bmatrix} 8 \\ 10 \\ 9 \end{bmatrix} \).
Step 2: Express the matrix equation as \( AX = B \). This means \( \begin{bmatrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 8 \\ 10 \\ 9 \end{bmatrix} \).
Step 3: Use the inverse of matrix A, denoted as \( A^{-1} \), to solve for X by multiplying both sides of the equation by \( A^{-1} \), giving \( X = A^{-1}B \).
Step 4: Substitute the given inverse matrix \( A^{-1} = \begin{bmatrix} \frac{7}{2} & 0 & -3 \\ 1 & 0 & 0 \\ 0 & -1 & 1 \end{bmatrix} \) and the constants matrix B into the equation \( X = A^{-1}B \).
Step 5: Multiply the inverse matrix \( A^{-1} \) by the constants matrix B to find the values of \( x, y, \) and \( z \). This involves performing matrix multiplication step-by-step without calculating the final numeric values.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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.
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Matrix Inverse and Its Role in Solving Systems
If the coefficient matrix A is invertible, the system AX = B can be solved by multiplying both sides by A's inverse, yielding X = A⁻¹B. This method provides a direct way to find the solution vector X when the inverse matrix is known.
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Matrix Multiplication
Matrix multiplication involves combining rows of the first matrix with columns of the second to produce a new matrix. Understanding this operation is essential to compute X = A⁻¹B correctly, ensuring each variable's value is accurately determined.
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