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. w - x + 2y = - 3 x - y + z = 4 - w + x - y + 2z = 2 - x + y - 2z = - 4 The inverse of is

Verified step by step guidance

1. 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. From the system, identify A, X, and B as follows: A = \begin{bmatrix} 1 & -1 & 2 & 0 \\ 0 & 1 & -1 & 1 \\ -1 & 1 & -1 & 2 \\ 0 & -1 & 1 & -2 \end{bmatrix}, X = \begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} -3 \\ 4 \\ 2 \\ -4 \end{bmatrix}
2. Step 2: Express the matrix equation as AX = B, which represents the system of linear equations in matrix form.
3. Step 3: To solve for X, use the inverse of matrix A, denoted as A^{-1}. Multiply both sides of the equation by A^{-1} to get X = A^{-1}B.
4. Step 4: Use the given inverse matrix A^{-1}: A^{-1} = \begin{bmatrix} 0 & 0 & -1 & -1 \\ 1 & 4 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 0 & -1 & 0 & -1 \end{bmatrix} Multiply A^{-1} by B to find the values of the variables in X.
5. Step 5: Perform the matrix multiplication A^{-1}B step-by-step by multiplying each row of A^{-1} by the column matrix B and summing the products to find each variable w, x, y, and z.

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 Solving Systems

If the coefficient matrix A is invertible, the system AX = B can be solved by multiplying both sides by A's inverse, giving X = A⁻¹B. This method provides a direct way to find the solution vector X when the inverse matrix is known.

Interpreting and Using Given Inverse Matrices

When the inverse of the coefficient matrix is provided, it must be correctly applied to the constants matrix B to find the solution. Understanding how to multiply matrices and use the given inverse is essential for efficiently solving the system.