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.
$\(\begin{cases}\)w - x + 2y \(\quad\]\quad\) = -3 \(\quad\[\quad\) x - y + z = 4 \\-w + x - y + 2z = 2 \(\quad\]\quad\) -x + y - 2z = -4\(\end{cases}\[\text{The inverse of }\]\begin{bmatrix}\)1 & -1 & 2 & 0 \\0 & 1 & -1 & 1 \\-1 & 1 & -1 & 2 \\0 & -1 & 1 & -2\(\end{bmatrix}\[\text{ is }\]\begin{bmatrix}\)0 & 0 & -1 & -1 \\1 & 4 & 1 & 3 \\1 & 2 & 1 & 2 \\0 & -1 & 0 & -1\(\end{bmatrix}\)$

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: \[A = \begin{bmatrix} 1 & -1 & 2 & 0 \\ 0 & 1 & -1 & 1 \\ -1 & 1 & -1 & 2 \\ 0 & -1 & 1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} -3 \\ 4 \\ 2 \\ -4 \end{bmatrix}\]
2. Step 2: Confirm that the inverse matrix \(A^{-1}\) is given as: \[A^{-1} = \begin{bmatrix} 0 & 0 & -1 & -1 \\ 1 & 4 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 0 & -1 & 0 & -2 \end{bmatrix}\]
3. Step 3: Use the matrix equation \(X = A^{-1}B\) to find the solution vector \(X\). This means multiplying the inverse matrix \(A^{-1}\) by the constants matrix \(B\).
4. Step 4: Perform the matrix multiplication by multiplying each row of \(A^{-1}\) by the column matrix \(B\) and summing the products to find each variable \(w\), \(x\), \(y\), and \(z\).
5. Step 5: Write the solution vector \(X\) explicitly as: \[X = \begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix} = A^{-1}B\] where each element corresponds to the value of the variables in 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

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 solution when the inverse matrix is known or can be computed.

Interpreting and Using the Given Inverse Matrix

The problem provides the inverse of the coefficient matrix, which allows immediate calculation of the solution vector X by multiplying this inverse with B. Understanding how to apply the given inverse matrix is essential for efficiently solving the system.