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. x - y + z = 8 2y - z = - 7 2x + 3y = 1 The inverse of is

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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. For the system: \[ \begin{cases} x - y + z = 8 \\ 0x + 2y - z = -7 \\ 2x + 3y + 0z = 1 \end{cases} \] The coefficient matrix A is: \[ A = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \end{bmatrix} \] The variable matrix X is: \[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \] The constants matrix B is: \[ B = \begin{bmatrix} 8 \\ -7 \\ 1 \end{bmatrix} \]
2. Step 2: Express the system as the matrix equation \( AX = B \). This means multiplying the coefficient matrix A by the variable matrix X equals the constants matrix B.
3. Step 3: Use the inverse of matrix A, denoted \( A^{-1} \), to solve for X. Multiply both sides of the equation \( AX = B \) by \( A^{-1} \) to get \( X = A^{-1}B \).
4. Step 4: Substitute the given inverse matrix \( A^{-1} = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \) and the constants matrix B into the equation \( X = A^{-1}B \).
5. Step 5: Perform the matrix multiplication \( A^{-1}B \) by multiplying each row of \( A^{-1} \) by the column matrix B to find the values of \( x, y, \) 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 Its Role in Solving Systems

The inverse of a square matrix A, denoted A⁻¹, is a matrix such that A⁻¹A = I, where I is the identity matrix. If A is invertible, the system AX = B can be solved by multiplying both sides by A⁻¹, giving X = A⁻¹B, which provides the solution vector.

Performing Matrix Multiplication to Find Solutions

To solve for the variable matrix X, multiply the inverse matrix A⁻¹ by the constants matrix B. This involves multiplying rows of A⁻¹ by columns of B and summing the products, resulting in the values of variables x, y, and z that satisfy the system.