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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
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Tutti i libri di testoBlitzer 8th EditionCh. 6 - Matrices and DeterminantsProblema 37
Capitolo 7, Problema 37

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=82x+7y+6z=102x+7y+7z=9The inverse of [266276277] is [720−3−1000−11].\(\begin{cases}\) 2x + 6y + 6z = 8 \\ 2x + 7y + 6z = 10 \\ 2x + 7y + 7z = 9 \(\end{cases}\) \\ \(\text{The inverse of }\) \(\begin{bmatrix}\) 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \(\end{bmatrix}\) \(\text{ is }\) \(\begin{bmatrix}\) \(\frac{7}{2}\) & 0 & -3 \\ -1 & 0 & 0 \\ 0 & -1 & 1 \(\end{bmatrix}\)\(\text{.}\)

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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 \(B = \begin{bmatrix} 8 \\ 10 \\ 9 \end{bmatrix}\).
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 \(X = A^{-1}B\).
Set up the multiplication \(X = \begin{bmatrix} \frac{7}{2} & 0 & -3 \\ 1 & 0 & 0 \\ 0 & -1 & 1 \end{bmatrix} \times \begin{bmatrix} 8 \\ 10 \\ 9 \end{bmatrix}\).
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 \(\left( \frac{7}{2} \times 8 \right) + (0 \times 10) + (-3 \times 9)\).
Write the resulting expressions for \(x\), \(y\), and \(z\) from the multiplication and simplify each to find the solution to the system.

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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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Introduction to Systems of Linear Equations

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.
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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.
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