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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 6 - Matrices and DeterminantsProblem 37
Chapter 7, Problem 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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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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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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Solving Systems of Equations - Substitution

Matrix Multiplication for Solution Computation

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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Related Practice
Textbook Question

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

{3w−4x+y+z=9w+x−y−z=02w+x+4y−2z=3−w+2x+y−3z=3\(\begin{cases}\)3w - 4x + y + z = 9 \(\w\) + x - y - z = 0 \\2w + x + 4y - 2z = 3 \\-w + 2x + y - 3z = 3\(\end{cases}\)

Textbook Question

Perform the indicated matrix operations given that A, B and C are defined as follows. If an operation is not defined, state the reason.

A=[40−3501],B=[51−2−2],C=[1−1−11]A=\(\begin{bmatrix}\)4 & 0\\ -3 & 5\\ 0 & 1\(\end{bmatrix}\),B=\(\begin{bmatrix}\)5 & 1\\ -2 & -2\(\end{bmatrix}\),C=\(\begin{bmatrix}\)1 & -1\\ -1 & 1\(\end{bmatrix}\)

4B - 3C

Textbook Question

In Exercises 37–44, use Cramer's Rule to solve each system. {x+y+z=02x−y+z=−1−x+3y−z=−8\(\begin{cases}\)x + y + z = 0 \\2x - y + z = -1 \\-x + 3y - z = -8\(\end{cases}\)

Textbook Question

Perform the indicated matrix operations given that A, B and C are defined as follows. If an operation is not defined, state the reason.

A=[40−3501],B=[51−2−2],C=[1−1−11]A=\(\begin{bmatrix}\)4 & 0\\ -3 & 5\\ 0 & 1\(\end{bmatrix}\),B=\(\begin{bmatrix}\)5 & 1\\ -2 & -2\(\end{bmatrix}\),C=\(\begin{bmatrix}\)1 & -1\\ -1 & 1\(\end{bmatrix}\)

BC + CB

Textbook Question

In Exercises 37–38, find the products and to determine whether B is the multiplicative inverse of A.

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Textbook Question

In Exercises 27 - 36, find (if possible) the following matrices: a. AB b. BA 1 2 2 - 3 1 - 1 - 1 1 A = B = 1 1 - 2 1 5 4 10 5