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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Blitzer 8th EditionCh. 6 - Matrices and Determinants문제 33
7장, 문제 33

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
{2x+2y+7z=−12x+y+2z=24x+6y+z=15\(\begin{cases}\)2x + 2y + 7z = -1 \\2x + y + 2z = 2 \\4x + 6y + z = 15\(\end{cases}\)

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Write the system of equations as an augmented matrix. For the system: \[\begin{cases} 2x + 2y + 7z = -1 \\ 2x + y + 2z = 2 \\ 4x + 6y + z = 15 \end{cases}\] The augmented matrix is: \[\left[ \begin{array}{ccc|c} 2 & 2 & 7 & -1 \\ 2 & 1 & 2 & 2 \\ 4 & 6 & 1 & 15 \end{array} \right]\]
Use Gaussian elimination to transform the matrix into an upper triangular form. Start by using the first row to eliminate the \(x\)-terms in the second and third rows. For example, subtract the first row from the second row, and subtract twice the first row from the third row.
Continue the elimination process to get zeros below the pivot in the second column. Use the second row to eliminate the \(y\)-term in the third row by appropriate row operations.
Once the matrix is in upper triangular form, write the corresponding system of equations from the matrix. This system will have the form: \[\begin{cases} a_{11}x + a_{12}y + a_{13}z = b_1 \\ 0 + a_{22}y + a_{23}z = b_2 \\ 0 + 0 + a_{33}z = b_3 \end{cases}\]
Use back-substitution to solve for \(z\) from the third equation, then substitute \(z\) into the second equation to solve for \(y\), and finally substitute \(y\) and \(z\) into the first equation to solve for \(x\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Systems of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. The goal is to find values for the variables that satisfy all equations simultaneously. Understanding how to represent and interpret these systems is fundamental before applying matrix methods.
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가이드 코스
4:27
Introduction to Systems of Linear Equations

Matrix Representation of Systems

Systems of linear equations can be expressed 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. This representation simplifies the use of matrix operations to solve the system efficiently.
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가이드 코스
6:19
Systems of Inequalities

Gaussian Elimination and Gauss-Jordan Elimination

Gaussian elimination transforms the augmented matrix into an upper triangular form to solve via back-substitution, while Gauss-Jordan elimination reduces it further to reduced row-echelon form for direct solution. Both methods use row operations to systematically solve linear systems.
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가이드 코스
6:48
Solving Systems of Equations - Elimination
관련 실천
교과서 질문

In Exercises 31–36, use the alternative method for evaluating third-order determinants on here to evaluate each determinant. ∣1561451910∣\(\begin{vmatrix}\)1 & 5 & 6 \\1 & 4 & 5 \\1 & 9 & 10\(\end{vmatrix}\)

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교과서 질문

Write each matrix equation as a system of linear equations without matrices.

[20−1030110][xyz]=[695]\(\begin{bmatrix}\)2 & 0 & -1 \\0 & 3 & 0 \\1 & 1 & 0\(\end{bmatrix}\[\begin{bmatrix}\)x \(\y\) \(\z\]\end{bmatrix}\)=\(\begin{bmatrix}\)6 \\9 \\5\(\end{bmatrix}\)

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교과서 질문

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

{w+x+y+z=42w+x−2y−z=0w−2x−y−2z=−23w+2x+y+3z=4\(\begin{cases}\)w + x + y + z = 4 \\2w + x - 2y - z = 0 \(\w\) - 2x - y - 2z = -2 \\3w + 2x + y + 3z = 4\(\end{cases}\)

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교과서 질문

In Exercises 27 - 36, find (if possible) the following matrices: a. AB b. BA 1 - 1 4 1 1 0 A = 4 - 1 3 B = 1 2 4 2 0 - 2 1 - 1 3

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교과서 질문

In Exercises 27 - 36, find (if possible) the following matrices: a. AB b. BA 4 2 2 3 4 A = 6 1 B = 3 5 - 1 - 2 0

920
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교과서 질문

Write each matrix equation as a system of linear equations without matrices.

[4−72−3][xy]=[−31]\(\begin{bmatrix}\)4 & -7 \\2 & -3\(\end{bmatrix}\[\begin{bmatrix}\)x \(\y\]\end{bmatrix}\)=\(\begin{bmatrix}\)-3 \\1\(\end{bmatrix}\)

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