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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 23
7장, 문제 23

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
{x+3y=0x+y+z=13x−y−z=11\(\begin{cases}\)x + 3y = 0 \(\x\) + y + z = 1 \\3x - y - z = 11\(\end{cases}\)

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Write the system of equations as an augmented matrix. The system is: \[\begin{cases} x + 3y + 0z = 0 \\ x + y + z = 1 \\ 3x - y - z = 11 \end{cases}\] The augmented matrix is: \[\left[\begin{array}{ccc|c} 1 & 3 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 3 & -1 & -1 & 11 \end{array}\right]\]
Use Gaussian elimination to create zeros below the leading 1 in the first column. Subtract the first row from the second row, and subtract 3 times the first row from the third row: - Row 2 = Row 2 - Row 1 - Row 3 = Row 3 - 3 * Row 1
After these operations, the matrix will have zeros below the first pivot. Next, focus on the second row and create a leading 1 in the second column if necessary, then eliminate the entry below it in the third row by appropriate row operations.
Continue the elimination process to get the matrix into upper triangular form (or reduced row echelon form if using Gauss-Jordan). This involves making zeros above and below pivots and scaling rows to have leading 1s.
Once the matrix is in upper triangular or reduced row echelon form, use back-substitution to solve for the variables \(z\), then \(y\), and finally \(x\).

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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 23–30, use expansion by minors to evaluate each determinant. ∣30021−525−1∣\(\begin{vmatrix}\)3 & 0 & 0 \\2 & 1 & -5 \\2 & 5 & -1\(\end{vmatrix}\)

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교과서 질문

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. {w+2x+3y−z=72x−3y+z=4w−4x+y=3\(\begin{cases}\)w + 2x + 3y - z = 7 \\2x - 3y + z = 4 \(\w\) - 4x + y = 3\(\end{cases}\)

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교과서 질문

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

{x+y−z=−22x−y+z=5−x+2y+2z=1\(\begin{cases}\)x + y - z = -2 \\2x - y + z = 5 \\-x + 2y + 2z = 1\(\end{cases}\)

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교과서 질문

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

A=[2−1253−1]B=[0−2321−5]C=[123−112−121]D=[−2313−24]A=\(\begin{bmatrix}\)2 & -1 & 2\\ 5 & 3 & -1\(\end{bmatrix}\[\quad\) B=\(\begin{bmatrix}\)0 & -2\\ 3 & 2\\ 1 & -5\(\end{bmatrix}\)C=\(\begin{bmatrix}\)1 & 2 & 3\\ -1 & 1 & 2\\ -1 & 2 & 1\(\end{bmatrix}\]\quad\) D=\(\begin{bmatrix}\)-2 & 3 & 1\\ 3 & -2 & 4\(\end{bmatrix}\)

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교과서 질문

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

{2x−y−z=4x+y−5z=−4x−2y=4\(\begin{cases}\)2x - y - z = 4 \(\x\) + y - 5z = -4 \(\x\) - 2y = 4\(\end{cases}\)

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교과서 질문

Let A=[−3−72−950]A = \(\begin{bmatrix}\)-3 & -7 \\2 & -9 \\5 & 0\(\end{bmatrix}\) and B=[−5−1003−4]B = \(\begin{bmatrix}\)-5 & -1 \\0 & 0 \\3 & -4\(\end{bmatrix}\). Solve each matrix equation for X. B - X = 4A

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