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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBlitzer 8th EditionCh. 6 - Matrices and DeterminantsProblema 5
Capitolo 7, Problema 5

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
{3x1+5x2−8x3+5x4=−8x1+2x2−3x3+x4=−72x1+3x2−7x3+3x4=−114x1+8x2−10x3+7x4=−10\(\begin{cases}\) 3x_1 + 5x_2 - 8x_3 + 5x_4 = -8 \\ x_1 + 2x_2 - 3x_3 + x_4 = -7 \\ 2x_1 + 3x_2 - 7x_3 + 3x_4 = -11 \\ 4x_1 + 8x_2 - 10x_3 + 7x_4 = -10 \(\end{cases}\)

Guida verificata passo dopo passo
1
Write the system of equations in matrix form as an augmented matrix \([A|\mathbf{b}]\), where \(A\) is the coefficient matrix and \(\mathbf{b}\) is the constants column vector.
Use Gaussian elimination to transform the augmented matrix into an upper triangular form by applying row operations: swapping rows, multiplying a row by a nonzero scalar, and adding multiples of one row to another.
Once the matrix is in upper triangular form, use back-substitution to solve for the variables starting from the last row and moving upwards.
Alternatively, use Gauss-Jordan elimination to reduce the augmented matrix to reduced row echelon form (RREF), where the coefficient matrix becomes the identity matrix.
From the RREF matrix, directly read off the solutions for the variables, as each variable corresponds to a leading 1 in the identity matrix.

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Systems of Linear Equations

A system of linear equations consists of two or more linear equations with 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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Introduction to Systems of Linear Equations

Gaussian Elimination with Back-Substitution

Gaussian elimination is a method to solve systems by transforming the augmented matrix into an upper triangular form using row operations. Back-substitution then solves for variables starting from the last equation upward. This stepwise approach simplifies complex systems into manageable forms.
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Solving Systems of Equations - Substitution

Gauss-Jordan Elimination

Gauss-Jordan elimination extends Gaussian elimination by reducing the matrix further into reduced row-echelon form, where each leading coefficient is 1 and is the only nonzero entry in its column. This method directly provides the solution without needing back-substitution.
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Solving Systems of Equations - Elimination
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Domanda del libro di testo

Write the augmented matrix for each system of linear equations.

{5x−2y−3z=0x+y=52x−3z=4\(\begin{cases}\) 5x - 2y - 3z = 0 \\ x + y = 5 \\ 2x - 3z = 4 \(\end{cases}\)

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In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. {3x+4y+2z=34x−2y−8z=−4x+y−z=3\(\begin{cases}\) 3x + 4y + 2z = 3 \\ 4x - 2y - 8z = -4 \\ x + y - z = 3 \(\end{cases}\)

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In Exercises 5 - 8, find values for the variables so that the matrices in each exercise are equal. [x4]=[6y]\(\begin{bmatrix}\) x \\ 4 \(\end{bmatrix}\) = \(\begin{bmatrix}\) 6 \\ y \(\end{bmatrix}\)

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In Exercises 5 - 8, find values for the variables so that the matrices in each exercise are equal. [x2yz9]=[41239]\(\begin{bmatrix}\) x & 2y \\ z & 9 \(\end{bmatrix}\) = \(\begin{bmatrix}\) 4 & 12 \\ 3 & 9 \(\end{bmatrix}\)

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Find the products AB and BA to determine whether B is the multiplicative inverse of A.

A=[−2132−12],B=[1234]A = \(\begin{bmatrix}\) -2 & 1 \\ \(\frac{3}{2}\) & -\(\frac{1}{2}\) \(\end{bmatrix}\), \(\quad\) B = \(\begin{bmatrix}\) 1 & 2 \\ 3 & 4 \(\end{bmatrix}\)

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Evaluate each determinant in Exercises 1–10.

∣−5−1−2−7∣\(\begin{vmatrix}\) -5 & -1 \\ -2 & -7 \(\end{vmatrix}\)

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