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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 35
Chapter 7, Problem 35

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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Write the system of equations as an augmented matrix. The coefficients of the variables \( w, x, y, z \) and the constants on the right side form the matrix: \[\left[\begin{array}{cccc|c} 1 & 1 & 1 & 1 & 4 \\ 2 & 1 & -2 & -1 & 0 \\ 1 & -2 & -1 & -2 & -2 \\ 3 & 2 & 1 & 3 & 4 \end{array}\right]\]
Use Gaussian elimination to transform the matrix into an upper triangular form (row echelon form). Start by using the first row to eliminate the \( w \)-terms in rows 2, 3, and 4 by performing row operations such as: - Replace row 2 with (row 2) - 2*(row 1) - Replace row 3 with (row 3) - (row 1) - Replace row 4 with (row 4) - 3*(row 1)
Next, use the second row to eliminate the \( x \)-terms in rows 3 and 4. This involves finding a suitable multiplier to subtract a multiple of row 2 from rows 3 and 4 to create zeros below the pivot in the second column.
Then, use the third row to eliminate the \( y \)-term in row 4 by subtracting a multiple of row 3 from row 4 to get a zero in the third column of the fourth row.
Once the matrix is in upper triangular form, use back-substitution to solve for the variables starting from the last row and moving upwards. This means solving for \( z \) first, then \( y \), then \( x \), and finally \( w \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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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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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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Related Practice
Textbook Question

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\(\end{bmatrix}\)

Textbook Question

In Exercises 31–36, use the alternative method for evaluating third-order determinants on here to evaluate each determinant. ∣0.5750.5390.513∣\(\begin{vmatrix}\)0.5 & 7 & 5 \\0.5 & 3 & 9 \\0.5 & 1 & 3\(\end{vmatrix}\)

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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

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

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Textbook Question

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\(\end{bmatrix}\)

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