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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 35
Capitolo 7, Problema 35

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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Step 1: Write down the given 3x3 determinant: \[\left| \begin{array}{ccc} 0.5 & 7 & 5 \\ 0.5 & 3 & 9 \\ 0.5 & 1 & 3 \end{array} \right|\]
Step 2: Use the alternative method for evaluating third-order determinants, which involves expanding along the first row or using the rule of Sarrus. Here, we will use the rule of Sarrus: - Repeat the first two columns to the right of the matrix: \[\begin{array}{ccc|cc} 0.5 & 7 & 5 & 0.5 & 7 \\ 0.5 & 3 & 9 & 0.5 & 3 \\ 0.5 & 1 & 3 & 0.5 & 1 \end{array}\]
Step 3: Calculate the sum of the products of the diagonals from top-left to bottom-right: \[ (0.5 \times 3 \times 3) + (7 \times 9 \times 0.5) + (5 \times 0.5 \times 1) \]
Step 4: Calculate the sum of the products of the diagonals from bottom-left to top-right: \[ (0.5 \times 3 \times 5) + (1 \times 9 \times 0.5) + (3 \times 0.5 \times 7) \]
Step 5: Subtract the sum from Step 4 from the sum in Step 3 to find the value of the determinant: \[ \text{Determinant} = \left[ (0.5 \times 3 \times 3) + (7 \times 9 \times 0.5) + (5 \times 0.5 \times 1) \right] - \left[ (0.5 \times 3 \times 5) + (1 \times 9 \times 0.5) + (3 \times 0.5 \times 7) \right] \]

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Third-Order Determinants

A third-order determinant is a scalar value calculated from a 3x3 matrix. It helps determine properties like matrix invertibility and solutions to systems of equations. The determinant is computed using specific formulas or methods, such as expansion by minors or the alternative method.
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Determinants of 3×3 Matrices

Alternative Method for Evaluating Determinants

The alternative method, often called the Rule of Sarrus, is a shortcut for finding the determinant of a 3x3 matrix. It involves summing the products of diagonals from left to right and subtracting the products of diagonals from right to left, simplifying the calculation process.
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Determinants of 3×3 Matrices

Matrix Representation of Systems of Equations

Matrices can represent systems of linear equations compactly, where each row corresponds to an equation and each column to a variable or constant. Evaluating the determinant of the coefficient matrix helps determine if the system has a unique solution, infinite solutions, or none.
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Introduction to Systems of Linear Equations
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Domanda del libro di testo

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

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Domanda del libro di testo

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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Domanda del libro di testo

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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Domanda del libro di testo

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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Domanda del libro di testo

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

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Domanda del libro di testo

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

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