Skip to main content
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 13
Capitolo 7, Problema 13

Perform each matrix row operation and write the new matrix.
[2−64∣1015−5∣0304∣7]12R1\(\begin{bmatrix}\) 2 & -6 & 4 & \(\vert\) & 10 \\ 1 & 5 & -5 & \(\vert\) & 0 \\ 3 & 0 & 4 & \(\vert\) & 7 \(\end{bmatrix}\) \(\quad\) \(\frac{1}{2}\)R_1

Guida verificata passo dopo passo
1
Identify the given matrix and the row operation to be performed. The matrix is: \[\left[ \begin{array}{ccc|c} 2 & -6 & 4 & 10 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \end{array} \right]\] and the operation is \[\frac{1}{2} R_1\], which means multiply every element in row 1 by \[\frac{1}{2}\].
Apply the operation to row 1 by multiplying each element in the first row by \[\frac{1}{2}\]: - First element: \[2 \times \frac{1}{2} = 1\] - Second element: \[-6 \times \frac{1}{2} = -3\] - Third element: \[4 \times \frac{1}{2} = 2\] - Augmented element: \[10 \times \frac{1}{2} = 5\]
Write the new matrix with the updated first row and the unchanged rows 2 and 3: \[\left[ \begin{array}{ccc|c} 1 & -3 & 2 & 5 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \end{array} \right]\]
Double-check that only row 1 has changed and rows 2 and 3 remain the same.
This completes the row operation \[\frac{1}{2} R_1\] on the matrix.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
2m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Matrix Row Operations

Matrix row operations are techniques used to manipulate the rows of a matrix to simplify or solve systems of linear equations. These include row swapping, scaling a row by a nonzero constant, and adding a multiple of one row to another. They preserve the solution set of the system.
Video consigliato:
Percorso guidato
8:38
Performing Row Operations on Matrices

Scalar Multiplication of a Row

Scalar multiplication involves multiplying every element in a row by the same nonzero constant. This operation changes the row but keeps the system equivalent. For example, multiplying row 1 by 1/2 scales all its entries by 0.5, simplifying the row for further operations.
Video consigliato:
Percorso guidato
8:38
Performing Row Operations on Matrices

Augmented Matrix Representation

An augmented matrix combines the coefficients of variables and constants from a system of linear equations into one matrix. It helps visualize and perform row operations efficiently to solve the system. The vertical bar separates coefficients from constants.
Video consigliato:
Percorso guidato
4:35
Introduction to Matrices
Pratica correlata
Domanda del libro di testo

Find the following matrices: A+BA + B

A=[2−41],B=[−53−1]A = \(\begin{bmatrix}\) 2 \\ -4 \\ 1 \(\end{bmatrix}\) , B = \(\begin{bmatrix}\) -5 \\ 3 \\ -1 \(\end{bmatrix}\)

99
views
Domanda del libro di testo

Use the fact that if A=[abcd]A = \(\begin{bmatrix}\) a & b \\ c & d \(\end{bmatrix}\), then A−1=1ad−bc[d−b−ca]A^{-1} = \(\frac{1}{ad-bc}\) \(\begin{bmatrix}\) d & -b \\ -c & a \(\end{bmatrix}\) to find the inverse of each matrix, if possible. Check that AA−1=I2AA^{-1} = I_2 and A−1A=I2A^{-1} A = I_2.

A=[23−12]A = \(\begin{bmatrix}\) 2 & 3 \\ -1 & 2 \(\end{bmatrix}\)

622
views
Domanda del libro di testo

For Exercises 11–22, use Cramer's Rule to solve each system. {12x+3y=152x−3y=13\(\begin{cases}\) 12x + 3y = 15 \\ 2x - 3y = 13 \(\end{cases}\)

742
views
Domanda del libro di testo

In Exercises 9 - 16, find the following matrices: b. A - B

114
views
Domanda del libro di testo

In Exercises 9 - 16, find the following matrices: c. - 4A

89
views
Domanda del libro di testo

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. {w−3x+y−4z=4−2w+x+2y=−23w−2x+y−6z=2−w+3x+2y−z=−6\(\begin{cases}\) w - 3x + y - 4z = 4 \\ -2w + x + 2y = -2 \\ 3w - 2x + y - 6z = 2 \\ -w + 3x + 2y - z = -6 \(\end{cases}\)

673
views