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1 Perform the row operation R₁R₁ on the matrix below. - -4 6-8 - 4 5 3 4 6 2 [::::)-1888 -4 5 3 (Simplify your answer.)

1 Perform the row operation R₁R₁ on the matrix below. - -4 6-8 - 4 5 3 4 6 2 [::::)-1888 -4 5 3 (Simplify your answer.)

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Perform the Row Operation on the Given Matrix** **Objective:** Perform the row operation \( \frac{1}{2}R_1 \rightarrow R_1 \) on the matrix below. **Given Matrix:** \[ \begin{bmatrix} -4 & 6 & \mid & -8 \\ -4 & 5 & \mid & 3 \\ \end{bmatrix} \] **Operation:** \[ \frac{1}{2}R_1 \rightarrow R_1 \] This means the first row \( R_1 \) of the matrix will be multiplied by \( \frac{1}{2} \). **Step-by-Step Solution:** 1. **Multiply Each Element in \( R_1 \) by \( \frac{1}{2} \):** - First element: \( -4 \times \frac{1}{2} = -2 \) - Second element: \( 6 \times \frac{1}{2} = 3 \) - Third element (after the vertical line): \( -8 \times \frac{1}{2} = -4 \) 2. **Form the New Matrix:** \[ \begin{bmatrix} -2 & 3 & \mid & -4 \\ -4 & 5 & \mid & 3 \\ \end{bmatrix} \] **Simplified Answer:** \[ \begin{bmatrix} -2 & 3 & \mid & -4 \\ -4 & 5 & \mid & 3 \\ \end{bmatrix} \] This completes the row operation on the given matrix.
Transcribed Image Text:**Perform the Row Operation on the Given Matrix** **Objective:** Perform the row operation \( \frac{1}{2}R_1 \rightarrow R_1 \) on the matrix below. **Given Matrix:** \[ \begin{bmatrix} -4 & 6 & \mid & -8 \\ -4 & 5 & \mid & 3 \\ \end{bmatrix} \] **Operation:** \[ \frac{1}{2}R_1 \rightarrow R_1 \] This means the first row \( R_1 \) of the matrix will be multiplied by \( \frac{1}{2} \). **Step-by-Step Solution:** 1. **Multiply Each Element in \( R_1 \) by \( \frac{1}{2} \):** - First element: \( -4 \times \frac{1}{2} = -2 \) - Second element: \( 6 \times \frac{1}{2} = 3 \) - Third element (after the vertical line): \( -8 \times \frac{1}{2} = -4 \) 2. **Form the New Matrix:** \[ \begin{bmatrix} -2 & 3 & \mid & -4 \\ -4 & 5 & \mid & 3 \\ \end{bmatrix} \] **Simplified Answer:** \[ \begin{bmatrix} -2 & 3 & \mid & -4 \\ -4 & 5 & \mid & 3 \\ \end{bmatrix} \] This completes the row operation on the given matrix.
### Matrix Row Operation Example **Problem Statement:** Perform the row operation \( R_1 + R_2 \rightarrow R_1 \) on the matrix below. \[ \begin{bmatrix} 4 & -6 & 7 \\ 3 & -5 & 2 \end{bmatrix} \] **Application of the Row Operation:** \[ \begin{bmatrix} 4 & -6 & 7 \\ 3 & -5 & 2 \end{bmatrix} \begin{array}{c} R_1 + R_2 \rightarrow R_1 \end{array} \sim \begin{bmatrix} [ \boxed{} & \boxed{} & \boxed{} ] \\ [ 3 & -5 & 2 ] \end{bmatrix} \] **Instructions:** Simplify your answer by calculating the new entries in the first row after applying the row operation \( R_1 + R_2 \rightarrow R_1 \). Replace the boxed placeholders with the calculated values. **Detailed Steps:** 1. Add the corresponding elements of \( R_1 \) and \( R_2 \): - First element: \( 4 + 3 = 7 \) - Second element: \( -6 + (-5) = -11 \) - Third element: \( 7 + 2 = 9 \) 2. Update the first row with the new values: \[ \begin{bmatrix} [ 7 & -11 & 9 ] \\ [ 3 & -5 & 2 ] \end{bmatrix} \] The final matrix after performing the row operation \( R_1 + R_2 \rightarrow R_1 \) is: \[ \begin{bmatrix} 7 & -11 & 9 \\ 3 & -5 & 2 \end{bmatrix} \] **Summary:** Applying the row operation helps in simplifying matrices for further mathematical operations, especially in solving systems of linear equations using methods like Gaussian elimination.
Transcribed Image Text:### Matrix Row Operation Example **Problem Statement:** Perform the row operation \( R_1 + R_2 \rightarrow R_1 \) on the matrix below. \[ \begin{bmatrix} 4 & -6 & 7 \\ 3 & -5 & 2 \end{bmatrix} \] **Application of the Row Operation:** \[ \begin{bmatrix} 4 & -6 & 7 \\ 3 & -5 & 2 \end{bmatrix} \begin{array}{c} R_1 + R_2 \rightarrow R_1 \end{array} \sim \begin{bmatrix} [ \boxed{} & \boxed{} & \boxed{} ] \\ [ 3 & -5 & 2 ] \end{bmatrix} \] **Instructions:** Simplify your answer by calculating the new entries in the first row after applying the row operation \( R_1 + R_2 \rightarrow R_1 \). Replace the boxed placeholders with the calculated values. **Detailed Steps:** 1. Add the corresponding elements of \( R_1 \) and \( R_2 \): - First element: \( 4 + 3 = 7 \) - Second element: \( -6 + (-5) = -11 \) - Third element: \( 7 + 2 = 9 \) 2. Update the first row with the new values: \[ \begin{bmatrix} [ 7 & -11 & 9 ] \\ [ 3 & -5 & 2 ] \end{bmatrix} \] The final matrix after performing the row operation \( R_1 + R_2 \rightarrow R_1 \) is: \[ \begin{bmatrix} 7 & -11 & 9 \\ 3 & -5 & 2 \end{bmatrix} \] **Summary:** Applying the row operation helps in simplifying matrices for further mathematical operations, especially in solving systems of linear equations using methods like Gaussian elimination.
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