Consider the following system of linear equations. -z = -1 3y -z=3 4x + 5y 3z = 5 2x Solve the system by completing the steps below to produce a reduced row-echelon form. R₁, R₂, and R3 denote the first, second, and third rows, respectively. The arrow notation (→) stands for "replaces," where the expression on the left of the arrow replaces the expression on the right.
Consider the following system of linear equations. -z = -1 3y -z=3 4x + 5y 3z = 5 2x Solve the system by completing the steps below to produce a reduced row-echelon form. R₁, R₂, and R3 denote the first, second, and third rows, respectively. The arrow notation (→) stands for "replaces," where the expression on the left of the arrow replaces the expression on the right.
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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Question
:
\[
\begin{align*}
2x & - z = -1 \\
3y & - z = 3 \\
4x + 5y & - 3z = 5
\end{align*}
\]
Solve the system by completing the steps below to produce a [reduced row-echelon form](#).
\( R_1, R_2, \) and \( R_3 \) denote the first, second, and third rows, respectively.
The arrow notation (\(\rightarrow\)) stands for "replaces," where the expression on the left of the arrow replaces the expression on the right.
Here is the [augmented matrix](#):
\[
\begin{bmatrix}
2 & 0 & -1 & | & -1 \\
0 & 3 & -1 & | & 3 \\
4 & 5 & -3 & | & 5
\end{bmatrix}
\]
Enter the missing coefficients for the row operations.
1. \( \square \cdot R_1 \rightarrow R_1 : \)
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & | & -\frac{1}{2} \\
0 & 3 & -1 & | & 3 \\
4 & 5 & -3 & | & 5
\end{bmatrix}
\]
2. \( \square \cdot R_1 + R_3 \rightarrow R_3 : \)
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & | & -\frac{1}{2} \\
0 & 3 & -1 & | & 3 \\
0 & 5 & -1 & | & 7
\end{bmatrix}
\]
3. \( \square \cdot R_2 \rightarrow R_2 : \)
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & | & -\frac{1}{2} \\
0 & 1 & -\frac{1}{3} & | & 1 \\
0 & 5 & -1 & | & 7
\end{bmatrix}
\]
4. \( \square \cd](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F50ad5a28-9438-427f-8fb0-25e8d5617a6d%2Faf7913e8-d84d-429c-adf1-d52f0516b99b%2Fd4lbeg9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the following [system of linear equations](#):
\[
\begin{align*}
2x & - z = -1 \\
3y & - z = 3 \\
4x + 5y & - 3z = 5
\end{align*}
\]
Solve the system by completing the steps below to produce a [reduced row-echelon form](#).
\( R_1, R_2, \) and \( R_3 \) denote the first, second, and third rows, respectively.
The arrow notation (\(\rightarrow\)) stands for "replaces," where the expression on the left of the arrow replaces the expression on the right.
Here is the [augmented matrix](#):
\[
\begin{bmatrix}
2 & 0 & -1 & | & -1 \\
0 & 3 & -1 & | & 3 \\
4 & 5 & -3 & | & 5
\end{bmatrix}
\]
Enter the missing coefficients for the row operations.
1. \( \square \cdot R_1 \rightarrow R_1 : \)
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & | & -\frac{1}{2} \\
0 & 3 & -1 & | & 3 \\
4 & 5 & -3 & | & 5
\end{bmatrix}
\]
2. \( \square \cdot R_1 + R_3 \rightarrow R_3 : \)
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & | & -\frac{1}{2} \\
0 & 3 & -1 & | & 3 \\
0 & 5 & -1 & | & 7
\end{bmatrix}
\]
3. \( \square \cdot R_2 \rightarrow R_2 : \)
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & | & -\frac{1}{2} \\
0 & 1 & -\frac{1}{3} & | & 1 \\
0 & 5 & -1 & | & 7
\end{bmatrix}
\]
4. \( \square \cd
![The image presents a series of matrix operations aimed at solving a linear system using row operations. The focus is on finding the appropriate coefficients for each operation to transform the system into its solution form.
1. **Initial Matrix:**
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & \vline & -\frac{1}{2} \\
0 & 3 & -1 & \vline & 3 \\
0 & 5 & -1 & \vline & 7
\end{bmatrix}
\]
2. **Step 2: Missing Coefficient for Row Operation \( \Box \cdot R_1 + R_3 \to R_3 \):**
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & \vline & -\frac{1}{2} \\
0 & 3 & -1 & \vline & 3 \\
0 & 5 & -1 & \vline & 7
\end{bmatrix}
\]
3. **Step 3: Missing Coefficient for Row Operation \( \Box \cdot R_2 \to R_2 \):**
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & \vline & -\frac{1}{2} \\
0 & 1 & -\frac{1}{3} & \vline & 1 \\
0 & 5 & -1 & \vline & 7
\end{bmatrix}
\]
4. **Step 4: Missing Coefficient for Row Operation \( \Box \cdot R_2 + R_3 \to R_3 \):**
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & \vline & -\frac{1}{2} \\
0 & 1 & -\frac{1}{3} & \vline & 1 \\
0 & 0 & \frac{2}{3} & \vline & 2
\end{bmatrix}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F50ad5a28-9438-427f-8fb0-25e8d5617a6d%2Faf7913e8-d84d-429c-adf1-d52f0516b99b%2Fwrcgx2c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image presents a series of matrix operations aimed at solving a linear system using row operations. The focus is on finding the appropriate coefficients for each operation to transform the system into its solution form.
1. **Initial Matrix:**
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & \vline & -\frac{1}{2} \\
0 & 3 & -1 & \vline & 3 \\
0 & 5 & -1 & \vline & 7
\end{bmatrix}
\]
2. **Step 2: Missing Coefficient for Row Operation \( \Box \cdot R_1 + R_3 \to R_3 \):**
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & \vline & -\frac{1}{2} \\
0 & 3 & -1 & \vline & 3 \\
0 & 5 & -1 & \vline & 7
\end{bmatrix}
\]
3. **Step 3: Missing Coefficient for Row Operation \( \Box \cdot R_2 \to R_2 \):**
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & \vline & -\frac{1}{2} \\
0 & 1 & -\frac{1}{3} & \vline & 1 \\
0 & 5 & -1 & \vline & 7
\end{bmatrix}
\]
4. **Step 4: Missing Coefficient for Row Operation \( \Box \cdot R_2 + R_3 \to R_3 \):**
\[
\begin{bmatrix}
1 & 0 & -\frac{1}{2} & \vline & -\frac{1}{2} \\
0 & 1 & -\frac{1}{3} & \vline & 1 \\
0 & 0 & \frac{2}{3} & \vline & 2
\end{bmatrix}
\]
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