The following system has an infinite number of solutions. Form an augmented matrix, then write the matrix in the reduced form. Write the reduced form of the matrix below and then write the solution in terms of z. The required augmented matrix is: where a11 = 1 a21 = a31 = 0 0 and x = 1 1 , a12 = 0 , a22 = , a32 = 1 0 1x 2x 3x , a 13 = ,a23 = , Q33 = +11 7/4 ly + 3z 2y + 1z ly + 5/4 0 a11 a12 a13 b1 a21 a22 a23 b2 a31 a32 a33 b3 b₁ = , b2 = b3 = " 2 = || || || 4z = = , y = 5/2 1/2 0 " 3 4 7
The following system has an infinite number of solutions. Form an augmented matrix, then write the matrix in the reduced form. Write the reduced form of the matrix below and then write the solution in terms of z. The required augmented matrix is: where a11 = 1 a21 = a31 = 0 0 and x = 1 1 , a12 = 0 , a22 = , a32 = 1 0 1x 2x 3x , a 13 = ,a23 = , Q33 = +11 7/4 ly + 3z 2y + 1z ly + 5/4 0 a11 a12 a13 b1 a21 a22 a23 b2 a31 a32 a33 b3 b₁ = , b2 = b3 = " 2 = || || || 4z = = , y = 5/2 1/2 0 " 3 4 7
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter1: Systems Of Linear Equations
Section1.2: Guassian Eliminationa And Guass-jordan Elimination
Problem 67E
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![### Infinite Solutions in Linear Systems
The following system of linear equations has an infinite number of solutions. Our goal is to form an augmented matrix, reduce the matrix, and then write the solution in terms of \( z \).
\[
\begin{align*}
1x + 1y + 3z &= 3 \\
2x - 2y + 1z &= 4 \\
3x - 1y + 4z &= 7 \\
\end{align*}
\]
#### Step 1: Form the Augmented Matrix
An augmented matrix is formed by appending the columns of the constants from the right-hand side of the system to the coefficient matrix:
\[
\left( \begin{array}{ccc|c}
a_{11} & a_{12} & a_{13} & b_1 \\
a_{21} & a_{22} & a_{23} & b_2 \\
a_{31} & a_{32} & a_{33} & b_3
\end{array} \right)
\]
#### Step 2: Reduced Matrix
We then row-reduce this matrix to its reduced form:
\[
\left( \begin{array}{ccc|c}
1 & 0 & \frac{7}{4} & \frac{5}{2} \\
0 & 1 & \frac{5}{4} & \frac{1}{2} \\
0 & 0 & 0 & 0 \\
\end{array} \right)
\]
where the elements of the reduced matrix are given as:
\[
\begin{align*}
a_{11} &= 1, & a_{12} &= 0, & a_{13} &= \frac{7}{4}, & b_1 &= \frac{5}{2} \\
a_{21} &= 0, & a_{22} &= 1, & a_{23} &= \frac{5}{4}, & b_2 &= \frac{1}{2} \\
a_{31} &= 0, & a_{32} &= 0, & a_{33} &= 0, & b_3 &= 0 \\
\end{align*}
\]
#### Step 3: Solution in Terms of \( z \)
Given the reduced form, the system indicates dependency among the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30a42a85-c58f-45ac-a4af-faeed1a599e1%2F8eec3d51-b3d2-4d5a-8184-e46955ab937f%2F0b26yy4_processed.png&w=3840&q=75)
Transcribed Image Text:### Infinite Solutions in Linear Systems
The following system of linear equations has an infinite number of solutions. Our goal is to form an augmented matrix, reduce the matrix, and then write the solution in terms of \( z \).
\[
\begin{align*}
1x + 1y + 3z &= 3 \\
2x - 2y + 1z &= 4 \\
3x - 1y + 4z &= 7 \\
\end{align*}
\]
#### Step 1: Form the Augmented Matrix
An augmented matrix is formed by appending the columns of the constants from the right-hand side of the system to the coefficient matrix:
\[
\left( \begin{array}{ccc|c}
a_{11} & a_{12} & a_{13} & b_1 \\
a_{21} & a_{22} & a_{23} & b_2 \\
a_{31} & a_{32} & a_{33} & b_3
\end{array} \right)
\]
#### Step 2: Reduced Matrix
We then row-reduce this matrix to its reduced form:
\[
\left( \begin{array}{ccc|c}
1 & 0 & \frac{7}{4} & \frac{5}{2} \\
0 & 1 & \frac{5}{4} & \frac{1}{2} \\
0 & 0 & 0 & 0 \\
\end{array} \right)
\]
where the elements of the reduced matrix are given as:
\[
\begin{align*}
a_{11} &= 1, & a_{12} &= 0, & a_{13} &= \frac{7}{4}, & b_1 &= \frac{5}{2} \\
a_{21} &= 0, & a_{22} &= 1, & a_{23} &= \frac{5}{4}, & b_2 &= \frac{1}{2} \\
a_{31} &= 0, & a_{32} &= 0, & a_{33} &= 0, & b_3 &= 0 \\
\end{align*}
\]
#### Step 3: Solution in Terms of \( z \)
Given the reduced form, the system indicates dependency among the
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