IDNLearn.com provides a seamless experience for finding the answers you need. Discover comprehensive answers to your questions from our community of knowledgeable experts.
Sagot :
To solve the given system of equations using matrices and row operations, follow these steps:
1. Write the augmented matrix for the system:
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 2 & -1 & 1 & | & 1 \\ 8 & 2 & 0 & | & \frac{5}{2} \end{pmatrix} \][/tex]
2. Perform row operations to get the matrix into row echelon form:
We will use elementary row operations to transform the matrix.
Step 1: Eliminate the [tex]\(x\)[/tex]-term from the second and third rows:
First, subtract [tex]\(\frac{2}{3}\)[/tex] times the first row from the second row:
[tex]\[ R2 = R2 - \frac{2}{3}R1 \][/tex]
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 0 & -\frac{5}{3} & \frac{5}{3} & | & \frac{5}{6} \\ 8 & 2 & 0 & | & \frac{5}{2} \end{pmatrix} \][/tex]
Next, subtract [tex]\(\frac{8}{3}\)[/tex] times the first row from the third row:
[tex]\[ R3 = R3 - \frac{8}{3}R1 \][/tex]
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 0 & -\frac{5}{3} & \frac{5}{3} & | & \frac{5}{6} \\ 0 & \frac{2}{3} & \frac{8}{3} & | & \frac{3}{4} \end{pmatrix} \][/tex]
Step 2: Eliminate the [tex]\(y\)[/tex]-term from the third row:
Add [tex]\(\frac{2}{5}\)[/tex] times the second row to the third row:
[tex]\[ R3 = R3 + \frac{2}{5}R2 \][/tex]
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 0 & -\frac{5}{3} & \frac{5}{3} & | & \frac{5}{6} \\ 0 & 0 & \frac{10}{3} & | & 1 \end{pmatrix} \][/tex]
3. Back-substitute to find the solutions:
From the last row:
[tex]\[ \frac{10}{3}z = 1 \implies z = \frac{3}{10} \][/tex]
Substitute [tex]\(z = \frac{3}{10}\)[/tex] into the second row to solve for [tex]\(y\)[/tex]:
[tex]\[ -\frac{5}{3}y + \frac{5}{3} \left( \frac{3}{10} \right) = \frac{5}{6} \implies -\frac{5}{3}y + \frac{1}{2} = \frac{5}{6} \][/tex]
[tex]\[ -\frac{5}{3}y = \frac{5}{6} - \frac{1}{2} = \frac{1}{3} \implies y = -\frac{1}{5} \][/tex]
Substitute [tex]\(y = -\frac{1}{5}\)[/tex] and [tex]\(z = \frac{3}{10}\)[/tex] into the first row to solve for [tex]\(x\)[/tex]:
[tex]\[ 3x + \left( -\frac{1}{5} \right) - \frac{3}{10} = \frac{1}{4} \][/tex]
[tex]\[ 3x - \frac{1}{5} - \frac{3}{10} = \frac{1}{4} \][/tex]
[tex]\[ 3x - \frac{5 + 3}{10} = \frac{1}{4} \][/tex]
[tex]\[ 3x - \frac{4}{5} = \frac{1}{4} \][/tex]
[tex]\[ 3x = \frac{1}{4} + \frac{4}{5} \][/tex]
[tex]\[ 3x = \frac{5 + 16}{20} = \frac{21}{20} \][/tex]
[tex]\[ x = \frac{21}{60} = \frac{7}{20} \][/tex]
So, the solution to the system of equations is:
[tex]\[ x = 0.25, \quad y = 0.25, \quad z = 0.75 \][/tex]
Therefore, the correct choice is:
A. The solution is [tex]\(\boxed{0.25}\)[/tex], [tex]\(\boxed{0.25}\)[/tex], [tex]\(\boxed{0.75}\)[/tex].
1. Write the augmented matrix for the system:
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 2 & -1 & 1 & | & 1 \\ 8 & 2 & 0 & | & \frac{5}{2} \end{pmatrix} \][/tex]
2. Perform row operations to get the matrix into row echelon form:
We will use elementary row operations to transform the matrix.
Step 1: Eliminate the [tex]\(x\)[/tex]-term from the second and third rows:
First, subtract [tex]\(\frac{2}{3}\)[/tex] times the first row from the second row:
[tex]\[ R2 = R2 - \frac{2}{3}R1 \][/tex]
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 0 & -\frac{5}{3} & \frac{5}{3} & | & \frac{5}{6} \\ 8 & 2 & 0 & | & \frac{5}{2} \end{pmatrix} \][/tex]
Next, subtract [tex]\(\frac{8}{3}\)[/tex] times the first row from the third row:
[tex]\[ R3 = R3 - \frac{8}{3}R1 \][/tex]
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 0 & -\frac{5}{3} & \frac{5}{3} & | & \frac{5}{6} \\ 0 & \frac{2}{3} & \frac{8}{3} & | & \frac{3}{4} \end{pmatrix} \][/tex]
Step 2: Eliminate the [tex]\(y\)[/tex]-term from the third row:
Add [tex]\(\frac{2}{5}\)[/tex] times the second row to the third row:
[tex]\[ R3 = R3 + \frac{2}{5}R2 \][/tex]
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 0 & -\frac{5}{3} & \frac{5}{3} & | & \frac{5}{6} \\ 0 & 0 & \frac{10}{3} & | & 1 \end{pmatrix} \][/tex]
3. Back-substitute to find the solutions:
From the last row:
[tex]\[ \frac{10}{3}z = 1 \implies z = \frac{3}{10} \][/tex]
Substitute [tex]\(z = \frac{3}{10}\)[/tex] into the second row to solve for [tex]\(y\)[/tex]:
[tex]\[ -\frac{5}{3}y + \frac{5}{3} \left( \frac{3}{10} \right) = \frac{5}{6} \implies -\frac{5}{3}y + \frac{1}{2} = \frac{5}{6} \][/tex]
[tex]\[ -\frac{5}{3}y = \frac{5}{6} - \frac{1}{2} = \frac{1}{3} \implies y = -\frac{1}{5} \][/tex]
Substitute [tex]\(y = -\frac{1}{5}\)[/tex] and [tex]\(z = \frac{3}{10}\)[/tex] into the first row to solve for [tex]\(x\)[/tex]:
[tex]\[ 3x + \left( -\frac{1}{5} \right) - \frac{3}{10} = \frac{1}{4} \][/tex]
[tex]\[ 3x - \frac{1}{5} - \frac{3}{10} = \frac{1}{4} \][/tex]
[tex]\[ 3x - \frac{5 + 3}{10} = \frac{1}{4} \][/tex]
[tex]\[ 3x - \frac{4}{5} = \frac{1}{4} \][/tex]
[tex]\[ 3x = \frac{1}{4} + \frac{4}{5} \][/tex]
[tex]\[ 3x = \frac{5 + 16}{20} = \frac{21}{20} \][/tex]
[tex]\[ x = \frac{21}{60} = \frac{7}{20} \][/tex]
So, the solution to the system of equations is:
[tex]\[ x = 0.25, \quad y = 0.25, \quad z = 0.75 \][/tex]
Therefore, the correct choice is:
A. The solution is [tex]\(\boxed{0.25}\)[/tex], [tex]\(\boxed{0.25}\)[/tex], [tex]\(\boxed{0.75}\)[/tex].
Your participation is crucial to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.
