Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To solve the given system of equations using Gauss-Jordan elimination, we should first write the augmented matrix corresponding to the system:
[tex]\[ \begin{pmatrix} 4 & 10 & -24 & -22 \\ 3 & 25 & -61 & -23 \\ 1 & 5 & -12 & -6 \end{pmatrix} \][/tex]
We now proceed with the Gauss-Jordan elimination method:
1. Make the leading coefficient of the first row 1 by dividing the first row by 4:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 3 & 25 & -61 & -23 \\ 1 & 5 & -12 & -6 \end{pmatrix} \][/tex]
2. Eliminate the x1 term from the second and third rows using the first row:
- For the second row: [tex]\( R2 = R2 - 3 \cdot R1 \)[/tex]
[tex]\[ R2 = \begin{pmatrix} 3 & 25 & -61 & -23 \end{pmatrix} - 3 \cdot \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 3 - 3 \cdot 1 & 25 - 3 \cdot 2.5 & -61 + 3 \cdot 6 & -23 + 3 \cdot 5.5 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 17.5 & -43 & -6.5 \end{pmatrix} \][/tex]
- For the third row: [tex]\( R3 = R3 - R1 \)[/tex]
[tex]\[ R3 = \begin{pmatrix} 1 & 5 & -12 & -6 \end{pmatrix} - \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
The updated matrix is:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 0 & 17.5 & -43 & -6.5 \\ 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
3. Make the leading coefficient of the second row 1 by dividing the second row by 17.5:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 0 & 1 & -2.4571 & -0.3714 \\ 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
4. Eliminate the x2 term from the first and third rows using the second row:
- For the first row: [tex]\( R1 = R1 - 2.5 \cdot R2 \)[/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} - 2.5 \cdot \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} \][/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \end{pmatrix} \][/tex]
- For the third row: [tex]\( R3 = R3 - 2.5 \cdot R2 \)[/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 2.5 & -6 & -0.5 \end{pmatrix} - 2.5 \cdot \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 0 & 0.1425 & 0.4285 \end{pmatrix} \][/tex]
The updated matrix is:
[tex]\[ \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \\ 0 & 1 & -2.4571 & -0.3714 \\ 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
5. Make the leading coefficient of the third row 1 (it's already 1), and then we have to eliminate the x3 term from the first and second rows using the third row:
- For the first row: [tex]\( R1 = R1 - 0.1429 \cdot R3 \)[/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \end{pmatrix} - 0.1429 \cdot \begin{pmatrix} 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0 & -4 + (- 5) \end{pmatrix} \][/tex]
- For the second row: [tex]\( R2 = R2 + 2.4571 \cdot R3 \)[/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} + 2.4571 \cdot \begin{pmatrix} 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 1 & 0 & 7 \end{pmatrix} \][/tex]
So, we finally have:
[tex]\[ \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 3 \end{pmatrix} \][/tex]
So the solution to the system is:
[tex]\[ x_1 = -5, \quad x_2 = 7, \quad x_3 = 3. \][/tex]
Thus, the correct choice is:
A. The unique solution is [tex]\(x_1 = -5\)[/tex] , [tex]\(x_2 = 7\)[/tex], and [tex]\(x_3 = 3\)[/tex].
[tex]\[ \begin{pmatrix} 4 & 10 & -24 & -22 \\ 3 & 25 & -61 & -23 \\ 1 & 5 & -12 & -6 \end{pmatrix} \][/tex]
We now proceed with the Gauss-Jordan elimination method:
1. Make the leading coefficient of the first row 1 by dividing the first row by 4:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 3 & 25 & -61 & -23 \\ 1 & 5 & -12 & -6 \end{pmatrix} \][/tex]
2. Eliminate the x1 term from the second and third rows using the first row:
- For the second row: [tex]\( R2 = R2 - 3 \cdot R1 \)[/tex]
[tex]\[ R2 = \begin{pmatrix} 3 & 25 & -61 & -23 \end{pmatrix} - 3 \cdot \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 3 - 3 \cdot 1 & 25 - 3 \cdot 2.5 & -61 + 3 \cdot 6 & -23 + 3 \cdot 5.5 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 17.5 & -43 & -6.5 \end{pmatrix} \][/tex]
- For the third row: [tex]\( R3 = R3 - R1 \)[/tex]
[tex]\[ R3 = \begin{pmatrix} 1 & 5 & -12 & -6 \end{pmatrix} - \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
The updated matrix is:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 0 & 17.5 & -43 & -6.5 \\ 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
3. Make the leading coefficient of the second row 1 by dividing the second row by 17.5:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 0 & 1 & -2.4571 & -0.3714 \\ 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
4. Eliminate the x2 term from the first and third rows using the second row:
- For the first row: [tex]\( R1 = R1 - 2.5 \cdot R2 \)[/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} - 2.5 \cdot \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} \][/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \end{pmatrix} \][/tex]
- For the third row: [tex]\( R3 = R3 - 2.5 \cdot R2 \)[/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 2.5 & -6 & -0.5 \end{pmatrix} - 2.5 \cdot \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 0 & 0.1425 & 0.4285 \end{pmatrix} \][/tex]
The updated matrix is:
[tex]\[ \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \\ 0 & 1 & -2.4571 & -0.3714 \\ 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
5. Make the leading coefficient of the third row 1 (it's already 1), and then we have to eliminate the x3 term from the first and second rows using the third row:
- For the first row: [tex]\( R1 = R1 - 0.1429 \cdot R3 \)[/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \end{pmatrix} - 0.1429 \cdot \begin{pmatrix} 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0 & -4 + (- 5) \end{pmatrix} \][/tex]
- For the second row: [tex]\( R2 = R2 + 2.4571 \cdot R3 \)[/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} + 2.4571 \cdot \begin{pmatrix} 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 1 & 0 & 7 \end{pmatrix} \][/tex]
So, we finally have:
[tex]\[ \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 3 \end{pmatrix} \][/tex]
So the solution to the system is:
[tex]\[ x_1 = -5, \quad x_2 = 7, \quad x_3 = 3. \][/tex]
Thus, the correct choice is:
A. The unique solution is [tex]\(x_1 = -5\)[/tex] , [tex]\(x_2 = 7\)[/tex], and [tex]\(x_3 = 3\)[/tex].
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
