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Sagot :
To solve the system of equations using Gaussian elimination, we need to perform row operations to convert the matrix into its reduced row echelon form (RREF). The matrices provided correspond to different stages during this process. We need to arrange them in the correct order to understand the solution steps sequentially. Here are the detailed steps:
1. Initial System (Matrix II):
[tex]\[ \begin{bmatrix} -1 & -1 & 2 & 7 \\ 2 & 1 & 1 & 2 \\ -3 & 2 & 3 & 7 \\ \end{bmatrix} \][/tex]
This matrix represents the original system of equations:
[tex]\[ \begin{cases} -a - b + 2c = 7 \\ 2a + b + c = 2 \\ -3a + 2b + 3c = 7 \end{cases} \][/tex]
2. First Row Operation (Matrix III):
[tex]\[ \begin{bmatrix} 1 & 1 & -2 & -7 \\ 0 & 1 & -5 & -16 \\ 0 & 0 & 22 & 66 \\ \end{bmatrix} \][/tex]
The operations performed here likely involve swapping rows and scaling to get a leading 1 in the top-left corner followed by eliminating the first column below the pivot.
3. Second Row Operation (Matrix IV):
[tex]\[ \begin{bmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \\ \end{bmatrix} \][/tex]
Here, further row operations have been performed to achieve zeros below the leading 1s in the second and first columns, leading towards the RREF form.
4. Final Reduced Row Echelon Form (Matrix I):
[tex]\[ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \\ \end{bmatrix} \][/tex]
This is the final form where the system of equations is fully reduced:
[tex]\[ \begin{cases} a = 0 \\ b = -1 \\ c = 3 \end{cases} \][/tex]
Thus, the correct order of matrices from start to finish in terms of solving the system of equations is:
1. Matrix II
2. Matrix III
3. Matrix IV
4. Matrix I
1. Initial System (Matrix II):
[tex]\[ \begin{bmatrix} -1 & -1 & 2 & 7 \\ 2 & 1 & 1 & 2 \\ -3 & 2 & 3 & 7 \\ \end{bmatrix} \][/tex]
This matrix represents the original system of equations:
[tex]\[ \begin{cases} -a - b + 2c = 7 \\ 2a + b + c = 2 \\ -3a + 2b + 3c = 7 \end{cases} \][/tex]
2. First Row Operation (Matrix III):
[tex]\[ \begin{bmatrix} 1 & 1 & -2 & -7 \\ 0 & 1 & -5 & -16 \\ 0 & 0 & 22 & 66 \\ \end{bmatrix} \][/tex]
The operations performed here likely involve swapping rows and scaling to get a leading 1 in the top-left corner followed by eliminating the first column below the pivot.
3. Second Row Operation (Matrix IV):
[tex]\[ \begin{bmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \\ \end{bmatrix} \][/tex]
Here, further row operations have been performed to achieve zeros below the leading 1s in the second and first columns, leading towards the RREF form.
4. Final Reduced Row Echelon Form (Matrix I):
[tex]\[ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \\ \end{bmatrix} \][/tex]
This is the final form where the system of equations is fully reduced:
[tex]\[ \begin{cases} a = 0 \\ b = -1 \\ c = 3 \end{cases} \][/tex]
Thus, the correct order of matrices from start to finish in terms of solving the system of equations is:
1. Matrix II
2. Matrix III
3. Matrix IV
4. Matrix I
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