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Sagot :
To transform the given matrix into reduced row echelon form (RREF), we need to apply a series of row operations to make it conform to the rules of RREF, where:
1. The leading entry in each nonzero row is 1.
2. Each leading 1 is the only nonzero entry in its column.
3. The leading 1 in a given row is to the right of the leading 1 in the row above it.
4. Rows with all zero elements, if any, are below rows with nonzero elements.
Given the matrix:
[tex]\[ \left[\begin{array}{ccc|c} 2 & -1 & -4 & 7 \\ 0 & -30 & 20 & -50 \\ 0 & 0 & 44 & -44 \end{array}\right] \][/tex]
Let's denote this matrix as \( A \).
### Step-by-Step Transformation
1. Normalize the first row to have a leading 1:
To achieve this, divide the entire first row by 2:
[tex]\[ \frac{1}{2} \times \left[\begin{array}{ccc|c} 2 & -1 & -4 & 7 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -2 & \frac{7}{2} \end{array}\right] \][/tex]
So, our system becomes:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -2 & \frac{7}{2} \\ 0 & -30 & 20 & -50 \\ 0 & 0 & 44 & -44 \end{array}\right] \][/tex]
2. Transform the second row to make the first column zero (it is already zero, so we skip this step).
3. Normalize the second row to have a leading 1 in the second position:
Divide the second row by -30:
[tex]\[ \frac{1}{-30} \times \left[\begin{array}{ccc|c} 0 & -30 & 20 & -50 \end{array}\right] = \left[\begin{array}{ccc|c} 0 & 1 & -\frac{2}{3} & \frac{5}{3} \end{array}\right] \][/tex]
So, our updated system is:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -2 & \frac{7}{2} \\ 0 & 1 & -\frac{2}{3} & \frac{5}{3} \\ 0 & 0 & 44 & -44 \end{array}\right] \][/tex]
4. Transform the first row to ensure the second column is zero except for the leading one:
Subtract \(-\frac{1}{2}\) times the second row from the first row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -2 & \frac{7}{2} \end{array}\right] + \frac{1}{2} \times \left[\begin{array}{ccc|c} 0 & 1 & -\frac{2}{3} & \frac{5}{3} \end{array}\right] \][/tex]
Simplifying, we get:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & -\frac{10}{3} & \frac{31}{6} \end{array}\right] \][/tex]
Updated system:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & -\frac{10}{3} & \frac{31}{6} \\ 0 & 1 & -\frac{2}{3} & \frac{5}{3} \\ 0 & 0 & 44 & -44 \end{array}\right] \][/tex]
5. Normalize the third row to have a leading 1 in the third position by dividing the third row by 44:
[tex]\[ \frac{1}{44} \times \left[\begin{array}{ccc|c} 0 & 0 & 44 & -44 \end{array}\right] = \left[\begin{array}{ccc|c} 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
Updated system:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & -\frac{10}{3} & \frac{31}{6} \\ 0 & 1 & -\frac{2}{3} & \frac{5}{3} \\ 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
6. Transform the first and second rows to make the third column zero except for the leading one:
- First row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & -\frac{10}{3} & \frac{31}{6} \end{array}\right] + \frac{10}{3} \times \left[\begin{array}{ccc|c} 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
Simplifying, we get:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \end{array}\right] \][/tex]
- Second row:
[tex]\[ \left[\begin{array}{ccc|c} 0 & 1 & -\frac{2}{3} & \frac{5}{3} \end{array}\right] + \frac{2}{3} \times \left[\begin{array}{ccc|c} 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
Simplifying, we get:
[tex]\[ \left[\begin{array}{ccc|c} 0 & 1 & 0 & 1 \end{array}\right] \][/tex]
Finally, the transformed matrix in reduced row echelon form is:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
To complete the solution, we fill in the \(\square\) in the RREF form asked in the question:
[tex]\[ \left[\begin{array}{lll|l} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
So the values in the [tex]\(\square\)[/tex] are 2, 1, and -1 respectively.
1. The leading entry in each nonzero row is 1.
2. Each leading 1 is the only nonzero entry in its column.
3. The leading 1 in a given row is to the right of the leading 1 in the row above it.
4. Rows with all zero elements, if any, are below rows with nonzero elements.
Given the matrix:
[tex]\[ \left[\begin{array}{ccc|c} 2 & -1 & -4 & 7 \\ 0 & -30 & 20 & -50 \\ 0 & 0 & 44 & -44 \end{array}\right] \][/tex]
Let's denote this matrix as \( A \).
### Step-by-Step Transformation
1. Normalize the first row to have a leading 1:
To achieve this, divide the entire first row by 2:
[tex]\[ \frac{1}{2} \times \left[\begin{array}{ccc|c} 2 & -1 & -4 & 7 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -2 & \frac{7}{2} \end{array}\right] \][/tex]
So, our system becomes:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -2 & \frac{7}{2} \\ 0 & -30 & 20 & -50 \\ 0 & 0 & 44 & -44 \end{array}\right] \][/tex]
2. Transform the second row to make the first column zero (it is already zero, so we skip this step).
3. Normalize the second row to have a leading 1 in the second position:
Divide the second row by -30:
[tex]\[ \frac{1}{-30} \times \left[\begin{array}{ccc|c} 0 & -30 & 20 & -50 \end{array}\right] = \left[\begin{array}{ccc|c} 0 & 1 & -\frac{2}{3} & \frac{5}{3} \end{array}\right] \][/tex]
So, our updated system is:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -2 & \frac{7}{2} \\ 0 & 1 & -\frac{2}{3} & \frac{5}{3} \\ 0 & 0 & 44 & -44 \end{array}\right] \][/tex]
4. Transform the first row to ensure the second column is zero except for the leading one:
Subtract \(-\frac{1}{2}\) times the second row from the first row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -2 & \frac{7}{2} \end{array}\right] + \frac{1}{2} \times \left[\begin{array}{ccc|c} 0 & 1 & -\frac{2}{3} & \frac{5}{3} \end{array}\right] \][/tex]
Simplifying, we get:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & -\frac{10}{3} & \frac{31}{6} \end{array}\right] \][/tex]
Updated system:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & -\frac{10}{3} & \frac{31}{6} \\ 0 & 1 & -\frac{2}{3} & \frac{5}{3} \\ 0 & 0 & 44 & -44 \end{array}\right] \][/tex]
5. Normalize the third row to have a leading 1 in the third position by dividing the third row by 44:
[tex]\[ \frac{1}{44} \times \left[\begin{array}{ccc|c} 0 & 0 & 44 & -44 \end{array}\right] = \left[\begin{array}{ccc|c} 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
Updated system:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & -\frac{10}{3} & \frac{31}{6} \\ 0 & 1 & -\frac{2}{3} & \frac{5}{3} \\ 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
6. Transform the first and second rows to make the third column zero except for the leading one:
- First row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & -\frac{10}{3} & \frac{31}{6} \end{array}\right] + \frac{10}{3} \times \left[\begin{array}{ccc|c} 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
Simplifying, we get:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \end{array}\right] \][/tex]
- Second row:
[tex]\[ \left[\begin{array}{ccc|c} 0 & 1 & -\frac{2}{3} & \frac{5}{3} \end{array}\right] + \frac{2}{3} \times \left[\begin{array}{ccc|c} 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
Simplifying, we get:
[tex]\[ \left[\begin{array}{ccc|c} 0 & 1 & 0 & 1 \end{array}\right] \][/tex]
Finally, the transformed matrix in reduced row echelon form is:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
To complete the solution, we fill in the \(\square\) in the RREF form asked in the question:
[tex]\[ \left[\begin{array}{lll|l} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \end{array}\right] \][/tex]
So the values in the [tex]\(\square\)[/tex] are 2, 1, and -1 respectively.
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