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Sagot :
To find the rank of the matrix, we need to reduce it to its row echelon form (REF) or reduced row echelon form (RREF). The rank of the matrix is the number of non-zero rows in its row echelon form. We'll perform a series of row operations to achieve this. Here's the matrix we're starting with:
[tex]\[ \left[\begin{array}{ccc} 4 & 2 & -3 \\ 1 & 3 & -6 \\ -5 & 0 & -7 \end{array}\right] \][/tex]
Step 1: Make the element in the first row, first column a pivot (which is already 4) and make the elements below it zero by row operations.
- [tex]\(R2 \leftarrow R2 - \frac{1}{4}R1\)[/tex]
- [tex]\(R3 \leftarrow R3 + \frac{5}{4}R1\)[/tex]
[tex]\[ R2 \leftarrow R2 - \frac{1}{4}R1 = \left[\begin{array}{c} 1 \\ 3 \\ -6 \end{array}\right] - \left[\begin{array}{c} 1 \\ 0.5 \\ -0.75 \end{array}\right] = \left[\begin{array}{c} 0 \\ 2.5 \\ -5.25 \end{array}\right] = \left[\begin{array}{ccc} 0 & 2.5 & -5.25 \end{array}\right] \][/tex]
[tex]\[ R3 \leftarrow R3 + \frac{5}{4}R1 = \left[\begin{array}{c} -5 \\ 0 \\ -7 \end{array}\right] + \left[\begin{array}{c} 5 \\ 2.5 \\ -3.75 \end{array}\right] = \left[\begin{array}{c} 0 \\ 2.5 \\ -10.75 \end{array}\right] = \left[\begin{array}{ccc} 0 & 2.5 & -10.75 \end{array}\right] \][/tex]
Now the matrix looks like this:
[tex]\[ \left[\begin{array}{ccc} 4 & 2 & -3 \\ 0 & 2.5 & -5.25 \\ 0 & 2.5 & -10.75 \end{array}\right] \][/tex]
Step 2: Make the element in the second row, second column a pivot (which is 2.5) and make the elements below it zero by row operations.
- [tex]\(R3 \leftarrow R3 - R2\)[/tex]
[tex]\[ R3 \leftarrow R3 - R2 = \left[\begin{array}{c} 0 \\ 2.5 \\ -10.75 \end{array}\right] - \left[\begin{array}{c} 0 \\ 2.5 \\ -5.25 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ -5.5 \end{array}\right] \][/tex]
Now the matrix looks like this:
[tex]\[ \left[\begin{array}{ccc} 4 & 2 & -3 \\ 0 & 2.5 & -5.25 \\ 0 & 0 & -5.5 \end{array}\right] \][/tex]
By these operations, we clearly see that all three rows are non-zero and each row after row reduction shows new leading non-zero entries which means they are linearly independent.
Finally, the rank of the matrix is determined by the number of non-zero rows. Therefore, the rank of the given matrix is:
[tex]\[ \boxed{3} \][/tex]
[tex]\[ \left[\begin{array}{ccc} 4 & 2 & -3 \\ 1 & 3 & -6 \\ -5 & 0 & -7 \end{array}\right] \][/tex]
Step 1: Make the element in the first row, first column a pivot (which is already 4) and make the elements below it zero by row operations.
- [tex]\(R2 \leftarrow R2 - \frac{1}{4}R1\)[/tex]
- [tex]\(R3 \leftarrow R3 + \frac{5}{4}R1\)[/tex]
[tex]\[ R2 \leftarrow R2 - \frac{1}{4}R1 = \left[\begin{array}{c} 1 \\ 3 \\ -6 \end{array}\right] - \left[\begin{array}{c} 1 \\ 0.5 \\ -0.75 \end{array}\right] = \left[\begin{array}{c} 0 \\ 2.5 \\ -5.25 \end{array}\right] = \left[\begin{array}{ccc} 0 & 2.5 & -5.25 \end{array}\right] \][/tex]
[tex]\[ R3 \leftarrow R3 + \frac{5}{4}R1 = \left[\begin{array}{c} -5 \\ 0 \\ -7 \end{array}\right] + \left[\begin{array}{c} 5 \\ 2.5 \\ -3.75 \end{array}\right] = \left[\begin{array}{c} 0 \\ 2.5 \\ -10.75 \end{array}\right] = \left[\begin{array}{ccc} 0 & 2.5 & -10.75 \end{array}\right] \][/tex]
Now the matrix looks like this:
[tex]\[ \left[\begin{array}{ccc} 4 & 2 & -3 \\ 0 & 2.5 & -5.25 \\ 0 & 2.5 & -10.75 \end{array}\right] \][/tex]
Step 2: Make the element in the second row, second column a pivot (which is 2.5) and make the elements below it zero by row operations.
- [tex]\(R3 \leftarrow R3 - R2\)[/tex]
[tex]\[ R3 \leftarrow R3 - R2 = \left[\begin{array}{c} 0 \\ 2.5 \\ -10.75 \end{array}\right] - \left[\begin{array}{c} 0 \\ 2.5 \\ -5.25 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ -5.5 \end{array}\right] \][/tex]
Now the matrix looks like this:
[tex]\[ \left[\begin{array}{ccc} 4 & 2 & -3 \\ 0 & 2.5 & -5.25 \\ 0 & 0 & -5.5 \end{array}\right] \][/tex]
By these operations, we clearly see that all three rows are non-zero and each row after row reduction shows new leading non-zero entries which means they are linearly independent.
Finally, the rank of the matrix is determined by the number of non-zero rows. Therefore, the rank of the given matrix is:
[tex]\[ \boxed{3} \][/tex]
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