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Sagot :
To determine whether the given linear system has solutions, we need to analyze the properties of the coefficient matrix [tex]\( A \)[/tex] and the augmented matrix [tex]\([A|b]\)[/tex]. Here is the process step-by-step.
Step 1: Write down the matrices.
The coefficient matrix [tex]\( A \)[/tex] and the vector [tex]\( b \)[/tex] for the given linear system are:
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 6 & 4 \end{pmatrix}, \quad b = \begin{pmatrix} 6 \\ 12 \end{pmatrix} \][/tex]
Step 2: Calculate the determinant of the coefficient matrix [tex]\( A \)[/tex].
The determinant of a [tex]\( 2 \times 2 \)[/tex] matrix
[tex]\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \][/tex]
is calculated as [tex]\( ad - bc \)[/tex].
For matrix [tex]\( A \)[/tex]:
[tex]\[ \det(A) = (3 \cdot 4) - (2 \cdot 6) = 12 - 12 = 0 \][/tex]
Since the determinant of [tex]\( A \)[/tex] is 0, [tex]\( A \)[/tex] is singular, and it does not have an inverse. This initially suggests that the system might not have a unique solution.
Step 3: Determine the rank of the coefficient matrix [tex]\( A \)[/tex].
The rank of a matrix is the maximum number of linearly independent rows or columns in the matrix. Here, the rows of [tex]\( A \)[/tex] are:
[tex]\(\begin{pmatrix} 3 & 2 \end{pmatrix}\)[/tex] and
[tex]\(\begin{pmatrix} 6 & 4 \end{pmatrix}\)[/tex]
We observe that the second row is just a scalar multiple (2 times) of the first row, indicating that the rows are not linearly independent. Therefore, the rank of [tex]\( A \)[/tex] is 1.
Step 4: Form and determine the rank of the augmented matrix [tex]\([A | b]\)[/tex].
The augmented matrix [tex]\([A | b]\)[/tex] is:
[tex]\[ [A | b] = \begin{pmatrix} 3 & 2 & 6 \\ 6 & 4 & 12 \end{pmatrix} \][/tex]
Observe that the third column is also a scalar multiple (2 times) of the first column, indicating that the third column does not add an additional linearly independent vector. We again find that the augmented matrix also has a rank of 1.
Step 5: Compare the ranks and determine the existence of solutions.
We now compare the ranks:
- Rank of [tex]\( A \)[/tex] ([tex]\( \text{rank}(A) \)[/tex]) = 1
- Rank of augmented matrix [tex]\([A | b]\)[/tex] ([tex]\( \text{rank}([A | b]) \)[/tex]) = 1
The rank of [tex]\( A \)[/tex] is equal to the rank of the augmented matrix [tex]\([A | b]\)[/tex], and this rank is less than the number of variables (2). When the rank of the coefficient matrix [tex]\( A \)[/tex] is equal to the rank of the augmented matrix [tex]\([A | b]\)[/tex], and less than the number of variables, the system has infinitely many solutions.
Conclusion:
The given linear system has solutions. Specifically, it has infinitely many solutions.
Step 1: Write down the matrices.
The coefficient matrix [tex]\( A \)[/tex] and the vector [tex]\( b \)[/tex] for the given linear system are:
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 6 & 4 \end{pmatrix}, \quad b = \begin{pmatrix} 6 \\ 12 \end{pmatrix} \][/tex]
Step 2: Calculate the determinant of the coefficient matrix [tex]\( A \)[/tex].
The determinant of a [tex]\( 2 \times 2 \)[/tex] matrix
[tex]\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \][/tex]
is calculated as [tex]\( ad - bc \)[/tex].
For matrix [tex]\( A \)[/tex]:
[tex]\[ \det(A) = (3 \cdot 4) - (2 \cdot 6) = 12 - 12 = 0 \][/tex]
Since the determinant of [tex]\( A \)[/tex] is 0, [tex]\( A \)[/tex] is singular, and it does not have an inverse. This initially suggests that the system might not have a unique solution.
Step 3: Determine the rank of the coefficient matrix [tex]\( A \)[/tex].
The rank of a matrix is the maximum number of linearly independent rows or columns in the matrix. Here, the rows of [tex]\( A \)[/tex] are:
[tex]\(\begin{pmatrix} 3 & 2 \end{pmatrix}\)[/tex] and
[tex]\(\begin{pmatrix} 6 & 4 \end{pmatrix}\)[/tex]
We observe that the second row is just a scalar multiple (2 times) of the first row, indicating that the rows are not linearly independent. Therefore, the rank of [tex]\( A \)[/tex] is 1.
Step 4: Form and determine the rank of the augmented matrix [tex]\([A | b]\)[/tex].
The augmented matrix [tex]\([A | b]\)[/tex] is:
[tex]\[ [A | b] = \begin{pmatrix} 3 & 2 & 6 \\ 6 & 4 & 12 \end{pmatrix} \][/tex]
Observe that the third column is also a scalar multiple (2 times) of the first column, indicating that the third column does not add an additional linearly independent vector. We again find that the augmented matrix also has a rank of 1.
Step 5: Compare the ranks and determine the existence of solutions.
We now compare the ranks:
- Rank of [tex]\( A \)[/tex] ([tex]\( \text{rank}(A) \)[/tex]) = 1
- Rank of augmented matrix [tex]\([A | b]\)[/tex] ([tex]\( \text{rank}([A | b]) \)[/tex]) = 1
The rank of [tex]\( A \)[/tex] is equal to the rank of the augmented matrix [tex]\([A | b]\)[/tex], and this rank is less than the number of variables (2). When the rank of the coefficient matrix [tex]\( A \)[/tex] is equal to the rank of the augmented matrix [tex]\([A | b]\)[/tex], and less than the number of variables, the system has infinitely many solutions.
Conclusion:
The given linear system has solutions. Specifically, it has infinitely many solutions.
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