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Sagot :
To solve the system of equations
[tex]\[ \begin{array}{rr} 3 x_1 - x_2 & = -1 \\ -4 x_1 + 3 x_2 & = 8 \end{array} \][/tex]
using the augmented matrix method, we can follow these steps:
1. Write the augmented matrix:
The system of equations can be represented by an augmented matrix as follows:
[tex]\[ \left[\begin{array}{ccc} 3 & -1 & | & -1 \\ -4 & 3 & | & 8 \end{array}\right] \][/tex]
2. Perform row operations to convert the matrix into row-echelon form:
We aim to simplify the matrix to get leading 1s (if possible) through row operations.
The starting matrix is:
[tex]\[ \left[\begin{array}{ccc} 3 & -1 & | & -1 \\ -4 & 3 & | & 8 \end{array}\right] \][/tex]
Let's perform the following operations:
- Multiply the first row by [tex]\( \frac{1}{3} \)[/tex] to make the leading coefficient in the first row a 1:
[tex]\[ \left[\begin{array}{ccc} 1 & -\frac{1}{3} & | & -\frac{1}{3} \\ -4 & 3 & | & 8 \end{array}\right] \][/tex]
- Add 4 times the first row to the second row to eliminate the [tex]\( -4 \)[/tex] in the second row first column:
[tex]\[ -4R1 + R2 \to R2 \][/tex]
[tex]\[ \left[\begin{array}{ccc} 1 & -\frac{1}{3} & | & -\frac{1}{3} \\ 0 & \frac{5}{3} & | & \frac{20}{3} \end{array}\right] \][/tex]
3. Convert to reduced row-echelon form (RREF):
- Multiply the second row by [tex]\( \frac{3}{5} \)[/tex]:
[tex]\[ \left[\begin{array}{ccc} 1 & -\frac{1}{3} & | & -\frac{1}{3} \\ 0 & 1 & | & 4 \end{array}\right] \][/tex]
- Add [tex]\( \frac{1}{3} \)[/tex] times the second row to the first row to eliminate the [tex]\( -\frac{1}{3} \)[/tex] in the first row second column:
[tex]\[ \frac{1}{3}R2 + R1 \to R1 \][/tex]
[tex]\[ \left[\begin{array}{ccc} 1 & 0 & | & 1 \\ 0 & 1 & | & 4 \end{array}\right] \][/tex]
4. Interpret the final matrix:
The matrix now represents the following system of equations:
[tex]\[ \begin{array}{rr} x_1 & = 1 \\ x_2 & = 4 \end{array} \][/tex]
Thus, the solution to the system is [tex]\( x_1 = 1 \)[/tex] and [tex]\( x_2 = 4 \)[/tex].
So, the correct choice is:
[tex]$\square$[/tex] A. The unique solution is [tex]\( x_1 = 1 \)[/tex] and [tex]\( x_2 = 4 \)[/tex].
[tex]\[ \begin{array}{rr} 3 x_1 - x_2 & = -1 \\ -4 x_1 + 3 x_2 & = 8 \end{array} \][/tex]
using the augmented matrix method, we can follow these steps:
1. Write the augmented matrix:
The system of equations can be represented by an augmented matrix as follows:
[tex]\[ \left[\begin{array}{ccc} 3 & -1 & | & -1 \\ -4 & 3 & | & 8 \end{array}\right] \][/tex]
2. Perform row operations to convert the matrix into row-echelon form:
We aim to simplify the matrix to get leading 1s (if possible) through row operations.
The starting matrix is:
[tex]\[ \left[\begin{array}{ccc} 3 & -1 & | & -1 \\ -4 & 3 & | & 8 \end{array}\right] \][/tex]
Let's perform the following operations:
- Multiply the first row by [tex]\( \frac{1}{3} \)[/tex] to make the leading coefficient in the first row a 1:
[tex]\[ \left[\begin{array}{ccc} 1 & -\frac{1}{3} & | & -\frac{1}{3} \\ -4 & 3 & | & 8 \end{array}\right] \][/tex]
- Add 4 times the first row to the second row to eliminate the [tex]\( -4 \)[/tex] in the second row first column:
[tex]\[ -4R1 + R2 \to R2 \][/tex]
[tex]\[ \left[\begin{array}{ccc} 1 & -\frac{1}{3} & | & -\frac{1}{3} \\ 0 & \frac{5}{3} & | & \frac{20}{3} \end{array}\right] \][/tex]
3. Convert to reduced row-echelon form (RREF):
- Multiply the second row by [tex]\( \frac{3}{5} \)[/tex]:
[tex]\[ \left[\begin{array}{ccc} 1 & -\frac{1}{3} & | & -\frac{1}{3} \\ 0 & 1 & | & 4 \end{array}\right] \][/tex]
- Add [tex]\( \frac{1}{3} \)[/tex] times the second row to the first row to eliminate the [tex]\( -\frac{1}{3} \)[/tex] in the first row second column:
[tex]\[ \frac{1}{3}R2 + R1 \to R1 \][/tex]
[tex]\[ \left[\begin{array}{ccc} 1 & 0 & | & 1 \\ 0 & 1 & | & 4 \end{array}\right] \][/tex]
4. Interpret the final matrix:
The matrix now represents the following system of equations:
[tex]\[ \begin{array}{rr} x_1 & = 1 \\ x_2 & = 4 \end{array} \][/tex]
Thus, the solution to the system is [tex]\( x_1 = 1 \)[/tex] and [tex]\( x_2 = 4 \)[/tex].
So, the correct choice is:
[tex]$\square$[/tex] A. The unique solution is [tex]\( x_1 = 1 \)[/tex] and [tex]\( x_2 = 4 \)[/tex].
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