Answer :
Certainly! Let's solve the given system of equations based on the provided row-echelon form of the augmented matrix. The augmented matrix already shows the solutions for each variable in a simplified form. Here is the augmented matrix:
[tex]\[ \left( \begin{array}{cccc:c} 1 & 0 & 0 & 0 & \frac{9}{4} \\ 0 & 1 & 0 & 0 & \frac{3}{5} \\ 0 & 0 & 1 & 0 & \frac{2}{3} \\ 0 & 0 & 0 & 1 & -\frac{9}{5} \end{array} \right) \][/tex]
To find the solutions, we can read the values directly from the augmented matrix. Each row represents an equation with the variables [tex]\( x_1, x_2, x_3, \)[/tex] and [tex]\( x_4 \)[/tex]:
1. From the first row:
[tex]\[ 1 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 + 0 \cdot x_4 = \frac{9}{4} \][/tex]
Therefore,
[tex]\( x_1 = \frac{9}{4} \)[/tex]
2. From the second row:
[tex]\[ 0 \cdot x_1 + 1 \cdot x_2 + 0 \cdot x_3 + 0 \cdot x_4 = \frac{3}{5} \][/tex]
Therefore,
[tex]\( x_2 = \frac{3}{5} \)[/tex]
3. From the third row:
[tex]\[ 0 \cdot x_1 + 0 \cdot x_2 + 1 \cdot x_3 + 0 \cdot x_4 = \frac{2}{3} \][/tex]
Therefore,
[tex]\( x_3 = \frac{2}{3} \)[/tex]
4. From the fourth row:
[tex]\[ 0 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 + 1 \cdot x_4 = -\frac{9}{5} \][/tex]
Therefore,
[tex]\( x_4 = -\frac{9}{5} \)[/tex]
Summarizing, the solutions to the system of equations are:
[tex]\[ x_1 = \frac{9}{4}, \quad x_2 = \frac{3}{5}, \quad x_3 = \frac{2}{3}, \quad x_4 = -\frac{9}{5} \][/tex]
Converting these fractions to decimal form, we get:
[tex]\[ x_1 = 2.25, \quad x_2 = 0.6, \quad x_3 = 0.6666666666666666, \quad x_4 = -1.8 \][/tex]
Thus, the final results are:
[tex]\[ (x_1, x_2, x_3, x_4) = (2.25, 0.6, 0.6666666666666666, -1.8) \][/tex]
These values represent the solutions to the given system of equations based on the provided augmented matrix in row-echelon form.
[tex]\[ \left( \begin{array}{cccc:c} 1 & 0 & 0 & 0 & \frac{9}{4} \\ 0 & 1 & 0 & 0 & \frac{3}{5} \\ 0 & 0 & 1 & 0 & \frac{2}{3} \\ 0 & 0 & 0 & 1 & -\frac{9}{5} \end{array} \right) \][/tex]
To find the solutions, we can read the values directly from the augmented matrix. Each row represents an equation with the variables [tex]\( x_1, x_2, x_3, \)[/tex] and [tex]\( x_4 \)[/tex]:
1. From the first row:
[tex]\[ 1 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 + 0 \cdot x_4 = \frac{9}{4} \][/tex]
Therefore,
[tex]\( x_1 = \frac{9}{4} \)[/tex]
2. From the second row:
[tex]\[ 0 \cdot x_1 + 1 \cdot x_2 + 0 \cdot x_3 + 0 \cdot x_4 = \frac{3}{5} \][/tex]
Therefore,
[tex]\( x_2 = \frac{3}{5} \)[/tex]
3. From the third row:
[tex]\[ 0 \cdot x_1 + 0 \cdot x_2 + 1 \cdot x_3 + 0 \cdot x_4 = \frac{2}{3} \][/tex]
Therefore,
[tex]\( x_3 = \frac{2}{3} \)[/tex]
4. From the fourth row:
[tex]\[ 0 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 + 1 \cdot x_4 = -\frac{9}{5} \][/tex]
Therefore,
[tex]\( x_4 = -\frac{9}{5} \)[/tex]
Summarizing, the solutions to the system of equations are:
[tex]\[ x_1 = \frac{9}{4}, \quad x_2 = \frac{3}{5}, \quad x_3 = \frac{2}{3}, \quad x_4 = -\frac{9}{5} \][/tex]
Converting these fractions to decimal form, we get:
[tex]\[ x_1 = 2.25, \quad x_2 = 0.6, \quad x_3 = 0.6666666666666666, \quad x_4 = -1.8 \][/tex]
Thus, the final results are:
[tex]\[ (x_1, x_2, x_3, x_4) = (2.25, 0.6, 0.6666666666666666, -1.8) \][/tex]
These values represent the solutions to the given system of equations based on the provided augmented matrix in row-echelon form.
