Respuesta :
Answer:
The solutions are [tex](x_1, x_2, x_3)=(1,1,0)[/tex].
Step-by-step explanation:
To find the solution of the the system of linear equations represented by this augmented matrix
[tex]\left[ \begin{array}{cccc} 2 & 1 & -1 & 3 \\\\ 1 & -1 & 1 & 0 \\\\ 0 & 1 & 2 & 1 \end{array} \right][/tex]
First, transform the augmented matrix to the reduced row echelon form. Any matrix can be transformed into its echelon forms, using a series of elementary row operations.
There are three kinds of elementary matrix operations.
- Interchange two rows (or columns).
- Multiply each element in a row (or column) by a non-zero number.
- Multiply a row (or column) by a non-zero number and add the result to another row (or column).
Applying the following elementary matrix operations:
Row Operation 1: Divide row 1 by 2 [tex]\left(R_1=\frac{R_1}{2}\right)[/tex]
Row Operation 2: Subtract row 1 from row 2 [tex]\left(R_2=R_2-R_1\right)[/tex]
Row Operation 3: Multiply row 2 by −2/3 [tex]\left(R_2=\left(- \frac{2}{3}\right)R_2\right)[/tex]
Row Operation 4: Subtract row 2 multiplied by 1/2 from row 1 [tex]\left(R_1=R_1-\left(\frac{1}{2}\right)R_2\right[/tex]
Row Operation 5: Subtract row 2 from row 3 [tex]\left(R_3=R_3-R_2\right)[/tex]
Row Operation 6: Divide row 3 by 3 [tex]\left(R_3=\frac{R_3}{3}\right)[/tex]
Row Operation 7: Add row 3 to row 2 [tex]\left(R_2=R_2+R_3\right)[/tex]
We get that the reduced row echelon form of the augmented matrix is:
[tex]\left[ \begin{array}{cccc} 1 & 0 & 0 & 1 \\\\ 0 & 1 & 0 & 1 \\\\ 0 & 0 & 1 & 0 \end{array} \right][/tex]
which corresponds to the system
[tex]\begin{array}{cccc} x_1 & & & =1 \\\\ & x_2 & & =1 \\\\ & & x_3 & =0 \end{array}[/tex]
Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution:
[tex](x_1, x_2, x_3)=(1,1,0)[/tex]
