Respuesta :
This involves quite a lot of arithmetic to do manually.
The first thing you do is to make the first number in row 2 = to 0.
This is done by R2 = -3/2 R1 + R2
so the matrix becomes
( 2 1 1) ( -3 )
( 0 -13/2 3/2) (1/2 )
(5 -1 2) (-2)
Next step is to make the 5 in row 5 = 0
then the -1 must become zero
You aim for the form
( 1 0 0) (x)
(0 1 0) (y)
(0 0 1) ( z)
x , y and z will be the required solutions.
Answer:
Option (a) is correct.
The solution is (1, -1 , -4)
Step-by-step explanation:
Given:
A system of equation having 3 equations,
[tex]2x+y+z=-3\\\\ 3x-5y+3z=-4\\\\ 5x-y+2z=-2[/tex]
We have to solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.
Consider the given system
[tex]2x+y+z=-3\\\\ 3x-5y+3z=-4\\\\ 5x-y+2z=-2[/tex]
Write in matrix form as
[tex]\begin{pmatrix}2&1&1\\ \:3&-5&3\\ \:5&-1&2\end{pmatrix}\begin{pmatrix}x\\ \:y\\ \:z\end{pmatrix}=\begin{pmatrix}-3\\ \:-4\\ \:-2\end{pmatrix}[/tex]
⇒ AX = b
Writing in Augmented matrix form , [A | b]
[tex]\begin{pmatrix}2&1&1&-3\\ 3&-5&3&-4\\ 5&-1&2&-2\end{pmatrix}[/tex]
Apply row operations to make A an identity matrix.
[tex]R_1\:\leftrightarrow \:R_3[/tex]
[tex]=\begin{pmatrix}5&-1&2&-2\\ 3&-5&3&-4\\ 2&1&1&-3\end{pmatrix}[/tex]
[tex]R_2\:\leftarrow \:R_2-\frac{3}{5}\cdot \:R_1[/tex]
[tex]=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 2&1&1&-3\end{pmatrix}[/tex]
[tex]R_3\:\leftarrow \:R_3-\frac{2}{5}\cdot \:R_1[/tex]
[tex]=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 0&\frac{7}{5}&\frac{1}{5}&-\frac{11}{5}\end{pmatrix}[/tex]
[tex]R_3\:\leftarrow \:R_3+\frac{7}{22}\cdot \:R_2[/tex]
[tex]=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 0&0&\frac{17}{22}&-\frac{34}{11}\end{pmatrix}[/tex]
[tex]R_3\:\leftarrow \frac{22}{17}\cdot \:R_3[/tex]
[tex]=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 0&0&1&-4\end{pmatrix}[/tex]
[tex]R_2\:\leftarrow \:R_2-\frac{9}{5}\cdot \:R_3[/tex]
[tex]=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&0&\frac{22}{5}\\ 0&0&1&-4\end{pmatrix}[/tex]
[tex]R_1\:\leftarrow \:R_1-2\cdot \:R_3[/tex]
[tex]=\begin{pmatrix}5&-1&0&6\\ 0&-\frac{22}{5}&0&\frac{22}{5}\\ 0&0&1&-4\end{pmatrix}[/tex]
[tex]R_2\:\leftarrow \:-\frac{5}{22}\cdot \:R_2[/tex]
[tex]=\begin{pmatrix}5&-1&0&6\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}[/tex]
[tex]R_1\:\leftarrow \:R_1+1\cdot \:R_2[/tex]
[tex]=\begin{pmatrix}5&0&0&5\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}[/tex]
[tex]R_1\:\leftarrow \frac{1}{5}\cdot \:R_1[/tex]
[tex]=\begin{pmatrix}1&0&0&1\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}[/tex]
Thus, We obtained an identity matrix
Thus, The solution is (1, -1 , -4)
