Question

Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.

Answer 1

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 2

Answer:

Option (a) is correct.

The solution is (1, -1 , -4)

Step-by-step explanation:

Given:

A system of equation having 3 equations,

$2x+y+z=-3\\\\ 3x-5y+3z=-4\\\\ 5x-y+2z=-2$

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

$2x+y+z=-3\\\\ 3x-5y+3z=-4\\\\ 5x-y+2z=-2$

Write in matrix form as

$\begin{pmatrix}2&1&1\\ \:3&-5&3\\ \:5&-1&2\end{pmatrix}\begin{pmatrix}x\\ \:y\\ \:z\end{pmatrix}=\begin{pmatrix}-3\\ \:-4\\ \:-2\end{pmatrix}$

⇒  AX = b

Writing in Augmented matrix form , [A | b]

$\begin{pmatrix}2&1&1&-3\\ 3&-5&3&-4\\ 5&-1&2&-2\end{pmatrix}$

Apply row operations to make A an identity  matrix.

$R_1\:\leftrightarrow \:R_3$

$=\begin{pmatrix}5&-1&2&-2\\ 3&-5&3&-4\\ 2&1&1&-3\end{pmatrix}$

$R_2\:\leftarrow \:R_2-\frac{3}{5}\cdot \:R_1$

$=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 2&1&1&-3\end{pmatrix}$

$R_3\:\leftarrow \:R_3-\frac{2}{5}\cdot \:R_1$

$=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 0&\frac{7}{5}&\frac{1}{5}&-\frac{11}{5}\end{pmatrix}$

$R_3\:\leftarrow \:R_3+\frac{7}{22}\cdot \:R_2$

$=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 0&0&\frac{17}{22}&-\frac{34}{11}\end{pmatrix}$

$R_3\:\leftarrow \frac{22}{17}\cdot \:R_3$

$=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 0&0&1&-4\end{pmatrix}$

$R_2\:\leftarrow \:R_2-\frac{9}{5}\cdot \:R_3$

$=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&0&\frac{22}{5}\\ 0&0&1&-4\end{pmatrix}$

$R_1\:\leftarrow \:R_1-2\cdot \:R_3$

$=\begin{pmatrix}5&-1&0&6\\ 0&-\frac{22}{5}&0&\frac{22}{5}\\ 0&0&1&-4\end{pmatrix}$

$R_2\:\leftarrow \:-\frac{5}{22}\cdot \:R_2$

$=\begin{pmatrix}5&-1&0&6\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}$

$R_1\:\leftarrow \:R_1+1\cdot \:R_2$

$=\begin{pmatrix}5&0&0&5\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}$

$R_1\:\leftarrow \frac{1}{5}\cdot \:R_1$

$=\begin{pmatrix}1&0&0&1\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}$

Thus, We obtained an identity matrix

Thus, The solution is (1, -1 , -4)