Question

7. The solution of the system of equations whose augmented matrix
1 1 1 2
1 2 3 1
3 1 -5 4
is
1. x=3, y = 1, z =1
2. x = 4, y = -3, z = 1
3. x = 1, y =1, z =1
4. x=7,y=-10,z=4​

Answer 1

Answer:

4 x=7,y=-10,z=4 I think it is correct

Answer 2

Answer:

2. x=4,y=-3,z=1

Step-by-step explanation:

AX=B

$\left[\begin{array}{ccc}1&1&1\\1&2&3\\3&1&-5\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2\\1\\4\end{array}\right]$

A:B=>AUGMENTED MATRIX

$\left[\begin{array}{ccc}1&1&1\\1&2&3\\3&1&-5\end{array}\right \left\begin{array}{ccc}2\\1\\4\end{array}\right]$

$R_{3}-> 3R_{1}-R_{3}$

$\left[\begin{array}{ccc}3&3&3\\1&2&3\\0&2&8\end{array}\right \left\begin{array}{ccc}6\\1\\2\end{array}\right]$

$R_{2}-> R_{1}-3R_{2}$

$\left[\begin{array}{ccc}3&3&3\\0&3&6\\0&2&8\end{array}\right \left\begin{array}{ccc}6\\-3\\2\end{array}\right]$

$R_{3}-> 2 R_{3}\\R_{2}-> 3R_{2}$

$\left[\begin{array}{ccc}3&3&3\\0&6&12\\0&6&24\end{array}\right \left\begin{array}{ccc}6\\-6\\6\end{array}\right]$

$R_{3}-> 2 R_{3}-3R_{2}$

$\left[\begin{array}{ccc}3&3&3\\0&6&12\\0&0&12\end{array}\right \left\begin{array}{ccc}6\\-6\\12\end{array}\right]$

$R_{1} ->R_{1}/3$

$\left[\begin{array}{ccc}1&1&1\\0&6&12\\0&0&12\end{array}\right \left\begin{array}{ccc}2\\-6\\12\end{array}\right]$

$R_{2} ->R_{2}/6$

$\left[\begin{array}{ccc}1&1&1\\0&1&2\\0&0&12\end{array}\right \left\begin{array}{ccc}2\\-1\\12\end{array}\right]$

$R_{3} ->R_{3}/12$

$\left[\begin{array}{ccc}1&1&1\\0&1&2\\0&0&1\end{array}\right \left\begin{array}{ccc}2\\-1\\1\end{array}\right]$

AX=B

$\left[\begin{array}{ccc}1&1&1\\0&1&2\\0&0&1\end{array}\right ] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

x+y+z=2==>1

y+2z=-1==>2

z=1==>3

Substitute equation 3 in equation 2

y+2(1)=-1

y+2=-1

y=-1-2

y=-3==>4

Substitute equation 3 and 4 in equation 1

x-3+1=2

x-2=2

x=2+2

x=4==>5

x=4,y=-3,z=1