Question

What is the solution to the following system? mc016-1.jpg x = –53, y = –2, z = 7 x = –41, y = –2, z = –7 x = –11, y = –2, z = –7 x = 1, y = –2, z = 7

Answer 1

Solution:

$\left[\begin{array}{ccc}-53&-2&7\\-41&-2&-7\\-11&-2&-7\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]= \left[\begin{array}{ccc}1\\-2\\7\end{array}\right]$

Apply few Row elementary operations

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

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

$\left[\begin{array}{ccc}-12&0&14\\-30&0&0\\-11&-2&-7\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}1\\-2\\7\end{array}\right]$

So,

-30x = -2

x =1/15

And

-12x+14z =1

14z = 1+12/15

z= 27/15*14

z=9/70

and

-11x-2y-7z =7

-11/15 -2y -9/10 =7

-2y = 7+11/15+9/10

-2y =(105+110+135)/150

-2y= 350/150

y = -35/15*2

y=-7/6

Hope it helps you.

Answer 2

Answer:

             $\left[ \begin{matrix} -53 & -2 & 7 \\ -41 & -2 & -7 \\ -11 & -2 & -7 \end{matrix} \right] \left[ \begin{matrix} X \\ Y \\ Z \end{matrix} \right] = \left[ \begin{matrix} 1 \\ -2 \\ 7 \end{matrix} \right]$

Apply Row Elementary Operation

${ R }_{ 1 }\Longrightarrow { R }_{ 1 }-{ R }_{ 2 }$

${ R }_{ 2 }\Longrightarrow { R }_{ 2 }-{ R }_{ 3 }$

$\left[ \begin{matrix} -12 & 0 & 14 \\ -30 & 0 & 0 \\ -11 & -2 & -7 \end{matrix} \right] \left[ \begin{matrix} X \\ Y \\ Z \end{matrix} \right] = \left[ \begin{matrix} 1 \\ -2 \\ 7 \end{matrix} \right]$

 So, from the second row the equation is

 -30 X = -2

X = 1/15

And  

-12 X + 14 Z = 1

14 Z = 1 + 12(1/15) = (15 + 12)/15

Z = 27/ (15*14) = 9/70

-11 X -2 Y – 7 Z = 7

-11/15 – 2 Y – 9/10 = 7

- 2 Y = 11/15 + 9/10 + 7

Y = -35/ (15*2)

Y = - 7/6