Question

Discuss the consistency of the following system of equations.

X+2y+3z=0,3x+4y+4z=0 ,7x+10y+12z=0

Answer 1

Answer:

hope it helps to you friend

Answer 2

Given:

$x+2y+3x=0$

$3x+4y+4z=0$

$7x+10y+12z=0$

To find:

Consistency of the system of equations.

Solution:

For a system of equations,

$x+2y+3x=0$

$3x+4y+4z=0$

$7x+10y+12z=0$

let they be represented in matrix where $A$ is given by:

$A=\left[\begin{array}{ccc}1&2&3\\3&4&4\\7&10&12\end{array}\right]$

which are the coefficients of $x,y,z$

and $B$ is given by:

$B=\left[\begin{array}{c}x\\y\\z\end{array}\right]$

and $C$ is given by

$C=\left[\begin{array}{c}0\\0\\0\end{array}\right]$

Number of equations = 3

Number of unknowns = x,y,z = 3

number of equations = number of unknowns

Transforming $A$ into echelon form by Gaussian elimination where rows are reduced in a sequential way by performing operations on them to determine the rank of the matrix.

[A | C]

 $= \left[\begin{array}{cccc}1&2&3&0\\3&4&4&0\\7&10&12&0\end{array}\right]$

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

$R_{3}=R_{3}-7R_{1}$

$= \left[\begin{array}{cccc}1&2&3&0\\0&-2&-5&0\\0&-4&-9&0\end{array}\right]$

$R_{2}=R_{2}* (-1)$

$R_{3}=R_{3}*(-1)$

$= \left[\begin{array}{cccc}1&2&3&0\\0&2&5&0\\0&4&9&0\end{array}\right]$

$R_{3}=R_{3}- 2R_{2}$

$= \left[\begin{array}{cccc}1&2&3&0\\0&2&5&0\\0&0&-1&0\end{array}\right]$

$R_{3}=R_{3}*(-1)$

$= \left[\begin{array}{cccc}1&2&3&0\\0&2&5&0\\0&0&1&0\end{array}\right]$

Since, the last element of {A} is 1, it is now in echelon form and thus we need not to reduce further.

Rank of A = ρ(A) = 3

Rank of [A | C] = ρ(A | C) = 3

Since, ρ(A) = ρ(A | C) = 3 = no. of unknowns

Hence, the system has unique solution.

$x=0,y=0,z=0$ is the trivial solution to the system of equations.

The system of equations $x+2y+3x=0$, $3x+4y+4z=0$, $7x+10y+12z=0$ has unique solution.