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Given the following set of equations:

X - y - z = 0

x + y + z = 46

x - 2y +z = 16

-x - y + 2z = -7

which of the following is true ?

a) no solution

b) infinite many solution

c) exactly one solution

d) exactly one solution and atleast one of the x,y,z is negative

2 Answers

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1 votes

$[A \vdots B]$ =$\begin{bmatrix} 1 & -1 & -1& \vdots &0 \\ 1& 1& 1 & \vdots& 46\\ 1& -2& 1&\vdots &16 \\ -1& -1& 2& \vdots&-7\end{bmatrix}$

R2->R2-R1 , R3->R3-R1 and R4->R1+R4

=$\begin{bmatrix} 1 & -1 & -1&\vdots & 0\\ 0& 2& 2 & \vdots&46 \\ 0& -1& 2&\vdots & 16\\ 0& -2& 1& \vdots & -7 \end{bmatrix}$  R2->R2/2

=$\begin{bmatrix} 1 & -1 & -1& \vdots & 0 \\ 0& 1& 1 & \vdots& 23\\ 0& -1& 2&\vdots &16 \\ 0& -2& 1& \vdots & -7 \end{bmatrix}$R3->R2+R3   and  R4->2R2+R4

=$\begin{bmatrix} 1 & -1 & -1& \vdots &0 \\ 0& 1& 1 & \vdots & 23\\ 0& 0& 3& \vdots & 39\\ 0& 0& 3& \vdots & 39 \end{bmatrix}$    R4->R4-R3

=$\begin{bmatrix} 1 & -1 & -1& \vdots &0 \\ 0& 1& 1 & \vdots & 23\\ 0& 0& 3& \vdots & 39\\ 0& 0& 0& \vdots & 0 \end{bmatrix}$

Here

Rank[A]=Rank$[A \vdots B]$ = n(Number of variables)

Hence,Unique solution is possible.

Hence,Option(C)Exactly one solution is the correct choice.

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