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5 votes
5 votes

Consider the system of equations 

 

x+y+z=3

x+2y+3z=4

x+4y+kz=6

 

The above system of equations will not have a unique solution for k equals to___

4 Answers

Best answer
3 votes
3 votes

on solving the above equations we will reach to a point where

$$X = \left[ \begin{array}{ccc|c} 1 & 0 & -1 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & (K-7) & 0 \\ \end{array} \right] $$

 

Now when k=7 the above system of equations will not have a unique solution.

 

Answer: K=7

edited by
3 votes
3 votes

For solving above system of equation , we simplify our matrix 

1 votes
1 votes
K=7 is the answer;

Use the Gauss elimination method and apply R2 ← R2 – R1, R3←R3  – R1, and then R3 ← R3 – 3R2.

You will get k-7 = 5 so when k=7 the system will become inconsistent and will have no solution.
1 votes
1 votes
The LS has unique solution iff the matrix $M=\begin{bmatrix}1&1 &1 \\ 1 & 2 & 3\\ 1 & 4 & k\end{bmatrix}$ is a full rank matrix, i.e., $det(M)=k-7\neq 0$

 

$ \implies$ does not have unique solution when $k=7$.
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