Applied Test Series

Question

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___

Comments

for unique solution  k value  equal to 7

Answer 1

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

Answer 2

For solving above system of equation , we simplify our matrix 

Comments

If k equals to 7 then infinite solution exists, when k not equals to 7 then no solution exist.

Answer 3

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.

Comments

k-7 = 5

How come?

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Apply the row operations as mentioned and you will get the last row as [0 0 k-7 | 5].

Answer 4

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 $k=7$.