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___ Linear Algebra test-series engineering-mathematics linear-algebra system-of-equations + – LRU asked Nov 5, 2021 LRU 981 views answer comment Share Follow See 1 comment See all 1 1 comment reply Chandrabhan Vishwa 1 commented Nov 5, 2021 reply Follow Share for unique solution k value equal to 7 1 votes 1 votes Please log in or register to add a comment.
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 M_eight answered Nov 13, 2021 edited Jan 5, 2022 by M_eight M_eight comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes For solving above system of equation , we simplify our matrix Isha_99 answered Dec 17, 2021 Isha_99 comment Share Follow See 1 comment See all 1 1 comment reply Sungoku commented Apr 4, 2022 reply Follow Share If k equals to 7 then infinite solution exists, when k not equals to 7 then no solution exist. 0 votes 0 votes Please log in or register to add a comment.
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. Vicky Panchal answered Nov 17, 2021 Vicky Panchal comment Share Follow See all 2 Comments See all 2 2 Comments reply LRU commented Nov 18, 2021 reply Follow Share k-7 = 5 How come? 0 votes 0 votes Vicky Panchal commented Nov 18, 2021 reply Follow Share Apply the row operations as mentioned and you will get the last row as [0 0 k-7 | 5]. 1 votes 1 votes Please log in or register to add a comment.
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 $k=7$. sandipanUMBC answered Nov 17, 2021 edited Nov 17, 2021 by sandipanUMBC sandipanUMBC comment Share Follow See all 0 reply Please log in or register to add a comment.