12 votes 12 votes $$a\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 4 \\ 5\end{array}\right]+b\left[\begin{array}{c}-1 \\ 2 \\ -3 \\ 4 \\ -5\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0 \\ 0 \\ 0\end{array}\right]$$How many number of pairs $(a, b)$ are there, that satisfy the above equation?$0$$1$Infinite$2$ Linear Algebra goclasses2025_csda_wq3 goclasses linear-algebra vector-space 1-mark + – GO Classes asked Mar 14, 2023 • edited Mar 13 by shadymademe GO Classes 853 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply DEBANJAN DAS2k commented Mar 16, 2023 reply Follow Share It can be observed that the given system of equation is reduced to $a=-b$ $a=b$ so only $a=0,b=0$ is possible. 0 votes 0 votes Srken commented Apr 29 reply Follow Share There are more chances to choose option A here because we can think that it just asked about number of solutions possible as the two vectors are linearly independent so trivial solution is always possible and we can choose number of pairs as 1. 0 votes 0 votes Please log in or register to add a comment.
18 votes 18 votes Linear combination of $\mathrm{LI}$ vectors can be $0$ vector, only if, all coefficients are $0 \mathrm{~s}$. Hence, only $1$ pair $(0,0)$ exists for $(a, b)$ Option B is correct. GO Classes answered Mar 14, 2023 • edited Mar 14, 2023 by Lakshman Bhaiya GO Classes comment Share Follow See all 0 reply Please log in or register to add a comment.
4 votes 4 votes The vector is of Span 5. The system of equations we get is: a – b = 0 2a + 2b = 0 => a + b = 0 3a – 3b = 0 => a – b = 0 4a + 4b = 0 => a + b = 0 5a – 5b = 0 => a – b = 0 Therefore there are 2 unique equations to construct a row picture: Caption From the Graph there is Only One point of Intersection between the lines => Only One Solution satisfying the equations i.e. ( 0 , 0 ) => a = 0, b = 0. P Jahnavi answered Mar 16, 2023 P Jahnavi comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes OPTION B SEEM TO BE CORRECT ONE Gutley>> answered Mar 17, 2023 Gutley>> comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes We can take( a,b ) many values like (1,1)(2,2)(3,3)etc The given condition is true for a==b anmol.gupta16392 answered Mar 16, 2023 anmol.gupta16392 comment Share Follow See 1 comment See all 1 1 comment reply KG commented Mar 16, 2023 reply Follow Share Suppose (a,b) is (1,1). 1*[1 2 3 4 5]$^{T}$ + 1*[-1 2 -3 4 -5]$^{T}$ = [0 2 0 4 0]$^{T}$ != [0 0 0 0 0]$^{T}$. So, (a,b) != (1,1). See [1 2 3 4 5]$^{T}$ & [-1 2 -3 4 -5]$^{T}$ are LI vectors. (you can see it because one is not the multiple of other) So, for what values of a & b, we can satisfy above equation? Only possibility is a=0 and b=0. 1 votes 1 votes Please log in or register to add a comment.