GO Classes CS/DA 2025 | Weekly Quiz 3 | Fundamental Course and Linear Algebra | Question: 8

Answer 1

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.

Answer 2

The vector is of Span 5. The system of equations we get is:

1. a – b = 0
2. 2a + 2b = 0 => a + b = 0
3. 3a – 3b = 0 => a – b = 0
4. 4a + 4b = 0 => a + b = 0
5. 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.

Answer 3

OPTION B SEEM TO BE CORRECT ONE 

Answer 4

We can take( a,b ) many values like (1,1)(2,2)(3,3)etc

The given condition is true for a==b

Comments

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.