Question

Suppose \(S\) and \(T\) are sets of vectors, where \(S = \{(1,0,0), (0, 0, -5), (0, 3, 4)\}\) and \(T = \{(5, 2, 3), (5, -3, 4)\}\), then:

• A. S and T both sets are linearly independent vectors
• B. S is a set of linearly independent vectors, but T is not
• C. T is a set of linearly independent vectors, but S is not
• D. Neither S nor T is a set of linearly independent vectors

Hint:

Check if the determinant formed by the three vectors of S is nonzero, and check whether either vector of T is a scalar multiple of the other.

Answer 1

The Correct Option is A

Here is a different way to reach the same result, by solving the defining equations directly instead of computing a determinant.

1. Set up the homogeneous system for S: for $S = \{(1,0,0), (0,0,-5), (0,3,4)\}$, write $a(1,0,0) + b(0,0,-5) + c(0,3,4) = (0,0,0)$. Adding the components gives the vector $(a, 3c, -5b+4c)$, so the equations are $a=0$, $3c=0$, and $-5b+4c=0$.
2. Solve this system: from $a=0$ and $3c=0$ we get $a=0, c=0$. Putting $c=0$ into the third equation gives $-5b=0$, so $b=0$. The only solution is $a=b=c=0$, the trivial one, which means S is linearly independent.
3. Check T using ratios: for $T = \{(5,2,3), (5,-3,4)\}$, two vectors are dependent only when every pair of matching coordinates is in the same ratio. Compare $5/5=1$, $2/(-3)\approx -0.67$, and $3/4=0.75$. Since these three ratios are not all equal, the vectors cannot be scalar multiples of each other.
4. Conclude for T: because no single scalar relates the two vectors, T is also linearly independent.

Both the system-of-equations check for S and the ratio check for T confirm that neither set contains a vector that depends on the others. So both S and T qualify as linearly independent sets.

Let's summarize:

• Solving $a(1,0,0)+b(0,0,-5)+c(0,3,4)=0$ forces $a=b=c=0$, so S is independent.
• The coordinate ratios of the two vectors in T are not all equal, so T is independent too.

So the correct choice is that both S and T are linearly independent, matching option (A).