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

Step 1: Recall the test for linear independence.
A set of vectors is linearly independent if the only way to write \(c_1v_1+c_2v_2+\dots=0\) is with every \(c_i=0\). For three vectors in \(\mathbb{R}^3\), we can check this by forming a 3x3 matrix out of the vectors and checking that its determinant is not zero. For a set of just two vectors, they are independent unless one is a scalar multiple of the other.

Step 2: Test the set S.
\(S = \{(1,0,0), (0,0,-5), (0,3,4)\}\). Write these as rows of a matrix and find the determinant.
\[ \begin{vmatrix} 1 & 0 & 0 \\ 0 & 0 & -5 \\ 0 & 3 & 4 \end{vmatrix} \] Expanding along the first row, only the first entry survives, since the other two are multiplied by 0:
\[ = 1 \times \begin{vmatrix} 0 & -5 \\ 3 & 4 \end{vmatrix} = 1 \times (0 \times 4 - (-5)\times 3) = 1 \times 15 = 15 \] The determinant is 15, which is not zero. So the three vectors in S are linearly independent.

Step 3: Test the set T.
\(T = \{(5,2,3), (5,-3,4)\}\) has only two vectors, so they can only be dependent if one is a scalar multiple of the other. Suppose \((5,-3,4) = k(5,2,3)\) for some scalar \(k\). Matching the first coordinates gives \(k=1\). But then the second coordinates would need \(2=-3\), which is false. So no such \(k\) exists, and the two vectors in T are linearly independent.

Step 4: Combine the results.
S is a linearly independent set of three vectors, since its determinant is nonzero, and T is a linearly independent set of two vectors, since neither is a scalar multiple of the other. So both S and T are linearly independent sets.

Step 5: Why the other options are wrong.
Option (B) claims T fails to be independent, but we showed the two vectors of T cannot be written as multiples of each other. Option (C) claims S fails, but its determinant is a clear nonzero 15. Option (D) claims both fail, which contradicts both checks above.

Final Answer:
Both S and T are sets of linearly independent vectors. \[ \boxed{\text{Both S and T are linearly independent}} \]