#### Answer

$ \begin{bmatrix} \blacksquare&*&*\\ 0&\blacksquare&*\\ 0&0&\blacksquare\\ 0&0&0 \end{bmatrix} $

#### Work Step by Step

If the first two vectors in the set are linearly independent, and the third vector is not a linear combination of the first two, then by Theorem 7, the three vectors must be linearly independent. Hence, the only possibility is to have a pivot in each column, leaving the final row with all zeros.