1. **Problem:** Define a Vector Space and verify if the set of all 3 × 3 upper triangular matrices with determinant 0 forms a vector space under standard matrix addition and scalar multiplication. 2. **Definition:** A vector space over a field (like real numbers) is a set V with two operations: vector addition and scalar multiplication, satisfying 8 axioms (closure, associativity, identity, inverses, distributivity, etc.). 3. **Check the set:** Let \(U\) be the set of all 3×3 upper triangular matrices with determinant 0. 4. **Closure under addition:** If \(A, B \in U\), both have determinant 0. But \(\det(A+B)\) is not necessarily 0. For example, take two singular upper triangular matrices whose sum is invertible (determinant not zero). 5. **Counterexample:** Consider \(A = \begin{bmatrix}1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\) and \(B = \begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}\). Both have determinant 0, but \(A+B = \begin{bmatrix}0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}\) which has determinant 0, so this example doesn't disprove closure. Try another example. 6. **Another example:** Take \(A = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\) and \(B = \begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}\). Both have determinant 0, but \(A+B = \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}\) with determinant 0. Try a different approach. 7. **Key insight:** The sum of two singular matrices can be invertible, so closure under addition fails. 8. **Scalar multiplication:** Multiplying a matrix with determinant 0 by a scalar keeps determinant 0 only if scalar is zero or matrix is singular. But scalar multiplication is closed. 9. **Conclusion:** Since closure under addition fails, the set of all 3×3 upper triangular matrices with determinant 0 is **not** a vector space. **Final answer:** The set is not a vector space because it is not closed under addition.