1. **State the problem:** We need to show that the determinant of the matrix $$\begin{vmatrix} 2 & 3 & -1 \\ 1 & 1 & 0 \\ 2 & -3 & 5 \end{vmatrix} = 0$$ without expanding it. 2. **Recall the property of determinants:** If one row (or column) of a matrix can be expressed as a linear combination of the other rows (or columns), then the determinant is zero. 3. **Check for linear dependence:** Observe the rows: - Row 1: $[2, 3, -1]$ - Row 2: $[1, 1, 0]$ - Row 3: $[2, -3, 5]$ Try to express Row 3 as a combination of Row 1 and Row 2: Let $a$ and $b$ be scalars such that $$a \times \text{Row 1} + b \times \text{Row 2} = \text{Row 3}$$ This gives the system: $$2a + 1b = 2$$ $$3a + 1b = -3$$ $$-1a + 0b = 5$$ From the third equation: $$-a = 5 \implies a = -5$$ Substitute $a = -5$ into the first two equations: $$2(-5) + b = 2 \implies -10 + b = 2 \implies b = 12$$ $$3(-5) + b = -3 \implies -15 + b = -3 \implies b = 12$$ Both equations give $b = 12$, consistent. 4. **Conclusion:** Since Row 3 = $-5 \times$ Row 1 + $12 \times$ Row 2, the rows are linearly dependent. Therefore, the determinant is zero without expansion. **Final answer:** $$\boxed{0}$$