1. **State the problem:** We are given a matrix $D$ that is row equivalent to a matrix $A$ after a sequence of row operations: 46 row replacements, 1 row scale by $-\frac{1}{2}$, 1 row scale by $\frac{1}{2}$, and 27 row swaps. We need to find $\det(D)$ given $A$. 2. **Recall determinant properties for row operations:** - Row replacement ($r_j \to r_j + c r_i$) does not change the determinant. - Row scaling ($r_i \to k r_i$) multiplies the determinant by $k$. - Row swapping ($r_i \leftrightarrow r_j$) multiplies the determinant by $-1$. 3. **Calculate $\det(A)$:** Matrix $A$ is lower-triangular: $$ A = \begin{bmatrix} -2 & 0 & 0 & 0 & 0 \\ -4 & -3 & 0 & 0 & 0 \\ 5 & 10 & -1 & 0 & 0 \\ 1 & 6 & -7 & -1 & 0 \\ -4 & 8 & 3 & 1 & 8 \end{bmatrix} $$ The determinant of a triangular matrix is the product of its diagonal entries: $$ \det(A) = (-2) \times (-3) \times (-1) \times (-1) \times 8 = $$ Calculate stepwise: $$ (-2) \times (-3) = 6 $$ $$ 6 \times (-1) = -6 $$ $$ -6 \times (-1) = 6 $$ $$ 6 \times 8 = 48 $$ So, $\det(A) = 48$. 4. **Apply the effect of row operations to $\det(D)$:** - 46 row replacements: no change to determinant. - Row scale by $-\frac{1}{2}$: multiply determinant by $-\frac{1}{2}$. - Row scale by $\frac{1}{2}$: multiply determinant by $\frac{1}{2}$. - 27 row swaps: multiply determinant by $(-1)^{27} = -1$. 5. **Combine all effects:** $$ \det(D) = \det(A) \times (-\frac{1}{2}) \times \frac{1}{2} \times (-1) = 48 \times (-\frac{1}{2}) \times \frac{1}{2} \times (-1) $$ Simplify stepwise: $$ 48 \times (-\frac{1}{2}) = -24 $$ $$ -24 \times \frac{1}{2} = -12 $$ $$ -12 \times (-1) = 12 $$ 6. **Final answer:** $$ \boxed{12} $$