Let $$A$$ be a $$3 \times 3$$ matrix with det$$(A) = 4$$. Let $$R_i$$ denote the $$i^{th}$$ row of $$A$$. If a matrix $$B$$ is obtained by performing the operation $$R_2 \to 2R_2 + 5R_3$$ on $$2A$$, then det$$(B)$$ is equal to:

We are given that $$\det(A) = 4$$ for a $$3 \times 3$$ matrix $$A$$.

First, we compute $$\det(2A)$$. For a $$3 \times 3$$ matrix, $$\det(kA) = k^3 \det(A)$$, so $$\det(2A) = 2^3 \cdot 4 = 32$$.

The matrix $$B$$ is obtained from $$2A$$ by the row operation $$R_2 \to 2R_2 + 5R_3$$. This operation can be decomposed as: first $$R_2 \to 2R_2$$ (which multiplies the determinant by 2), then $$R_2 \to R_2 + 5R_3$$ (adding a multiple of one row to another does not change the determinant).

Therefore, $$\det(B) = 2 \cdot \det(2A) = 2 \cdot 32 = 64$$.