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Step-by-Step Solution
Step 1: Note the Given Information
We have a 3×3 matrix A with determinant 4, i.e.
$ \det(A) = 4 $.
We are given that matrix B is obtained from the matrix 2A by performing the row operation
$ R_2 \to 2R_2 + 5R_3 $.
We need to determine $ \det(B).$
Step 2: Determine $ \det(2A) $
The matrix 2A is formed by multiplying every entry of A by 2. In terms of determinants, if we multiply each row of a 3×3 matrix A by 2, we get:
$ \det(2A) = 2^3 \,\det(A) = 8 \,\det(A). $
Since $ \det(A) = 4, $
we have
$ \det(2A) = 8 \times 4 = 32. $
Step 3: Analyze the Row Operation $ R_2 \to 2R_2 + 5R_3 $
A row operation of the form $ R_2 \to R_2 + kR_3 $ does not change the determinant. However, $ R_2 \to 2R_2 $ multiplies the determinant by 2. So the operation $ R_2 \to 2R_2 + 5R_3 $ can be viewed as two successive steps:
$ R_2 \to 2R_2 \, $ (this multiplies the determinant by 2),
$ R_2 \to R_2 + 5R_3 \, $ (this step does not change the determinant).
Hence the net effect is that the determinant is multiplied by 2 from these row operations on 2A.
Step 4: Compute $ \det(B) $
Since we started from 2A (which had determinant 32) and the row operation effectively multiplies the result by 2, we get
$ \det(B) = 2 \times \det(2A) = 2 \times 32 = 64. $
Step 5: Final Answer
The determinant of B is
$ 64. $
