Let A be a 3 x 3 matrix with det(A) = 4. Let R<sub>i</sub> denote the i<sup>th</sup> row of A. If a matrix B is obtained by performing the operation R<sub>2</sub> -> 2R<sub>2</sub> + 5R<sub>3</sub> on 2A, then det(B) is equal to :
Step-by-Step Solution
Key Concept: Use properties of determinants: det(kA) = k^n * det(A) for an n x n matrix, and row operations. Specifically, if R2 -> 2R2 + 5R3, the determinant is multiplied by 2.
Given that $A$ is a $3 \times 3$ matrix with $\det(A) = 4$.
Step 1: Determine the determinant of the matrix $2A$.
For an $n \times n$ matrix $A$ and a scalar $k$, $\det(kA) = k^n \det(A)$.
In this case, $n=3$ and $k=2$.
$$ \det(2A) = 2^3 \det(A) = 8 \times 4 = 32 $$
Step 2: Analyze the effect of the row operation on the determinant.
A matrix $B$ is obtained from $2A$ by performing the operation $R_2 \to 2R_2 + 5R_3$.
The properties of determinants under elementary row operations are:
1. If a row is multiplied by a scalar $c$, the determinant is multiplied by $c$.
2. If a multiple of one row is added to another row, the determinant remains unchanged.
The operation $R_2 \to 2R_2 + 5R_3$ can be viewed as two sequential operations:
* First, $R_2 \to 2R_2$. This multiplies the determinant by a factor of 2.
* Second, $R_2 \to R_2 + 5R_3$ (where $R_2$ now represents the modified row). This operation does not change the determinant.
Therefore, the net effect of the operation $R_2 \to 2R_2 + 5R_3$ is to multiply the determinant by 2.
Step 3: Calculate $\det(B)$.
Since the operation $R_2 \to 2R_2 + 5R_3$ is performed on $2A$, the determinant of $B$ is 2 times the determinant of $2A$.
$$ \det(B) = 2 \times \det(2A) = 2 \times 32 = 64 $$
Correct Answer: 3