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.
Let $A$ be 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 $k$, the determinant of $kA$ is given by $\det(kA) = k^n \det(A)$.
Since $A$ is a $3 \times 3$ matrix, $n=3$.
$$ \det(2A) = 2^3 \det(A) = 8 \times 4 = 32 $$
Step 2: Determine the determinant of matrix $B$.
Matrix $B$ is obtained from $2A$ by performing the row operation $R_2 \to 2R_2 + 5R_3$.
The properties of determinants state:
1. If a matrix $M'$ is obtained from a matrix $M$ by multiplying a single row by a scalar $c$, then $\det(M') = c \det(M)$.
2. If a matrix $M'$ is obtained from a matrix $M$ by adding a multiple of one row to another row, then $\det(M') = \det(M)$.
The operation $R_2 \to 2R_2 + 5R_3$ on $2A$ can be viewed as two steps:
1. Multiplying the second row of $2A$ by $2$. This multiplies the determinant by $2$.
2. Adding $5$ times the third row to the modified second row. This operation does not change the determinant.
Therefore, the overall effect of the operation $R_2 \to 2R_2 + 5R_3$ is to multiply the determinant of $2A$ by a factor of $2$.
$$ \det(B) = 2 \times \det(2A) $$
Substituting the value of $\det(2A)$ from Step 1:
$$ \det(B) = 2 \times 32 = 64 $$
Correct Answer: 3