Key Concepts and Formulas

• Scalar Multiplication of a Matrix: For an $n \times n$ matrix $A$ and a scalar $k$, the determinant of the matrix $kA$ is given by: $\det(kA) = k^n \det(A)$ This property arises because multiplying a matrix by $k$ means every element is multiplied by $k$. When calculating the determinant, we can factor out $k$ from each of the $n$ rows (or $n$ columns), resulting in $k^n$.

• Effect of Elementary Row Operations on Determinants:

  • If a matrix $B$ is obtained from $A$ by multiplying a single row by a scalar $k$ (i.e., $R_i \to kR_i$), then $\det(B) = k \det(A)$.
  • If a matrix $B$ is obtained from $A$ by adding a scalar multiple of one row to another row (i.e., $R_i \to R_i + kR_j$), then $\det(B) = \det(A)$.
  • If a matrix has two identical rows, its determinant is 0.

• Linearity of Determinant with respect to a Row: The determinant is linear with respect to each row. For example, if $R_1, R_2, R_3$ are rows of a matrix, then: $\det\begin{pmatrix} R_1 \\ cR_2 + dR_3 \\ R_3 \end{pmatrix} = c \det\begin{pmatrix} R_1 \\ R_2 \\ R_3 \end{pmatrix} + d \det\begin{pmatrix} R_1 \\ R_3 \\ R_3 \end{pmatrix}$

Step-by-Step Solution

Step 1: Calculate the determinant of the matrix $2A$.

• What we are doing: We first determine the determinant of the matrix $2A$ using the property of scalar multiplication of a matrix.
• Why we are doing it: The problem specifies that the subsequent row operation is performed on $2A$, so we need its determinant as a starting point.
• Math: Given that $A$ is a $3 \times 3$ matrix, the dimension $n=3$. Given $\det(A) = 4$. Using the property $\det(kA) = k^n \det(A)$ with $k=2$: $\det(2A) = 2^3 \det(A)$ $\det(2A) = 8 \times 4$ $\det(2A) = 32$
• Reasoning: Since $A$ is a $3 \times 3$ matrix, multiplying the entire matrix by 2 is equivalent to multiplying each of its three rows by 2. Each such row multiplication factors out a 2 from the determinant, leading to $2^3$ as the overall scalar multiplier for the determinant.

Step 2: Apply the row operation $R_2 \to 2R_2 + 5R_3$ to the matrix $2A$ to obtain matrix $B$.

• What we are doing: We apply the given complex row operation to the matrix $2A$ and determine how this operation affects its determinant.
• Why we are doing it: This is the final step in constructing matrix $B$ from $2A$, and we need to find $\det(B)$.
• Math: Let $C = 2A$. The rows of $C$ are $R_1'$, $R_2'$, and $R_3'$. From Step 1, we know $\det(C) = \det\begin{pmatrix} R_1' \\ R_2' \\ R_3' \end{pmatrix} = 32$. Matrix $B$ is obtained from $C$ by the operation $R_2' \to 2R_2' + 5R_3'$. Therefore, the rows of $B$ are $R_1'$, $(2R_2' + 5R_3')$, and $R_3'$. We can write $\det(B)$ as: $\det(B) = \det\begin{pmatrix} R_1' \\ 2R_2' + 5R_3' \\ R_3' \end{pmatrix}$ Using the linearity property of determinants with respect to the second row: $\det(B) = 2 \det\begin{pmatrix} R_1' \\ R_2' \\ R_3' \end{pmatrix} + 5 \det\begin{pmatrix} R_1' \\ R_3' \\ R_3' \end{pmatrix}$ We know that if a matrix has two identical rows, its determinant is 0. In the second term, the second and third rows are both $R_3'$, so $\det\begin{pmatrix} R_1' \\ R_3' \\ R_3' \end{pmatrix} = 0$. Substituting this into the equation for $\det(B)$: $\det(B) = 2 \det\begin{pmatrix} R_1' \\ R_2' \\ R_3' \end{pmatrix} + 5 \times 0$ $\det(B) = 2 \det(C)$ Now, substitute the value of $\det(C)$ from Step 1: $\det(B) = 2 \times 32$ $\det(B) = 64$
• Reasoning: The row operation $R_2 \to 2R_2 + 5R_3$ is a combination. Due to the linearity of the determinant, we can split this into two terms. The first term, $2 \det(C)$, accounts for the scaling of $R_2'$. The second term, $5 \det(\text{matrix with } R_3' \text{ in both second and third rows})$, results in zero because a matrix with two identical rows has a determinant of zero. Thus, only the scaling factor of the affected row contributes to the change in the determinant.

Common Mistakes & Tips

• Incorrect Scalar Multiplication: A frequent error is to assume $\det(kA) = k \det(A)$. Remember to raise the scalar $k$ to the power of the matrix dimension $n$, i.e., $\det(kA) = k^n \det(A)$.
• Misinterpreting Complex Row Operations: An operation like $R_i \to kR_i + lR_j$ is not simply an "add a multiple of another row" operation. The $k$ multiplying $R_i$ itself will scale the determinant by $k$, while the $lR_j$ part contributes zero due to the property of identical rows.
• Order of Operations: Always perform matrix operations in the sequence specified. First, multiply the matrix by the scalar, then apply the row operation.

Summary

We began by calculating the determinant of $2A$. Since $A$ is a $3 \times 3$ matrix, $\det(2A)$ is $2^3$ times $\det(A)$, which yielded $32$. Next, we applied the row operation $R_2 \to 2R_2 + 5R_3$ to $2A$ (let's call it $C$) to obtain matrix $B$. Using the linearity property of determinants with respect to rows, we expressed $\det(B)$ as a sum of two determinants: $2 \det(C)$ and $5 \det(\text{a matrix with two identical rows})$. The second term evaluates to zero because a matrix with identical rows has a determinant of zero. Therefore, $\det(B)$ simplified to $2 \det(C)$, which is $2 \times 32 = 64$.

Final Answer

The final answer is \boxed{64}, which corresponds to option (A).