If `A=[(1, 2,3),(4, 5, 6)] and B=[(1, 4),(2, 5), (3, 6)]`, then the determinant value of BA is

Options

A. 8
B. 0
C. `-8`
D. 24

Answer

To find the determinant value of the product of matrices \( BA \), we will follow these steps:

### Step 1: Define the matrices A and B
Given:
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix} \]

### Step 2: Calculate the product \( BA \)
To compute \( BA \), we multiply matrix \( B \) with matrix \( A \):
\[ BA = B \cdot A = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \]

The resulting matrix will be a \( 3 \times 3 \) matrix. We calculate each element of the resulting matrix as follows:

- First row:
- \( (1 \cdot 1 + 4 \cdot 4) = 1 + 16 = 17 \)
- \( (1 \cdot 2 + 4 \cdot 5) = 2 + 20 = 22 \)
- \( (1 \cdot 3 + 4 \cdot 6) = 3 + 24 = 27 \)

- Second row:
- \( (2 \cdot 1 + 5 \cdot 4) = 2 + 20 = 22 \)
- \( (2 \cdot 2 + 5 \cdot 5) = 4 + 25 = 29 \)
- \( (2 \cdot 3 + 5 \cdot 6) = 6 + 30 = 36 \)

- Third row:
- \( (3 \cdot 1 + 6 \cdot 4) = 3 + 24 = 27 \)
- \( (3 \cdot 2 + 6 \cdot 5) = 6 + 30 = 36 \)
- \( (3 \cdot 3 + 6 \cdot 6) = 9 + 36 = 45 \)

Thus, the resulting matrix \( BA \) is:
\[ BA = \begin{pmatrix} 17 & 22 & 27 \\ 22 & 29 & 36 \\ 27 & 36 & 45 \end{pmatrix} \]

### Step 3: Calculate the determinant of \( BA \)
To find the determinant of the matrix \( BA \):
\[ \text{det}(BA) = \begin{vmatrix} 17 & 22 & 27 \\ 22 & 29 & 36 \\ 27 & 36 & 45 \end{vmatrix} \]

Using the property of determinants, we can perform column operations to simplify the calculation. We can modify the second and third columns:
- \( C_2 \rightarrow C_2 - C_1 \)
- \( C_3 \rightarrow C_3 - C_1 \)

This gives us:
\[ \begin{vmatrix} 17 & 22 - 17 & 27 - 17 \\ 22 & 29 - 22 & 36 - 22 \\ 27 & 36 - 27 & 45 - 27 \end{vmatrix} = \begin{vmatrix} 17 & 5 & 10 \\ 22 & 7 & 14 \\ 27 & 9 & 18 \end{vmatrix} \]

Next, we apply the operation \( C_3 \rightarrow C_3 - C_2 \):
\[ \begin{vmatrix} 17 & 5 & 10 - 5 \\ 22 & 7 & 14 - 7 \\ 27 & 9 & 18 - 9 \end{vmatrix} = \begin{vmatrix} 17 & 5 & 5 \\ 22 & 7 & 7 \\ 27 & 9 & 9 \end{vmatrix} \]

### Step 4: Identify identical columns
Now, we see that the second and third columns are identical:
\[ \begin{vmatrix} 17 & 5 & 5 \\ 22 & 7 & 7 \\ 27 & 9 & 9 \end{vmatrix} \]

Since two columns of the determinant are identical, the determinant is zero:
\[ \text{det}(BA) = 0 \]

### Final Answer
Thus, the value of the determinant of \( BA \) is:
\[ \boxed{0} \]