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 of the product of matrices \( BA \), where \( A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the product \( BA \)

We first need to multiply matrix \( B \) (which is \( 3 \times 2 \)) by matrix \( A \) (which is \( 2 \times 3 \)). The resulting matrix \( BA \) will be of size \( 3 \times 3 \).

The multiplication is done as follows:

\[ BA = B \cdot A = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \]

Calculating each element of \( BA \):

- 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 2: Calculate the determinant of \( BA \)

To find the determinant of the \( 3 \times 3 \) matrix \( BA \):

\[ \text{det}(BA) = \begin{vmatrix} 17 & 22 & 27 \\ 22 & 29 & 36 \\ 27 & 36 & 45 \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(BA) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where:
- \( a = 17, b = 22, c = 27 \)
- \( d = 22, e = 29, f = 36 \)
- \( g = 27, h = 36, i = 45 \)

Calculating the individual components:

1. \( ei - fh = 29 \cdot 45 - 36 \cdot 36 = 1305 - 1296 = 9 \)
2. \( di - fg = 22 \cdot 45 - 36 \cdot 27 = 990 - 972 = 18 \)
3. \( dh - eg = 22 \cdot 36 - 29 \cdot 27 = 792 - 783 = 9 \)

Now substituting back into the determinant formula:

\[ \text{det}(BA) = 17 \cdot 9 - 22 \cdot 18 + 27 \cdot 9 \]
\[ = 153 - 396 + 243 \]
\[ = 153 + 243 - 396 = 396 - 396 = 0 \]

### Conclusion

Thus, the determinant of \( BA \) is:

\[ \text{det}(BA) = 0 \]