To find the element in row 2, column 3 of the matrix product [tex]\( B \cdot A \)[/tex], let's go through the steps of calculating the matrix multiplication:
Given matrices:
[tex]\[
A = \begin{bmatrix}
1 & 2 & -4 \\
0 & 3 & -1
\end{bmatrix}
\][/tex]
[tex]\[
B = \begin{bmatrix}
-1 & 0 \\
2 & 4 \\
7 & 2
\end{bmatrix}
\][/tex]
To multiply [tex]\( B \)[/tex] (a 3x2 matrix) by [tex]\( A \)[/tex] (a 2x3 matrix), the resulting matrix [tex]\( C = B \cdot A \)[/tex] will be a 3x3 matrix. The element in the resulting matrix [tex]\( C \)[/tex] at position [tex]\( (i, j) \)[/tex] is obtained by taking the dot product of the [tex]\( i \)[/tex]-th row of [tex]\( B \)[/tex] with the [tex]\( j \)[/tex]-th column of [tex]\( A \)[/tex].
To find the element in row 2, column 3 of [tex]\( C \)[/tex], we need the dot product of the 2nd row of [tex]\( B \)[/tex] and the 3rd column of [tex]\( A \)[/tex].
Let's denote the 2nd row of [tex]\( B \)[/tex] as [tex]\( B_{2} = [2, 4] \)[/tex] and the 3rd column of [tex]\( A \)[/tex] as [tex]\( A_{3} = [-4, -1] \)[/tex]. The dot product is calculated as follows:
[tex]\[
(2 \times -4) + (4 \times -1)
\][/tex]
[tex]\[
= (-8) + (-4)
\][/tex]
[tex]\[
= -12
\][/tex]
Thus, the element in row 2, column 3 of the matrix product [tex]\( B \cdot A \)[/tex] is:
[tex]\[
-12
\][/tex]
