Let's find [tex]\( A \times B \)[/tex] and [tex]\( B \times A \)[/tex] for the given matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
Given matrices:
[tex]\[ A = \begin{pmatrix} 1 & 7 \\ -3 & 2 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 6 & 0 \\ 1 & -2 \end{pmatrix} \][/tex]
### Part (a): Finding [tex]\( A \times B \)[/tex]
Let’s multiply matrix [tex]\( A \)[/tex] by matrix [tex]\( B \)[/tex]:
[tex]\[ A \times B = \begin{pmatrix} 1 & 7 \\ -3 & 2 \end{pmatrix} \times \begin{pmatrix} 6 & 0 \\ 1 & -2 \end{pmatrix} \][/tex]
To calculate this, we use the rule of matrix multiplication, where each element in the resulting matrix is calculated as the dot product of the corresponding row from the first matrix and the column from the second matrix.
[tex]\[
\begin{pmatrix}
(1 \times 6 + 7 \times 1) & (1 \times 0 + 7 \times -2) \\
(-3 \times 6 + 2 \times 1) & (-3 \times 0 + 2 \times -2)
\end{pmatrix}
= \begin{pmatrix}
(6 + 7) & (0 - 14) \\
(-18 + 2) & (0 - 4)
\end{pmatrix}
= \begin{pmatrix}
13 & -14 \\
-16 & -4
\end{pmatrix}
\][/tex]
So,
[tex]\[ A \times B = \begin{pmatrix} 13 & -14 \\ -16 & -4 \end{pmatrix} \][/tex]
### Part (b): Finding [tex]\( B \times A \)[/tex]
Now let’s multiply matrix [tex]\( B \)[/tex] by matrix [tex]\( A \)[/tex]:
[tex]\[ B \times A = \begin{pmatrix} 6 & 0 \\ 1 & -2 \end{pmatrix} \times \begin{pmatrix} 1 & 7 \\ -3 & 2 \end{pmatrix} \][/tex]
Following the same rule for matrix multiplication, we get:
[tex]\[
\begin{pmatrix}
(6 \times 1 + 0 \times -3) & (6 \times 7 + 0 \times 2) \\
(1 \times 1 + -2 \times -3) & (1 \times 7 + -2 \times 2)
\end{pmatrix}
= \begin{pmatrix}
(6 + 0) & (42 + 0) \\
(1 + 6) & (7 - 4)
\end{pmatrix}
= \begin{pmatrix}
6 & 42 \\
7 & 3
\end{pmatrix}
\][/tex]
So,
[tex]\[ B \times A = \begin{pmatrix} 6 & 42 \\ 7 & 3 \end{pmatrix} \][/tex]
### Final results:
(a) [tex]\( A \times B = \begin{pmatrix} 13 & -14 \\ -16 & -4 \end{pmatrix} \)[/tex]
(b) [tex]\( B \times A = \begin{pmatrix} 6 & 42 \\ 7 & 3 \end{pmatrix} \)[/tex]
