Option B:
[tex]BA=\left(\begin{array}{ccc}37 & 55 & -20 \\23 & 11 & -20 \\25 & 28 & -9\end{array}\right)[/tex]
Solution:
Given data:
[tex]A=\left[\begin{array}{ccc}1 & 7 & -3 \\5 & 2 & 1 \\-1 & 2 & -4\end{array}\right][/tex]
[tex]B=\left[\begin{array}{ccc}5 & 7 & 3 \\-3 & 7 & 9 \\2 & 5 & 2\end{array}\right][/tex]
To find BA:
Multiply the rows of the first matrix by the columns of the second matrix.
[tex]\left(\begin{array}{ccc}5 & 7 & 3\end{array}\right)\left(\begin{array}{c}1 \\5 \\-1\end{array}\right)=5 \cdot 1+7 \cdot 5+3(-1)[/tex]
[tex]\left(\begin{array}{lll}5 & 7 & 3\end{array}\right)\left(\begin{array}{l}7 \\2 \\2\end{array}\right)=5 \cdot 7+7 \cdot 2+3 \cdot 2[/tex]
[tex]\left(\begin{array}{lll}5 & 7 & 3\end{array}\right)\left(\begin{array}{c}-3 \\1 \\-4\end{array}\right)=5(-3)+7 \cdot 1+3(-4)[/tex]
[tex]\left(\begin{array}{ccc}-3 & 7 & 9\end{array}\right)\left(\begin{array}{c}1 \\5 \\-1\end{array}\right)=(-3) \cdot 1+7 \cdot 5+9(-1)[/tex]
[tex]\left(\begin{array}{lll}-3 & 7 & 9\end{array}\right)\left(\begin{array}{l}7 \\2 \\2\end{array}\right)=(-3) \cdot 7+7 \cdot 2+9 \cdot 2[/tex]
[tex]\left(\begin{array}{ccc}-3 & 7 & 9\end{array}\right)\left(\begin{array}{c}-3 \\1 \\-4\end{array}\right)=(-3)(-3)+7 \cdot 1+9(-4)[/tex]
[tex]\left(\begin{array}{lll}2 & 5 & 2\end{array}\right)\left(\begin{array}{c}1 \\5 \\-1\end{array}\right)=2 \cdot 1+5 \cdot 5+2(-1)[/tex]
[tex]\left(\begin{array}{lll}2 & 5 & 2\end{array}\right)\left(\begin{array}{l}7 \\2 \\2\end{array}\right)=2 \cdot 7+5 \cdot 2+2 \cdot 2[/tex]
[tex]\left(\begin{array}{lll}2 & 5 & 2\end{array}\right)\left(\begin{array}{c}-3 \\1 \\-4\end{array}\right)=2(-3)+5 \cdot 1+2(-4)[/tex]
[tex]BA=\left(\begin{array}{ccc}5 \cdot 1+7 \cdot 5+3(-1) & 5 \cdot 7+7 \cdot 2+3 \cdot 2 & 5(-3)+7 \cdot 1+3(-4) \\(-3) \cdot 1+7 \cdot 5+9(-1) & (-3) \cdot 7+7 \cdot 2+9 \cdot 2 & (-3)(-3)+7 \cdot 1+9(-4) \\2 \cdot 1+5 \cdot 5+2(-1) & 2 \cdot 7+5 \cdot 2+2 \cdot 2 & 2(-3)+5 \cdot 1+2(-4)\end{array}\right)[/tex]Simplify each element.
[tex]BA=\left[\begin{array}{ccc}5+35-3&35+14+6&-15+7-12\\-3+35-9&-21+14+18&9+7-36\\2+25-2&14+10+4&-6+5-8\end{array}\right][/tex]
[tex]BA=\left(\begin{array}{ccc}37 & 55 & -20 \\23 & 11 & -20 \\25 & 28 & -9\end{array}\right)[/tex]
Hence option C is the correct answer.
