Question

Suppose A and B are 3x3 matrices. Prove that (i) (A+B)^T = B^T +A^T
(ii) det (AB)^T= det(A).det(B)

Answer 1

(i) Let $A=\left(\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right), \ \ \ B=\left(\begin{array}{ccc} b_{11} & b_{12} & a_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array}\right).$

Then $(A+B)^T=\left(\begin{array}{ccc} a_{11}+b_{11} & a_{12}+b_{12} & a_{13}+b_{13} \\ a_{21}+b_{21} & a_{22}+b_{22} & a_{23}+b_{23} \\ a_{31}+b_{31} & a_{32}+b_{32} & a_{33}+b_{33} \end{array}\right)^T =$

$=\left(\begin{array}{ccc} a_{11}+b_{11} & a_{21}+b_{21} & a_{31}+b_{31} \\ a_{12}+b_{12} & a_{22}+b_{22} & a_{32}+b_{32} \\ a_{13}+b_{13} & a_{23}+b_{32} & a_{33}+b_{33} \end{array}\right)=$

$=\left(\begin{array}{ccc} b_{11}+a_{11} & b_{21}+a_{21} & b_{31}+a_{31} \\ a_{12}+b_{12} & b_{22}+ a_{22} & b_{32}+a_{32} \\ b_{13}+a_{13} & b_{32}+a_{23} & b_{33}+ a_{33} \end{array}\right)=$

$=\left(\begin{array}{ccc} b_{11} & b_{21} & b_{31} \\ b_{12} & b_{22} & b_{32} \\ b_{13} & b_{32} & b_{33} \end{array}\right)+\left(\begin{array}{ccc} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{array}\right)=B^T+A^T.$

(ii) Taking into account that $\det(M^T)=\det (M)$ and $\det(MN)=\det(M)\det(N)$ for any $3\times 3$ matrix $M$ and $N$, we conclude that $\det (AB)^T=\det(AB)= \det(A)\det(B).$