Question

Show that 
(i)\(\begin{bmatrix}5&-1\\6&7\end{bmatrix}\)\(\begin{bmatrix}2&1\\3&4\end{bmatrix}\)\(\neq \begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}\)

(ii)\(\begin{bmatrix}1&2&3\\0&1&0\\ 1&1&0\end{bmatrix}\)\(\begin{bmatrix}-1&1&0\\0&-1&1\\ 2&3&4\end{bmatrix}\)\(\neq \) \(\begin{bmatrix}-1&1&0\\0&-1&1\\ 2&3&4\end{bmatrix}\)\(\begin{bmatrix}1&2&3\\0&1&0\\ 1&1&0\end{bmatrix}\)

Answer 1

(i)\(\begin{bmatrix}5&-1\\6&7\end{bmatrix}\)\(\begin{bmatrix}2&1\\3&4\end{bmatrix}\)

=\(\begin{bmatrix}5(2)-1(3)&5(1)-1(4)\\6(2)+7(3)&6(1)+7(4)\end{bmatrix}\)

=\(\begin{bmatrix}10-3&5-4\\12+21&6+28\end{bmatrix}\)=\(\begin{bmatrix}7&1\\33&34\end{bmatrix}\)

\(\begin{bmatrix}2&1\\3&4\end{bmatrix}\)\(\begin{bmatrix}2&1\\3&4\end{bmatrix}\)

=\(\begin{bmatrix}2(5)+1(6)&2(-1)+1(7)\\3(5)+4(6)&3(-1)+4(7)\end{bmatrix}\)

=\(\begin{bmatrix}10+6&-2+7\\15+24&-3+28\end{bmatrix}\)

=[\(\begin{bmatrix}16&5\\39&25\end{bmatrix}\)

\(\therefore\) \(\begin{bmatrix}5&-1\\6&7\end{bmatrix}\)\(\begin{bmatrix}2&1\\3&4\end{bmatrix}\)≠\(\begin{bmatrix}2&1\\3&4\end{bmatrix}\)\(\begin{bmatrix}5&-1\\6&7\end{bmatrix}\)

---

(ii)\(\begin{bmatrix}1&2&3\\0&1&0\\ 1&1&0\end{bmatrix}\)\(\begin{bmatrix}-1&1&0\\0&-1&1\\ 2&3&4\end{bmatrix}\)

=\(\begin{bmatrix}1(-1)+2(0)+3(2)&1(1)+2(-1)+3(3)&1(0)+2(-1)+3(4)\\0(-1)+1(0)+0(2)&0(1)+1(-1)+0(3)&0(0)+1(1)+0(4)\\1(-1)+1(0)+0(2)&1(1)+1(-1)+0(3)&1(0)+1(1)+0(4)\end{bmatrix}\)

=\(\begin{bmatrix}5&8&14\\0&-1&1\\ -1&0&1\end{bmatrix}\)

\(\begin{bmatrix}-1&1&0\\0&-1&1\\ 2&3&4\end{bmatrix}\)\(\begin{bmatrix}1&2&3\\0&1&0\\ 1&1&0\end{bmatrix}\)

=\(\begin{bmatrix}-1(1)1(0)+0(1)&-1(2)+1(1)+0(1)&-1(3)+1(0)+0(0)\\0(1)+(-1(0)+1(1)&0(2)+(-1)(1)+1(1)&0(3)+(-1)(0)+1(0)\\2(1)+3(0)+4(1)&2(2)+3(1)+4(1)&2(3)+3(0)+4(0)\end{bmatrix}\)

=\(\begin{bmatrix}-1&-1&-3\\1&0&0\\ 6&11&6\end{bmatrix}\)

\(\therefore\)  \(\begin{bmatrix}1&2&3\\0&1&0\\ 1&1&0\end{bmatrix}\)\(\begin{bmatrix}-1&1&0\\0&-1&1\\ 2&3&4\end{bmatrix}\)≠ \(\begin{bmatrix}-1&1&0\\0&-1&1\\ 2&3&4\end{bmatrix}\)\(\begin{bmatrix}1&2&3\\0&1&0\\ 1&1&0\end{bmatrix}\)