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

This question has 2 parts and each part has been answered separately here.

Toolbox:
• Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B:
• $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
We need to prove that $LHS: \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \: \neq \: RHS: \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}$
$LHS: Multiplying \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$:
$\Rightarrow LHS: \begin{bmatrix}5\times 2+(-1)\times 3 & 5\times 1+(-1)\times 4\\6\times 2+7\times 3 & 6\times 1+7\times 4\end{bmatrix}$
$\Rightarrow LHS: \begin{bmatrix}10-3 & 5-4\\12+21 & 6+28\end{bmatrix}$
$\Rightarrow LHS: \begin{bmatrix}7 & 1\\33 & 34\end{bmatrix}$
$RHS: Multiplying$ $\begin{bmatrix}2 & 1\\ 3&4 \end{bmatrix}\begin{bmatrix}5 & -1\\ 6&7 \end{bmatrix}$:
$\Rightarrow RHS:$ $\begin{bmatrix}2\times 5+1\times 6 &2\times(- 1)+1\times 7\\ 3\times 5+4\times 6&3(-1)+4\times 7 \end{bmatrix}$
$\Rightarrow RHS:$ $\begin{bmatrix}10+6 & -2+7\\ 15+24&-3+28 \end{bmatrix}$
$\Rightarrow RHS: \begin{bmatrix}16 & 5\\ 39&25\end{bmatrix}$
We can see that $LHS \neq RHS$