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# If $A=\begin{bmatrix}3 & -4\\1 & 1\\2 & 0\end{bmatrix}\;and\;B=\begin{bmatrix}2 & 1 & 2\\1 & 2 & 4\end{bmatrix} \;$ then verify that $(BA)^2\neq B^2A^2$

Toolbox:
• 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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• The product of two matrices is defined if the number of column of A is equal to the number of rows of B
Step1:
Consider the LHS:-
$BA=\begin{bmatrix}2 & 1 & 2\\1 & 2 & 4\end{bmatrix}\begin{bmatrix}3 & -4\\1 & 1\\2 & 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}2(3)+1(1)+2(2) & 2(-4)+1(1)+2(0) \\1(3)+2(1)+4(2) & 1(-4)+2(1)+4(0) \end{bmatrix}$
$\;\;\;=\begin{bmatrix}6+1+4 & -8+1+0 \\3+2+8 & -4+2+0 \end{bmatrix}$
$\;\;\;=\begin{bmatrix}11 & -7 \\13 & -2 \end{bmatrix}$
$(BA)^2=BA.BA$
$\;\;\;=\begin{bmatrix}11 & -7 \\13 & -2 \end{bmatrix}\begin{bmatrix}11 & -7 \\13 & -2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix}11(11)+(-7)(13) &11( -7 )+(-7)(-2)\\13(11)+(-2)(13) & 13(-7)+(-2)(-2) \end{bmatrix}$
$\;\;\;=\begin{bmatrix}121-91 & -77+14 \\143 -26& -91+4 \end{bmatrix}$
$\;\;\;=\begin{bmatrix}30 & -63 \\117 & -87 \end{bmatrix}$
Step2:
RHS:-
$B^2A^2$
$\Rightarrow (B.B)(A.A).$
$B.B=\begin{bmatrix}2 & 1& 2 \\1 & 2 & 4\end{bmatrix}\begin{bmatrix}2 & 1& 2 \\1 & 2 & 4\end{bmatrix}$
A.A=$\begin{bmatrix}3 & -4\\1 & 1\\2 & 0\end{bmatrix}\begin{bmatrix}3 & -4\\1 & 1\\2 & 0\end{bmatrix}$
We cannot obtain $B^2A^2$ as the property of matrix multiplication says the no of rows of a matrix should be equal to the number of column of the $2^{nd}$ matrix.
edited Mar 21, 2013