# If A=begin{bmatrix}1 & 2&0 1 & 1&01&4&0 end{bmatrix}, B=begin{bmatrix}1 & 2&3 1 & 1&-12&2&2 end{bmatrix} text{ and } C=begin{bmatrix}1 & 2&3 1 & 1&-11&1&1end{bmatrix}. Show that AB=AC text{ but } B neq C

If . Show that
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Step 1
Given data:
The first matrix is $A=\left[\begin{array}{ccc}1& 2& 0\\ 1& 1& 0\\ 1& 4& 0\end{array}\right]$
The second matrix is $B=\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& -1\\ 2& 2& 2\end{array}\right]$
The third matrix is $C=\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& -1\\ 1& 1& 1\end{array}\right]$
Step 2
The multiplication of AB matrices is,
$AB=\left[\begin{array}{ccc}1& 2& 0\\ 1& 1& 0\\ 1& 4& 0\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& -1\\ 2& 2& 2\end{array}\right]$
$=\left[\begin{array}{ccc}\left(1+2\right)& \left(2+2\right)& \left(3-2\right)\\ \left(1+1\right)& \left(2+1\right)& \left(3-1\right)\\ \left(1+4\right)& \left(2+4\right)& \left(3-4\right)\end{array}\right]$
$=\left[\begin{array}{ccc}3& 4& 1\\ 2& 3& 2\\ 5& 6& -1\end{array}\right]$
The multiplication of AC matrices is,
$AC=\left[\begin{array}{ccc}1& 2& 0\\ 1& 1& 0\\ 1& 4& 0\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& -1\\ 1& 1& 1\end{array}\right]$
$=\left[\begin{array}{ccc}\left(1+2\right)& \left(2+2\right)& \left(3-2\right)\\ \left(1+1\right)& \left(2+1\right)& \left(3-1\right)\\ \left(1+4\right)& \left(2+4\right)& \left(3-4\right)\end{array}\right]$
$=\left[\begin{array}{ccc}3& 4& 1\\ 2& 3& 2\\ 5& 6& -1\end{array}\right]$
The multiplication of AB and AC is equal.
Thus, AB=AC, but B not equal to C.