# verify that (AB)C = A(BC) If A=begin{bmatrix}2 & 4 1 & 3 end{bmatrix}, B=begin{bmatrix}-2 & 1 0 & 4 end{bmatrix},C=begin{bmatrix}3 & 1 2 & 1 end{bmatrix}

verify that (AB)C = A(BC)
If $A=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right],B=\left[\begin{array}{cc}-2& 1\\ 0& 4\end{array}\right],C=\left[\begin{array}{cc}3& 1\\ 2& 1\end{array}\right]$
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Bentley Leach
Step 1
Given matrices:
$A=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right],B=\left[\begin{array}{cc}-2& 1\\ 0& 4\end{array}\right],C=\left[\begin{array}{cc}3& 1\\ 2& 1\end{array}\right]$ Step 2 Calculate (AB)C:
$AB=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]\left[\begin{array}{cc}-2& 1\\ 0& 4\end{array}\right]$
$=\left[\begin{array}{cc}2×\left(-2\right)+4×0& 2×1+4×4\\ 1×\left(-2\right)+3×0& 1×1+3×4\end{array}\right]$
$=\left[\begin{array}{cc}-4& 18\\ -2& 13\end{array}\right]$
$AB\left(C\right)=\left[\begin{array}{cc}-4& 18\\ -2& 13\end{array}\right]\left[\begin{array}{cc}3& 1\\ 2& 1\end{array}\right]$
$=\left[\begin{array}{cc}\left(-4\right)×3+18×2& \left(-4\right)×1+18×1\\ \left(-2\right)×3+13×2& \left(-2\right)×1+13×1\end{array}\right]$
$=\left[\begin{array}{cc}24& 14\\ 20& 11\end{array}\right]$
Step 3
Calculate A(BC)
$BC=\left[\begin{array}{cc}-2& 1\\ 0& 4\end{array}\right]\left[\begin{array}{cc}3& 1\\ 2& 1\end{array}\right]$
$=\left[\begin{array}{cc}-2×3+1×2& -2×1+1×1\\ 0×3+4×2& 0×1+4×1\end{array}\right]$
$=\left[\begin{array}{cc}-4& -1\\ 8& 4\end{array}\right]$
$A\left(BC\right)=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]\left[\begin{array}{cc}-4& -1\\ 8& 4\end{array}\right]$
$=\left[\begin{array}{cc}2×\left(-4\right)+4×8& 2×\left(-1\right)+4×4\\ 1×\left(-4\right)+3×8& 1×\left(-1\right)+3×4\end{array}\right]$
$=\left[\begin{array}{cc}24& 14\\ 20& 11\end{array}\right]$
Step 4
Thus,
$A\left(BC\right)=\left(AB\right)C=\left[\begin{array}{cc}24& 14\\ 20& 11\end{array}\right]$
Jeffrey Jordon
Jeffrey Jordon