# verify that (AB)^T = B^TA^T If A=begin{bmatrix}2 & 1 6 & 3 -2&4 end{bmatrix} text{ and } B=begin{bmatrix}2 & 4 1 & 6 end{bmatrix}

verify that $\left(AB{\right)}^{T}={B}^{T}{A}^{T}$
If
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Step 1
Given matrices:
$A=\left[\begin{array}{cc}2& 1\\ 6& 3\\ -2& 4\end{array}\right]$
$B=\left[\begin{array}{cc}2& 4\\ 1& 6\end{array}\right]$
Now, Transpose of a matrix is given by turning rows into columns:
${A}^{T}={\left[\begin{array}{cc}2& 1\\ 6& 3\\ -2& 4\end{array}\right]}^{T}=\left[\begin{array}{ccc}2& 6& -2\\ 1& 3& 4\end{array}\right]$
${B}^{T}={\left[\begin{array}{cc}2& 4\\ 1& 6\end{array}\right]}^{T}=\left[\begin{array}{cc}2& 1\\ 4& 6\end{array}\right]$
Step 2
Product of given matrices:
$AB=\left[\begin{array}{cc}2& 1\\ 6& 3\\ -2& 4\end{array}\right]\left[\begin{array}{cc}2& 4\\ 1& 6\end{array}\right]$
$=\left[\begin{array}{cc}2\cdot 2+1\cdot 1& 2\cdot 4+1\cdot 6\\ 6\cdot 2+3\cdot 1& 6\cdot 4+3\cdot 6\\ \left(-2\right)\cdot 2+4\cdot 1& \left(-2\right)\cdot 4+4\cdot 6\end{array}\right]$
$=\left[\begin{array}{cc}5& 14\\ 15& 42\\ 0& 16\end{array}\right]$
And its transpose is:
$\left(AB{\right)}^{T}={\left[\begin{array}{cc}5& 14\\ 15& 42\\ 0& 16\end{array}\right]}^{T}=\left[\begin{array}{ccc}5& 15& 0\\ 14& 42& 16\end{array}\right]$
Step 3
Product of transpose of given matrices:
${B}^{T}{A}^{T}=\left[\begin{array}{cc}2& 1\\ 4& 6\end{array}\right]\left[\begin{array}{ccc}2& 6& -2\\ 1& 3& 4\end{array}\right]$
$=\left[\begin{array}{ccc}2\cdot 2+1\cdot 1& 2\cdot 6+1\cdot 3& 2\left(-2\right)+1\cdot 4\\ 4\cdot 2+6\cdot 1& 4\cdot 6+6\cdot 3& 4\left(-2\right)+6\cdot 4\end{array}\right]$
$=\left[\begin{array}{ccc}5& 15& 0\\ 14& 42& 16\end{array}\right]$
Therefore,
$\left(AB{\right)}^{T}={B}^{T}{A}^{T}=\left[\begin{array}{ccc}5& 15& 0\\ 14& 42& 16\end{array}\right]$
Hence, verified.