Problems from Transposed matrix

Development:

Now we just have to swap rows with columns again:

$B=\left(\begin{array}{ccc}1& 0& 0\\ 3& 1& 5\\ 5& 0& 4\\ 6& 2& 1\end{array}\right)$

If we now calculate the transpose of this matrix we see that it is $B^t$.

Solution:

$B=\left(\begin{array}{ccc}1& 0& 0\\ 3& 1& 5\\ 5& 0& 4\\ 6& 2& 1\end{array}\right)$

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Development:

Let's swap rows with columns in order:

$A^t=\left(\begin{array}{ccc}1& 2& 0\\ 8& 7& 5\\ 1& 0& 3\end{array}\right)$

Solution:

$A^t=\left(\begin{array}{ccc}1& 2& 0\\ 8& 7& 5\\ 1& 0& 3\end{array}\right)$

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