Question
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{A}^{\text{T}})^{\text{T}}=\text{A}.$

Answer

We have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2
$\text{A}^{\text{T}}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}^{\text{T}}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}$
Now, $(\text{A}^{\text{T}})^{\text{T}}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}^{\text{T}}$
$=\text{A}$
Hence proved.

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