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If \(\rm A = \left[ \begin{array}{cc} 1&-5\\-3&7\end{array} \right]\) and \(\rm B = \left[ \begin{array}{cc} 8&4\\1&3\end{array} \right]\) then the value of (AB)T 
1. \(\rm \left[ \begin{array}{cc} 3&-11\\-17&9\end{array} \right]\)
2. \(\rm \left[ \begin{array}{cc} 3&-17\\-11&9\end{array} \right]\)
3. \(\rm \left[ \begin{array}{cc} -3&17\\11&-9\end{array} \right]\)
4. \(\rm \left[ \begin{array}{cc} -6&17\\22&-9\end{array} \right]\)

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Correct Answer - Option 2 : \(\rm \left[ \begin{array}{cc} 3&-17\\-11&9\end{array} \right]\)

Concept:

The transpose of a matrix is simply a flipped version of the original matrix. We can get transpose by switching its rows with its columns. it is denoted by AT.

 

Calculation:​

Given:  \(\rm A = \left[ \begin{array}{cc} 1&-5\\-3&7\end{array} \right]\) and \(\rm B = \left[ \begin{array}{cc} 8&4\\1&3\end{array} \right]\) 

AB = \(\rm \left[ \begin{array}{cc} 1&-5\\-3&7\end{array} \right]\)\(\rm \left[ \begin{array}{cc} 8&4\\1&3\end{array} \right]\)

⇒ \(\rm \left[ \begin{array}{cc} (1\times8)+(-5\times1)&(1\times4)+(-5\times3)\\(-3\times8)+(7\times1)&(-3\times4)+(7\times3)\end{array} \right]\)

 \(\rm \left[ \begin{array}{cc} 3&-11\\-17&9\end{array} \right]\)

Now, we can get transpose of AB by switching its rows with its columns.

∴ (AB)T = \(\rm \left[ \begin{array}{cc} 3&-17\\-11&9\end{array} \right]\)

Hence, option (2) is correct. 

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