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  • Let A equals [1 2 - 1 3] and a = 4, b = -2. Show that: (AB)^T = B^T A^T

Let A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right] and a = 4, b = -2.

Show that: (AB)^T = B^T A^T

Answers (1)

To prove: (AB)^T = B^T A^T

By multiplying the matrices and taking the transpose, we get,
\therefore \mathrm{LHS}=\left(\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]\right)^{\mathrm{T}}$ \\$\Rightarrow \mathrm{LHS}=\left[\begin{array}{ll}1 \times 4+2 \times 1 & 1 \times 0+2 \times 5 \\ -1 \times 4+3 \times 1 & -1 \times 0+3 \times 5\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}6 & 10 \\ -1 & 15\end{array}\right]^{\mathrm{T}}$ \\$\therefore \mathrm{LHS}=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]
As \mathrm{RHS}=\mathrm{B}^{\top} \mathrm{A}^{\top}
By taking transpose of matrices and then multiplying, we get,
\mathrm{RHS}=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]^{\mathrm{T}}\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ll}4 & 1 \\ 0 & 5\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$ \\$\Rightarrow \mathrm{RHS}=\left[\begin{array}{ll}4 \times 1+1 \times(2) & 4 \times (-1)+1 \times 3 \\ 0 \times 1+5 \times(2) & 0 \times (-1)+5 \times 3\end{array}\right]=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]
We have, LHS = RHS = \left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]
Hence (A B)^{\top}=B^{\top} A^{\top}... proved

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