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Miscellaneous Exercise 4(B) · Q187

Q.If A=[2−33−2−14]A=\begin{bmatrix}2 & -3\\3 & -2\\-1 & 4\end{bmatrix}, B=[−3412−1−3]B=\begin{bmatrix}-3 & 4 & 1\\2 & -1 & -3\end{bmatrix}, Verify (A+BT)T=AT+2B(A+B^T)^T=A^T+2B.

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A=[2−33−2−14]A=\begin{bmatrix}2 & -3\\3 & -2\\-1 & 4\end{bmatrix} (3x2), B=[−3412−1−3]B=\begin{bmatrix}-3 & 4 & 1\\2 & -1 & -3\end{bmatrix} (2x3), so BT=[−324−11−3]B^T=\begin{bmatrix}-3 & 2\\4 & -1\\1 & -3\end{bmatrix} (3x2).

A+BT=[−1−17−301]A+B^T=\begin{bmatrix}-1 & -1\\7 & -3\\0 & 1\end{bmatrix}, so (A+BT)T=[−170−1−31](A+B^T)^T=\begin{bmatrix}-1 & 7 & 0\\-1 & -3 & 1\end{bmatrix}.

AT=[23−1−3−24]A^T=\begin{bmatrix}2 & 3 & -1\\-3 & -2 & 4\end{bmatrix}. Compare to the printed RHS AT+2BA^T+2B: 2B=[−6824−2−6]2B=\begin{bmatrix}-6 & 8 & 2\\4 & -2 & -6\end{bmatrix}, so AT+2B=[−41111−4−2]A^T+2B=\begin{bmatrix}-4 & 11 & 1\\1 & -4 & -2\end{bmatrix} — this does NOT match (A+BT)T(A+B^T)^T. …

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