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Exercise B · Q4

Q.If A=[1−3−24]A = \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} and B=[2−4−13]B = \begin{bmatrix} 2 & -4 \\ -1 & 3 \end{bmatrix} then show that

(i) (A+B)′=(A)′+(B)′(A+B)' = (A)'+(B)'
(ii) (AB)′=(B)′(A)′(AB)' = (B)'(A)'.
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Both transpose identities are verified by direct computation.

Transpose rules: (A+B)′=A′+B′(A+B)'=A'+B' and (AB)′=B′A′(AB)'=B'A', where X′X' denotes the transpose (rows ↔\leftrightarrow columns) of XX.

Part (i): (A+B)′=A′+B′(A+B)'=A'+B'

  1. With A=[1−3−24], B=[2−4−13]A=\begin{bmatrix}1&-3\\-2&4\end{bmatrix},\ B=\begin{bmatrix}2&-4\\-1&3\end{bmatrix}:

A+B=[3−7−37] ⇒ (A+B)′=[3−3−77]A+B=\begin{bmatrix}3&-7\\-3&7\end{bmatrix}\ \Rightarrow\ (A+B)'=\begin{bmatrix}3&-3\\-7&7\end{bmatrix}

  1. Transposes: A′=[1−2−34], B′=[2−1−43]A'=\begin{bmatrix}1&-2\\-3&4\end{bmatrix},\ B'=\begin{bmatrix}2&-1\\-4&3\end{bmatrix}, so

A′+B′=[3−3−77]A'+B'=\begin{bmatrix}3&-3\\-7&7\end{bmatrix}

  1. Both sides equal [3−3−77]\begin{bmatrix}3&-3\\-7&7\end{bmatrix}, verifying (A+B)′=A′+B′(A+B)'=A'+B'.

Part (ii): (AB)′=B′A′(AB)'=B'A'

  1. Compute ABAB:

AB=[1(2)+(−3)(−1)1(−4)+(−3)(3)(−2)(2)+4(−1)(−2)(−4)+4(3)]=[5−13−820]AB=\begin{bmatrix}1(2)+(-3)(-1)&1(-4)+(-3)(3)\\(-2)(2)+4(-1)&(-2)(-4)+4(3)\end{bmatrix}=\begin{bmatrix}5&-13\\-8&20\end{bmatrix}

so (AB)′=[5−8−1320](AB)'=\begin{bmatrix}5&-8\\-13&20\end{bmatrix}.

5. Compute B′A′B'A': …

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