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Exercise B · Q7

Q.Given A=[1−10234012]A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}, B=[22−4−42−42−15]B = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}, find:

(i) 2A−3B2A - 3B
(ii) ABAB
(iii) BABA
(iv) AB−BAAB - BA.
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Both ABAB and BABA equal 6I6I, so AB−BA=OAB-BA=O (here BB is 6A−16A^{-1}).

Linear combination (2A−3B)ij=2aij−3bij(2A-3B)_{ij}=2a_{ij}-3b_{ij}; product (AB)ij=∑kaikbkj(AB)_{ij}=\sum_k a_{ik}b_{kj}.

Given A=[1−10234012], B=[22−4−42−42−15]A=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix},\ B=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}.

(i) 2A−3B2A-3B

  1. 2A=[2−20468024], 3B=[66−12−126−126−315]2A=\begin{bmatrix}2&-2&0\\4&6&8\\0&2&4\end{bmatrix},\ 3B=\begin{bmatrix}6&6&-12\\-12&6&-12\\6&-3&15\end{bmatrix}.
  2. 2A−3B=[2−6−2−60+124+126−68+120−62+34−15]=[−4−81216020−65−11]2A-3B=\begin{bmatrix}2-6&-2-6&0+12\\4+12&6-6&8+12\\0-6&2+3&4-15\end{bmatrix}=\begin{bmatrix}-4&-8&12\\16&0&20\\-6&5&-11\end{bmatrix}.

(ii) ABAB

  1. Row 1: [1−10]\begin{bmatrix}1&-1&0\end{bmatrix} against columns of BB gives [2+4+0, 2−2+0, −4+4+0]=[6,0,0][2+4+0,\ 2-2+0,\ -4+4+0]=[6,0,0].
  2. Row 2: [234]\begin{bmatrix}2&3&4\end{bmatrix} gives [4−12+8, 4+6−4, −8−12+20]=[0,6,0][4-12+8,\ 4+6-4,\ -8-12+20]=[0,6,0].
  3. Row 3: [012]\begin{bmatrix}0&1&2\end{bmatrix} gives [0−4+4, 0+2−2, 0−4+10]=[0,0,6][0-4+4,\ 0+2-2,\ 0-4+10]=[0,0,6].

AB=[600060006]=6IAB=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=6I

(iii) BABA …

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