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Exercise 7.1 · Q10

Q.Give your own examples of matrices satisfying the following conditions in each case:

(i) AA and BB such that AB≠BAAB \ne BA.
(ii) AA and BB such that AB=O, BA=O, A≠OAB=O,\ BA=O,\ A \ne O and B≠OB \ne O.
(iii) AA and BB such that AB=OAB=O and BA≠OBA \ne O.
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Each part asks for a genuine (self-constructed) 2×22\times2 counterexample; we build one for each condition and verify it by direct multiplication.

Step 1. (i) Choose A,BA,B and test commutativity. Take A=(1101), B=(1011)A=\begin{pmatrix}1&1\\0&1\end{pmatrix},\ B=\begin{pmatrix}1&0\\1&1\end{pmatrix}.

AB=(1(1)+1(1)1(0)+1(1)0(1)+1(1)0(0)+1(1))=(2111)AB=\begin{pmatrix}1(1)+1(1)&1(0)+1(1)\\0(1)+1(1)&0(0)+1(1)\end{pmatrix}=\begin{pmatrix}2&1\\1&1\end{pmatrix}

BA=(1(1)+0(0)1(1)+0(1)1(1)+1(0)1(1)+1(1))=(1112)BA=\begin{pmatrix}1(1)+0(0)&1(1)+0(1)\\1(1)+1(0)&1(1)+1(1)\end{pmatrix}=\begin{pmatrix}1&1\\1&2\end{pmatrix}

Since AB=(2111)≠(1112)=BAAB=\begin{pmatrix}2&1\\1&1\end{pmatrix}\ne\begin{pmatrix}1&1\\1&2\end{pmatrix}=BA, this pair satisfies (i).

Step 2. (ii) Choose A,BA,B nonzero with both products zero. Take A=B=(0100)A=B=\begin{pmatrix}0&1\\0&0\end{pmatrix} (nonzero).

AB=(0100)(0100)=(0(0)+1(0)0(1)+1(0)00)=(0000)=OAB=\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0(0)+1(0)&0(1)+1(0)\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}=O.

Since A=BA=B here, BA=AB=OBA=AB=O too. Both A≠OA\ne O and B≠OB\ne O, so this pair satisfies (ii).

Step 3. (iii) Choose A,BA,B with AB=OAB=O but BA≠OBA\ne O. Take A=(1000), B=(0011)A=\begin{pmatrix}1&0\\0&0\end{pmatrix},\ B=\begin{pmatrix}0&0\\1&1\end{pmatrix}. …

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