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      <text>7.4.1.3 The inverse of a square matrix</text>
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      <text><![CDATA[<ul>
    <li><p>A matrix that has no inverse is said to be a <b>singular matrix</b>. A matrix which has an inverse is said to be <b>non-singular</b>.</p></li>
    <li>
        <p>If a matrix has an inverse then that inverse is unique.</p>
        <p>Suppose \(B\) and \(C\) are both inverses of \(A\). Then, by definition of the inverse,</p>
        <p style="text-align: center;">\(AB = BA = I\) &emsp; <b>and</b> &emsp; \(AC = CA = I\).</p>
        <p>Consider the two ways of forming the product \(CAB\).</p>
        <ol type="1">
            <li>\(CAB = C(AB) = CI = C\).</li>
            <li>\(CAB = (CA)B = IB = B\).</li>
        </ol>
        <p>Hence, \(B = C\) and the <b>inverse is unique</b>.</p>
    </li>
    <li>
        <p>There is no such operation as division in matrix algebra.</p>
        <p>We do not write \(\dfrac{B}{A}\) but rather \(A^{-1}B\) or \(BA^{-1}\), depending on the order required.</p>
    </li>
    <li>
        <p>Assuming that the square matrix \(A\) has an inverse \(A^{-1}\) then the solution of the systems of equations \(AX = B\) is found by pre-multiplying both sides by \(A^{-1}\).</p>
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            <tbody>
                <tr>
                    <td></td>
                    <td style="text-align: right;">\(AX\)</td>
                    <td style="text-align: center;">&ensp;\(=\)&ensp;</td>
                    <td style="text-align: left;">\(B\),</td>
                    <td></td>
                </tr>
                <tr>
                    <td>pre-multiplying by \(A^{-1}\): &emsp;</td>
                    <td style="text-align: right;">\(A^{-1}(AX)\)</td>
                    <td style="text-align: center;">\(=\)</td>
                    <td style="text-align: left;">\(A^{-1}B\),</td>
                    <td></td>
                </tr>
                <tr>
                    <td>using associativity:</td>
                    <td style="text-align: right;">\((A^{-1}A)X\)</td>
                    <td style="text-align: center;">\(=\)</td>
                    <td style="text-align: left;">\(B\),</td>
                    <td></td>
                </tr>
                <tr>
                    <td>using \(A^{-1}A = \):</td>
                    <td style="text-align: right;">\(IX\)</td>
                    <td style="text-align: center;">\(=\)</td>
                    <td style="text-align: left;">\(A^{-1}B\),</td>
                    <td></td>
                </tr>
                <tr>
                    <td>using property of \(I\):</td>
                    <td style="text-align: right;">\(X\)</td>
                    <td style="text-align: center;">\(=\)</td>
                    <td style="text-align: left;">\(A^{-1}B\)</td>
                    <td>&emsp; which is the solution we seek.</td>
                </tr>
            </tbody>
        </table>
    </li>
</ul>]]></text>
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