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      <text>7.4.2.2 Key point ─ The inverse of a 2 x 2 matrix</text>
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      <text><![CDATA[<p>You will see that, had we chosen</p>
\[C = \frac{1}{ad - bc} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\]
<p>instead of \(B\), then both products \(AC\) and \(CA\) will be equal to \(I\). This requires \(ad - bc \neq 0\). Hence this matrix \(C\) is the inverse of \(A\). However, if \(ad - bc = 0\) then \(A\) has <b>no inverse</b>.</p>
<p>Note that for the matrix</p>
\[A = \begin{bmatrix}1 & 0 \\ 2 & 0\end{bmatrix},\]
<p>which occurred in the last question, \(ad - bc = 1 \times 0 - 0 \times 2 = 0\), confirming as we found that \(A\) has no inverse.</p>

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<h4>Key Point<br></h4>
<p>If \(ad-bc\neq 0\) then the \(2\times 2\) matrix \(A=\left[\begin{array}{cc}a&amp;b\\ c&amp;d\end{array}\right]\) has a (unique) inverse given by</p>
\[A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d&amp; -b\\ -c &amp; a \end{array}\right]\]
<p>Note that \(ad-bc=|A|\), the determinant of the matrix \(A\).</p>
<p>In words: To find the inverse of a \(2\times 2\) matrix \(A\) we interchange the diagonal elements, change the sign of the other two elements, and then divide by the determinant of \(A\).
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