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Worked Examples · Example 13

Q.Find the adjoint and the inverse of A=(3152)A=\begin{pmatrix}3&1\\5&2\end{pmatrix}.

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First, ∣A∣=(3)(2)−(1)(5)=6−5=1|A|=(3)(2)-(1)(5)=6-5=1. Since ∣A∣=1≠0|A|=1\ne0, AA is non-singular and has an inverse.

For A=(abcd)=(3152)A=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}3&1\\5&2\end{pmatrix}, the adjoint shortcut gives

adj⁡(A)=(d−b−ca)=(2−1−53)\operatorname{adj}(A)=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}=\begin{pmatrix}2&-1\\-5&3\end{pmatrix}

Then A−1=1∣A∣adj⁡(A)=11(2−1−53)=(2−1−53)A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A)=\dfrac{1}{1}\begin{pmatrix}2&-1\\-5&3\end{pmatrix}=\begin{pmatrix}2&-1\\-5&3\end{pmatrix}.

Independent check — multiply AA by A−1A^{-1} and confirm it gives II: …

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