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Veronica's question at Yahoo! Answers (determinants)

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Chris L T521

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Jan 26, 2012
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Chris L T521

Well-known member
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Jan 26, 2012
995
Hi Veronica,

We can express the adjugate $\text{adj}(A)$ in terms of $A^{-1}$ by using the formula \[A^{-1}=\dfrac{1}{\det A}\text{adj}(A)\implies\text{adj}(A)=(\det A)A^{-1}\]

From here, we substitute this into $\det(A^{-1}+4\text{adj}(A))$ and use the fact that $\det A=2$ to get the following:

\[\begin{aligned}\det(A^{-1}+4\text{adj}(A)) &= \det(A^{-1}+4[(\det A)A^{-1}])\\ &=\det([4(\det A)+1]A^{-1})\\ &= \det([4(2)+1]A^{-1})\\ &= \det(9A^{-1}).\end{aligned}\]

Now, we use the fact that if $M$ is a $3\times 3$ matrix and $c$ is a constant, then $\det(cM)=c^3\det(M)$. We also recall that $\det (A^{-1})=\dfrac{1}{\det A}$.

Thus, we now see that
\[\begin{aligned}\det(9A^{-1}) &=9^3\det(A^{-1})\\ &=\frac{729}{\det A}\\ &=\frac{729}{2}.\end{aligned}\]

Therefore, if $A$ is a $3\times 3$ matrix with $\det A=2$, then $\det(A^{-1}+4\text{adj}(A))=\dfrac{729}{2}$.