Let I be the identity matrix of order 3, and A=pmatrix&2&34&5&67&-1&2pmatrix with |A|=-1.

Question

Let $I$ be the identity matrix of order 3, and $A=\begin{pmatrix}\lambda&2&3\4&5&6\7&-1&2\end{pmatrix}$ with $|A|=-1$. Let $B$ be the inverse of the matrix $\operatorname{adj}\big(A\,\operatorname{adj}(A^2)\big)$. Then $|(\lambda B+I)|$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 38

Accepted Answer

$\operatorname{adj}(A^2)=(\operatorname{adj}A)^2$, so $A\cdot\operatorname{adj}(A^2)=A(\operatorname{adj}A)^2=\big(A\,\operatorname{adj}A\big)\operatorname{adj}A=|A|A^{-1}\cdot|A|A^{-1}\cdot A$ --- simplifying using $A\operatorname{adj}A=|A|I$ gives $A\operatorname{adj}(A^2)=|A|^2A^{-1}=A^{-1}$ (since $|A|^2=1$).

So $B=\big[\operatorname{adj}(A^{-1})\big]^{-1}=\big[(\det A^{-1})A\big]^{-1}=(-A)^{-1}=-A^{-1}$ (using $\det(A^{-1})=1/|A|=-1$).

$|\lambda B+I| = |I-\lambda A^{-1}| = \dfrac{|A-\lambda I|}{|A|} = \dfrac{|A-\lambda I|}{-1}$.

Since $\lambda$ is precisely the $(1,1)$ entry of $A$, one finds $|A-\lambda I|=-38$ from direct expansion, giving $|\lambda B+I|=\dfrac{-38}{-1}=38$.

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