Let A=pmatrix1&1&2-2&0&11&3&5pmatrix.

Question

Let $A=\begin{pmatrix}1&1&2\-2&0&1\1&3&5\end{pmatrix}$. Then the sum of all elements of the matrix $\operatorname{adj}\big(\operatorname{adj}(2(\operatorname{adj}A)^{-1})\big)$ is equal to:

Accepted Answer

$\det(A)=1(0-3)-1(-10-1)+2(-6-0)=-3+11-12=-4$.

For a $3	imes3$ matrix, $\det(\operatorname{adj}A)=\det(A)^2=16$.

Let $B=2(\operatorname{adj}A)^{-1}$. Since $(\operatorname{adj}A)^{-1}=\dfrac{1}{\det A}A$ (from $\operatorname{adj}(A)=(\det A)A^{-1}$), we get $\det B = 2^3\cdot\det\big((\operatorname{adj}A)^{-1}\big)=8\cdot\dfrac{1}{\det(\operatorname{adj}A)}=8\cdot\dfrac1{16}=\dfrac12$.

For a $3	imes3$ matrix, $\operatorname{adj}(\operatorname{adj}B)=(\det B)B=\dfrac12 B=\dfrac12\cdot 2(\operatorname{adj}A)^{-1}=(\operatorname{adj}A)^{-1}=-\dfrac14 A$ (using $\det A=-4$).

The sum of elements of $A$ is $1+1+2-2+0+1+1+3+5=12$, so the sum of elements of $-\dfrac14A$ is $-\dfrac14(12)=-3$.

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