Let A=pmatrix-1&1&-11&0&10&0&1pmatrix.

Question

Let $A=\begin{pmatrix}-1&1&-1\1&0&1\0&0&1\end{pmatrix}$. Suppose $A^2+\alpha\big(\operatorname{adj}(\operatorname{adj}A)\big)+\beta\big(\operatorname{adj}(A)\operatorname{adj}(\operatorname{adj}A)\big)=\begin{pmatrix}2&-2&2\-2&0&-1\0&0&-1\end{pmatrix}$ for some $\alpha,\beta\in\mathbb{R}$. Then $(\alpha-\beta)^2$ is equal to \_\_\_\_.

Answer: 4

Accepted Answer

Expanding along the third row, $\det(A)=1\cdot\begin{vmatrix}-1&1\1&0\end{vmatrix}=(-1)(0)-(1)(1)=-1$, so $A$ is invertible.

For an invertible $3	imes3$ matrix, $\operatorname{adj}(\operatorname{adj}A)=(\det A)^{1}A=-A$.

Also $A\cdot\operatorname{adj}(A)=(\det A)I=-I$, so $\operatorname{adj}(A)\cdot\operatorname{adj}(\operatorname{adj}A)=\operatorname{adj}(A)(-A)=-(\operatorname{adj}(A)A)=-(-I)=I$.

So the equation becomes $A^2-\alpha A+\beta I = \begin{pmatrix}2&-2&2\-2&0&-1\0&0&-1\end{pmatrix}$.

Direct computation gives $A^2=\begin{pmatrix}2&-1&1\-1&1&0\0&0&1\end{pmatrix}$.

Comparing entries of $A^2-\alpha A+\beta I$ with the RHS: from $(1,3)$: $1+\alpha=2\Rightarrow\alpha=1$; from $(3,3)$: $1-\alpha+\beta=-1\Rightarrow\beta=-1$. (All other entries check out consistently.)

So $\alpha-\beta=1-(-1)=2$, and $(\alpha-\beta)^2=4$.

Solutions · 1