For some alpha, beta in R, let A=[\llalpha 2 \ 1 2\] and B=[\ll1 1 \ 1 beta\] be such that A^2-4 A+2 I=B^2-3 B+I=O. Then (det(adj( A^3- B^3)))^2 is equal to \_\_\_\_

Tr( A)=4 => alpha+2=4 => alpha=2 Tr( B)=3 => beta+1=3 => beta=2 A^2-4 A+2 I=O A^3=4 A^2-2 A=16 A-8 I-2 A 14A-8I =[\cc28 28 \ 14 28\]+[\cc-8 0 \ 0 -8\] = A^3=[\ll20 28 \ 14 20\] B^2-3 B+ I= O B^2=3 B- I => B^3=3 B^2- B=3(3 B- I)- B=8 B-3 I B^3=(\cc 8 8 \ 8 16 \)+(\cc -3 0 \ 0 -3 \)=(\cc 5 8 \ 8 13 \) \ A^3- B^3=(\ll 20 28 \ 14 20 \)-(\cc 5 8 \ 8 13 \)=(\cc 15 20 \ 6 7 \) \ => | A^3- B^3|=105-120=-15 \ => |adj( A^3- B^3)|=| A^3- B^3|=-15 \ => |adj( A^3- B^3)|^2=225

For some alpha beta in R let A beginarrayll alpha and 2... | null (Mathematics)

\operatorname{Tr}(\mathrm{A})=4 \Rightarrow \alpha+2=4 \Rightarrow \alpha=2\\

\operatorname{Tr}(\mathrm{B})=3 \Rightarrow \beta+1=3 \Rightarrow \beta=2\\

A^{2}-4 A+2 I=O\\

A^{3}=4 A^{2}-2 A=16 A-8 I-2 A\\

14A-8I

\\=\left[\begin{array}{cc}28 & 28 \\ 14 & 28\end{array}\right]+\left[\begin{array}{cc}-8 & 0 \\ 0 & -8\end{array}\right]\\

=\mathrm{A}^{3}=\left[\begin{array}{ll}20 & 28 \\ 14 & 20\end{array}\right]\\

\mathrm{B}^{2}-3 \mathrm{~B}+\mathrm{I}=\mathrm{O}\\

\mathrm{B}^{2}=3 \mathrm{~B}-\mathrm{I}\\

\Rightarrow \mathrm{B}^{3}=3 \mathrm{~B}^{2}-\mathrm{B}=3(3 \mathrm{~B}-\mathrm{I})-\mathrm{B}=8 \mathrm{~B}-3 \mathrm{I}\\

\begin{aligned} & \mathrm{B}^{3}=\left(\begin{array}{cc} 8 & 8 \\ 8 & 16 \end{array}\right)+\left(\begin{array}{cc} -3 & 0 \\ 0 & -3 \end{array}\right)=\left(\begin{array}{cc} 5 & 8 \\ 8 & 13 \end{array}\right) \\ & \mathrm{A}^{3}-\mathrm{B}^{3}=\left(\begin{array}{ll} 20 & 28 \\ 14 & 20 \end{array}\right)-\left(\begin{array}{cc} 5 & 8 \\ 8 & 13 \end{array}\right)=\left(\begin{array}{cc} 15 & 20 \\ 6 & 7 \end{array}\right) \\ & \Rightarrow\left|\mathrm{A}^{3}-\mathrm{B}^{3}\right|=105-120=-15 \\ & \Rightarrow\left|\operatorname{adj}\left(\mathrm{~A}^{3}-\mathrm{B}^{3}\right)\right|=\left|\mathrm{A}^{3}-\mathrm{B}^{3}\right|=-15 \\ & \Rightarrow\left|\operatorname{adj}\left(\mathrm{~A}^{3}-\mathrm{B}^{3}\right)\right|^{2}=225 \end{aligned}

Explanation

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