Animated Solution for Mathematics - Matrices and Determinants: For some $\alpha ,\beta \in \mathbb{R}$, let $A=\left[\begin{array}{cc}\alpha & 2\\ 1& 2\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 1\\ 1& \beta \end{array}\right]$ be such that $A^2-4A+2I=B^2-3B+I=O$. Then $(\det (\text{adj }(A^3-B^3)){)}^2$ is equal to .........

We are given matrices $A=\left[\begin{array}{cc}\alpha & 2\\ 1& 2\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 1\\ 1& \beta \end{array}\right]$ with the constraints:

The moment you see a matrix satisfying a quadratic equation, your mind should immediately invoke the Cayley-Hamilton Theorem. This theorem acts as the bridge between the abstract matrix and its scalar properties.

For any $2\times 2$ matrix $M$, the characteristic equation is defined as ${\lambda }^2-(\text{Tr }M)\lambda +\det M=0$. By comparing this to our given equations, we instantly unmask the trace and determinant of our matrices.

For matrix $A$, we identify $\text{Tr }A=4$ and $\det A=2$. For matrix $B$, we identify $\text{Tr }B=3$ and $\det B=1$.

Now that we have the trace and determinant, finding $\alpha$ and $\beta$ is straightforward. The trace of $A$ is the sum of its diagonal elements, $\alpha +2$.

Since $\text{Tr }A=4$, we find $\alpha =2$. Similarly, for $B$, the trace is $1+\beta$.

Since $\text{Tr }B=3$, we find $\beta =2$. We have successfully decoded the matrices $A=\left[\begin{array}{cc}2& 2\\ 1& 2\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right]$.

We need to compute $A^3-B^3$. Rather than performing direct matrix multiplication, we use the power reduction technique.

Given $A^2=4A-2I$, we multiply by $A$ to obtain $A^3=4A^2-2A$. Substituting the expression for $A^2$ back into this equation yields:

Applying the same logic to $B$, where $B^2=3B-I$, we multiply by $B$ to get $B^3=3B^2-B$. Substituting $B^2$:

Now, we compute the difference:

Substituting the matrices $A$ and $B$:

Let $M=\left[\begin{array}{cc}15& 20\\ 6& 7\end{array}\right]$. The determinant is $\det M=(15)(7)-(20)(6)=105-120=-15$.

Finally, we use the property $\det (\text{adj }M)=(\det M{)}^{n-1}$. Since $n=2$, the determinant of the adjoint is simply $\det M=-15$.

Squaring this result, we obtain the final answer: 225.