If  and A² − 4A + 2I = 0, B² − 2B + I = 0, then |adj (A³ − B³)| is equal to

The correct answer is 11.

From the given image, the matrices are defined as:
$A=\left[\begin{array}{cc}\alpha & 2\\ 1& 2\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 1\\ \beta & 1\end{array}\right]$

According to the Cayley-Hamilton theorem, any square matrix satisfies its own characteristic equation. For a 2 × 2 matrix $M$, the characteristic equation is given by:
$M^2-\text{tr}\left(M\right)M+\left|M\right|I=0$

Step 1: Determine Matrix A
We are given that $A$ satisfies the equation:
$A^2-4A+2I=0$
Comparing this with the characteristic equation of $A$:
$\text{tr}\left(A\right)=\alpha +2=4\Rightarrow \alpha =2$
The determinant of $A$ is:
$\left|A\right|=2\alpha -2=2\left(2\right)-2=2$
This is consistent with the coefficient of $I$ in the equation. Therefore, matrix $A$ is:
$A=\left[\begin{array}{cc}2& 2\\ 1& 2\end{array}\right]$

Step 2: Determine Matrix B
We are given that $B$ satisfies the equation:
$B^2-2B+I=0$
Comparing this with the characteristic equation of $B$:
$\text{tr}\left(B\right)=1+1=2$
The determinant of $B$ is:
$\left|B\right|=1-\beta =1\Rightarrow \beta =0$
Therefore, matrix $B$ is:
$B=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

Step 3: Calculate A³
From $A^2-4A+2I=0$, we have:
$A^2=4A-2I$
Multiplying by $A$:
$A^3=4A^2-2A$
Substituting $A^2$ into the equation:
$A^3=4(4A-2I)-2A=14A-8I$
Now, computing the elements of $A^3$:
$A^3=14\left[\begin{array}{cc}2& 2\\ 1& 2\end{array}\right]-8\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}28& 28\\ 14& 28\end{array}\right]-\left[\begin{array}{cc}8& 0\\ 0& 8\end{array}\right]=\left[\begin{array}{cc}20& 28\\ 14& 20\end{array}\right]$

Step 4: Calculate B³
For a matrix of the form $B=\left[\begin{array}{cc}1& d\\ 0& 1\end{array}\right]$, its power is given by $B^n=\left[\begin{array}{cc}1& nd\\ 0& 1\end{array}\right]$.
Thus:
$B^3=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$

Step 5: Calculate A³ - B³
Subtracting $B^3$ from $A^3$:
$A^3-B^3=\left[\begin{array}{cc}20& 28\\ 14& 20\end{array}\right]-\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}19& 25\\ 14& 19\end{array}\right]$

Step 6: Find the Determinant of adj(A³ - B³)
For any square matrix $X$ of order $n$, we have the property:
$\left|\text{adj}\right(X\left)\right|={\left|X\right|}^{n-1}$
Since $A^3-B^3$ is a matrix of order $n=2$:
$\left|\text{adj}\right(A^3-B^3\left)\right|=|A^3-B^3|$
Calculating the determinant:
$|A^3-B^3|=(19\times 19)-(25\times 14)$
$|A^3-B^3|=361-350=11$