Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& \alpha & \beta \\ 0& \beta & \alpha \end{array}\right]$ and $|2A{|}^3=2^{21}$ where $\alpha ,\beta \in \mathbb{Z}$, Then a value of $\alpha$ is

The given matrix is:

Because the first row contains two zeros, we expand the determinant along the first row to simplify the calculation. This bypasses tedious arithmetic and yields:

Thus, our core expression for the determinant is:

We are given the equation $|2A{|}^3=2^{21}$. Instead of modifying the matrix directly, we utilize the determinant property $|kA|=k^n|A|$, where $n$ is the order of the matrix.

Here, $k=2$ and $n=3$, so:

Substituting this into the original equation, we obtain:

Since $8=2^3$, we can rewrite the expression as:

Expanding the exponents gives:

Dividing both sides by $2^9$, we find:

Taking the cube root of both sides, we arrive at:

The Final Lock

We now equate our findings:

Given the constraint that $\alpha ,\beta \in \mathbb{Z}$, we factor the expression as a difference of squares:

We test integer factor pairs of 16, such as $(8,2)$, $(4,4)$, and $(16,1)$. For the pair $(8,2)$, we set:

Adding these two equations yields $2\alpha =10$, which results in $\alpha =5$.

This confirms a valid integer solution. By recognizing the structure of the matrix and applying determinant properties, we have successfully solved the puzzle.