Key Concepts and Formulas Used:

This problem requires a strong understanding of fundamental matrix operations and properties. The core ideas we will leverage are:

1. Matrix Multiplication with Column Vectors: If a matrix $B$ is composed of its column vectors, say $B = [B_1 \ B_2 \ B_3]$, then the product $AB$ can be expressed as $AB = [AB_1 \ AB_2 \ AB_3]$. This property is crucial for constructing the matrix $AB$ from the given information.
2. Determinant Property of Products: For any two square matrices $A$ and $B$ of the same order, the determinant of their product is the product of their individual determinants: $|AB| = |A||B|$. This property allows us to find $|B|$ without explicitly calculating $B$ first.
3. Matrix Inverse: For an invertible square matrix $A$, its inverse $A^{-1}$ exists and is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$, where $\text{adj}(A)$ is the adjoint of $A$ (transpose of the cofactor matrix). This is used to find matrix $B$ by premultiplying $AB$ by $A^{-1}$.
4. Trace of a Matrix: The trace of a square matrix $B$, denoted as $\text{Tr}(B)$, is the sum of its diagonal elements. This definition is essential for calculating $\beta$.

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Step-by-Step Solution:

Step 1: Construct the product matrix $AB$.

We are given that matrix $B$ is composed of its column vectors $B_1, B_2, B_3$. That is, $B = [B_1 \ B_2 \ B_3]$. According to the property of matrix multiplication with column vectors, the product $AB$ can be formed by multiplying $A$ with each column of $B$: $AB = [AB_1 \ AB_2 \ AB_3]$ We are provided with the results of these individual products: $AB_1=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right], \quad AB_2=\left[\begin{array}{l} 2 \\ 3 \\ 0 \end{array}\right], \quad AB_3=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right]$ Why this step? By directly assembling these column vectors, we can form the complete product matrix $AB$ without needing to first determine the matrix $B$. This simplifies the initial phase of the problem, especially for calculating its determinant.

Concatenating these column vectors, we obtain the matrix $AB$: $AB = \left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{array}\right]$

Step 2: Calculate the determinant of matrix $A$.

The given matrix $A$ is: $A=\left[\begin{array}{lll} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]$ Why this step? We need the determinant of $A$, $|A|$, to apply the determinant property $|AB| = |A||B|$. This property will allow us to find $|B|$ efficiently. We calculate $|A|$ by expanding along the second column, as it contains two zero elements, which simplifies the calculation: $|A| = (0) \cdot \text{Cofactor}(A_{12}) + (1) \cdot \text{Cofactor}(A_{22}) + (0) \cdot \text{Cofactor}(A_{32})$ $|A| = 1 \cdot (-1)^{2+2} \left| \begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array} \right|$ $|A| = 1 \cdot ( (2)(1) - (1)(1) ) = 1 \cdot (2 - 1) = 1$ So, the determinant of $A$ is $|A| = 1$.

Step 3: Calculate the determinant of matrix $AB$.

From Step 1, we have the matrix $AB$: $AB = \left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{array}\right]$ Why this step? We need $|AB|$ to use the determinant property $|AB| = |A||B|$ to find $|B|$. Observe that $AB$ is an upper triangular matrix. The determinant of an upper (or lower) triangular matrix is simply the product of its diagonal elements. $|AB| = (1)(3)(1) = 3$

Step 4: Determine $\alpha = |B|$ using the determinant property.

Now we apply the determinant property $|AB| = |A||B|$. Why this step? This is the most efficient way to find the determinant of $B$ ($\alpha$) without having to calculate the entire matrix $B$ first. Substituting the values we found for $|A|$ and $|AB|$: $3 = (1) \cdot |B|$ $|B| = 3$ Therefore, $\alpha = |B| = 3$.

Step 5: Find matrix $B$.

To determine $\beta$, which is the sum of the diagonal elements of $B$, we need to find the matrix $B$ itself. We know that $AB = \left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{array}\right]$. Since $|A|=1 \neq 0$, matrix $A$ is invertible. We can find $A^{-1}$ and then calculate $B$ as $B = A^{-1}(AB)$. Why this step? Calculating $B$ using its inverse is a systematic and often more efficient approach than solving three separate systems of linear equations ($A B_j = C_j$) when you need all columns of $B$.

Step 5a: Find the inverse of matrix $A$, $A^{-1}$.

First, we find the cofactor matrix of $A$, and then its transpose to get the adjoint matrix, $\text{adj}(A)$. Cofactors of $A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]$: $C_{11} = \left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right| = 1$ $C_{12} = - \left| \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right| = -(1) = -1$ $C_{13} = \left| \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right| = -1$

$C_{21} = - \left| \begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array} \right| = 0$ $C_{22} = \left| \begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array} \right| = 1$ $C_{23} = - \left| \begin{array}{cc} 2 & 0 \\ 1 & 0 \end{array} \right| = 0$

$C_{31} = \left| \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right| = -1$ $C_{32} = - \left| \begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array} \right| = -(-1) = 1$ $C_{33} = \left| \begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array} \right| = 2$

The cofactor matrix is: $\text{Cof}(A) = \left[\begin{array}{ccc} 1 & -1 & -1 \\ 0 & 1 & 0 \\ -1 & 1 & 2 \end{array}\right]$ The adjoint matrix is the transpose of the cofactor matrix: $\text{adj}(A) = (\text{Cof}(A))^T = \left[\begin{array}{ccc} 1 & 0 & -1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{array}\right]$ Since $A^{-1} = \frac{1}{|A|} \text{adj}(A)$ and $|A|=1$: $A^{-1} = \left[\begin{array}{ccc} 1 & 0 & -1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{array}\right]$ Tip: Always double-check your determinant and inverse calculations, as errors here propagate through the rest of the problem. A quick check: $A A^{-1} = I$. For example, $A_{11} = (2)(1)+(0)(-1)+(1)(-1) = 2-1=1$. This looks correct.

Step 5b: Calculate $B = A^{-1}(AB)$.

Now we multiply $A^{-1}$ by $AB$: $B = \left[\begin{array}{ccc} 1 & 0 & -1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{array}\right] \left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{array}\right]$ Performing the matrix multiplication: $B_{11} = (1)(1) + (0)(0) + (-1)(0) = 1$ $B_{12} = (1)(2) + (0)(3) + (-1)(0) = 2$ $B_{13} = (1)(3) + (0)(2) + (-1)(1) = 3 - 1 = 2$

$B_{21} = (-1)(1) + (1)(0) + (1)(0) = -1$ $B_{22} = (-1)(2) + (1)(3) + (1)(0) = -2 + 3 = 1$ $B_{23} = (-1)(3) + (1)(2) + (1)(1) = -3 + 2 + 1 = 0$

$B_{31} = (-1)(1) + (0)(0) + (2)(0) = -1$ $B_{32} = (-1)(2) + (0)(3) + (2)(0) = -2$ $B_{33} = (-1)(3) + (0)(2) + (2)(1) = -3 + 2 = -1$

Thus, matrix $B$ is: $B = \left[\begin{array}{ccc} 1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1 \end{array}\right]$

Step 6: Determine $\beta$, the sum of all diagonal elements of $B$.

The diagonal elements of matrix $B$ are $B_{11}, B_{22}, B_{33}$. Why this step? This directly applies the definition of the trace of a matrix. $\beta = \text{Tr}(B) = B_{11} + B_{22} + B_{33}$ $\beta = 1 + 1 + (-1) = 1$ Therefore, $\beta = 1$.

Step 7: Calculate $\alpha^3 + \beta^3$.

We have found $\alpha = 3$ and $\beta = 1$. Now we substitute these values into the expression $\alpha^3 + \beta^3$: $\alpha^3 + \beta^3 = (3)^3 + (1)^3$ $\alpha^3 + \beta^3 = 27 + 1$ $\alpha^3 + \beta^3 = 28$

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Summary and Key Takeaway:

This problem demonstrates the power of using matrix properties strategically. We efficiently found the determinant of $B$ ($\alpha$) by leveraging $|AB| = |A||B|$, avoiding the need to calculate $B$ directly at that stage. Subsequently, finding $B$ via $A^{-1}(AB)$ allowed us to calculate its trace ($\beta$). While finding $B$ explicitly can be calculation-intensive, understanding when to apply specific matrix properties (like the determinant of a product or the determinant of a triangular matrix) can significantly simplify the problem. Always ensure meticulous calculation of determinants, cofactors, and matrix products to avoid errors.

The final answer is $\boxed{28}$.