1. Key Concepts and Formulas

• Definition of an Inverse Matrix: For any square matrix $Y$, its inverse, denoted as $Y^{-1}$, satisfies the property: $Y \cdot Y^{-1} = I$ where $I$ is the identity matrix of the same dimension. For a $3 \times 3$ matrix, the identity matrix is: I = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]
• Matrix Scalar Multiplication and Addition: To multiply a matrix by a scalar, multiply each element by the scalar. To add matrices, add their corresponding elements.
• Matrix Multiplication: The element in the $i$-th row and $j$-th column of the product of two matrices $A$ and $B$ (i.e., $AB$) is obtained by taking the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.

2. Step-by-Step Solution

Step 1: Calculate Powers of Matrix $X$ and Construct Matrix $Y$

We are given the matrix $X$: X = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right] We need to calculate $X^2$ as it is a component of matrix $Y$. {X^2} = X \cdot X = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right] \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right] Note: It's useful to observe that $X^3 = X^2 \cdot X = O$ (the null matrix), indicating $X$ is a nilpotent matrix of order 3. This property simplifies higher powers of $X$.

Next, we construct matrix $Y$ using the given expression $Y = \alpha I + \beta X + \gamma {X^2}$. Y = \alpha \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] + \beta \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right] + \gamma \left[ {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right] Performing scalar multiplication and matrix addition: Y = \left[ {\matrix{ \alpha & 0 & 0 \cr 0 & \alpha & 0 \cr 0 & 0 & \alpha \cr } } \right] + \left[ {\matrix{ 0 & \beta & 0 \cr 0 & 0 & \beta \cr 0 & 0 & 0 \cr } } \right] + \left[ {\matrix{ 0 & 0 & \gamma \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right] = \left[ {\matrix{ \alpha & \beta & \gamma \cr 0 & \alpha & \beta \cr 0 & 0 & \alpha \cr } } \right]

Step 2: Apply the Inverse Matrix Definition ($Y \cdot Y^{-1} = I$)

We are given the matrix $Y^{-1}$: {Y^{ - 1}} = \left[ {\matrix{ {{1 \over 5}} & {{{ - 2} \over 5}} & {{1 \over 5}} \cr 0 & {{1 \over 5}} & {{{ - 2} \over 5}} \cr 0 & 0 & {{1 \over 5}} \cr } } \right] Now, we perform the matrix multiplication $Y \cdot Y^{-1}$ and equate the resulting matrix to the identity matrix $I$: \left[ {\matrix{ \alpha & \beta & \gamma \cr 0 & \alpha & \beta \cr 0 & 0 & \alpha \cr } } \right] \left[ {\matrix{ {{1 \over 5}} & {{{ - 2} \over 5}} & {{1 \over 5}} \cr 0 & {{1 \over 5}} & {{{ - 2} \over 5}} \cr 0 & 0 & {{1 \over 5}} \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] Let's compute the elements of the product matrix:

• $(1,1)$ element: $\alpha({1 \over 5}) + \beta(0) + \gamma(0) = {\alpha \over 5}$
• $(1,2)$ element: $\alpha({-2 \over 5}) + \beta({1 \over 5}) + \gamma(0) = {-2\alpha + \beta \over 5}$
• $(1,3)$ element: $\alpha({1 \over 5}) + \beta({-2 \over 5}) + \gamma({1 \over 5}) = {\alpha - 2\beta + \gamma \over 5}$
• $(2,1)$ element: $0({1 \over 5}) + \alpha(0) + \beta(0) = 0$
• $(2,2)$ element: $0({-2 \over 5}) + \alpha({1 \over 5}) + \beta(0) = {\alpha \over 5}$
• $(2,3)$ element: $0({1 \over 5}) + \alpha({-2 \over 5}) + \beta({1 \over 5}) = {-2\alpha + \beta \over 5}$
• $(3,1)$ element: $0({1 \over 5}) + 0(0) + \alpha(0) = 0$
• $(3,2)$ element: $0({-2 \over 5}) + 0({1 \over 5}) + \alpha(0) = 0$
• $(3,3)$ element: $0({1 \over 5}) + 0({-2 \over 5}) + \alpha({1 \over 5}) = {\alpha \over 5}$

So, the product matrix is: \left[ {\matrix{ {\alpha \over 5} & {{-2\alpha + \beta} \over 5} & {{\alpha - 2\beta + \gamma} \over 5} \cr 0 & {\alpha \over 5} & {{-2\alpha + \beta} \over 5} \cr 0 & 0 & {\alpha \over 5} \cr } } \right]

Step 3: Form and Solve a System of Equations

Equating the elements of the product matrix to the corresponding elements of the identity matrix $I$:

1. From the $(1,1)$ element (or $(2,2)$ or $(3,3)$): ${\alpha \over 5} = 1 \implies \alpha = 5$
2. From the $(1,2)$ element (or $(2,3)$): ${{-2\alpha + \beta} \over 5} = 0$ Substitute $\alpha = 5$: ${{-2(5) + \beta} \over 5} = 0 \implies -10 + \beta = 0 \implies \beta = 10$
3. From the $(1,3)$ element: ${{\alpha - 2\beta + \gamma} \over 5} = 0$ Substitute $\alpha = 5$ and $\beta = 10$: ${{5 - 2(10) + \gamma} \over 5} = 0 \implies {5 - 20 + \gamma} = 0 \implies -15 + \gamma = 0 \implies \gamma = 15$ Thus, we have found the values of the parameters: $\alpha = 5$, $\beta = 10$, and $\gamma = 15$.

Step 4: Calculate the Final Expression

The problem asks for the value of $(\alpha - \beta + \gamma) 2$. Substitute the values of $\alpha$, $\beta$, and $\gamma$: $(\alpha - \beta + \gamma) 2 = (5 - 10 + 15) \times 2$ $= (-5 + 15) \times 2$ $= (10) \times 2$ $= 20$

3. Common Mistakes & Tips

• Matrix Multiplication Accuracy: Matrix multiplication is a common source of errors. Always double-check each element calculation.
• Order of Operations: Remember that matrix multiplication is not commutative ($AB \neq BA$ in general). However, the definition $Y \cdot Y^{-1} = I$ is specific.
• Identity Matrix: Be familiar with the structure of the identity matrix for different dimensions.
• Distractors: The matrix $Z$ was provided in the problem statement but was not required for finding $\alpha$, $\beta$, $\gamma$ or the final expression. Focus on the information directly relevant to the question asked.

4. Summary

This problem tests the fundamental definition of a matrix inverse and proficiency in matrix algebra. By first constructing the matrix $Y$ based on the given parameters and matrix $X$, and then utilizing the property $Y \cdot Y^{-1} = I$, we formed a system of linear equations. Solving this system yielded the values $\alpha = 5$, $\beta = 10$, and $\gamma = 15$. Substituting these values into the required expression $(\alpha - \beta + \gamma) 2$ gives $(5 - 10 + 15) \times 2 = 20$.

The final answer is $\boxed{0}$.