Understanding the Core Concept: The Inverse Matrix

The problem revolves around the definition of an inverse matrix. For a square matrix $A$, its inverse, denoted as $A^{-1}$, is another square matrix such that their product is the identity matrix $I$. This fundamental relationship is expressed as: $A A^{-1} = A^{-1} A = I$ For a $3 \times 3$ matrix, the identity matrix $I$ is: I = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right) In this problem, we are given that matrix $B$ is the inverse of matrix $A$. Therefore, we can write $A B = I$. Our goal is to use this property to find the value of $\alpha$.

Step 1: Setting Up the Matrix Equation

We are given the matrices: A = \left( {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right) B = \left( {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right) Since $B = A^{-1}$, we must have $AB = I$. Let's write out this matrix multiplication: A B = \left( {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right) \left( {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right) = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right)

Step 2: Performing Matrix Multiplication and Identifying Relevant Elements

To find $\alpha$, we don't need to compute all nine elements of the product matrix $AB$. We only need to calculate the elements that involve $\alpha$ and then equate them to the corresponding elements in the identity matrix $I$. The variable $\alpha$ appears in the element $B_{23}$ (second row, third column of matrix $B$). Let $C = AB$. The elements of $C$ are calculated as $C_{ij} = \sum_{k=1}^3 A_{ik} B_{kj}$. The elements of $C$ that will involve $B_{23}$ (and thus $\alpha$) are $C_{13}$, $C_{23}$, and $C_{33}$. We can choose any one of these to solve for $\alpha$.

Let's choose to calculate $C_{13}$, the element in the first row and third column of the product matrix $AB$. To find $C_{13}$, we multiply the first row of $A$ by the third column of $B$: $C_{13} = (A_{11} \cdot B_{13}) + (A_{12} \cdot B_{23}) + (A_{13} \cdot B_{33})$ Substituting the values: $C_{13} = (1 \cdot 2) + ((-1) \cdot \alpha) + (1 \cdot 3)$ $C_{13} = 2 - \alpha + 3$ $C_{13} = 5 - \alpha$

Step 3: Equating to the Identity Matrix and Solving for $\alpha$

Since $AB = I$, the element $C_{13}$ must be equal to the element $I_{13}$ of the identity matrix. From the identity matrix, $I_{13} = 0$. Therefore, we set our calculated $C_{13}$ equal to $0$: $5 - \alpha = 0$ Solving for $\alpha$: $\alpha = 5$

Step 4: Verification (Optional but Recommended)

To confirm our answer, we can quickly calculate another element involving $\alpha$, for instance, $C_{23}$ (second row, third column of $AB$). $C_{23} = (A_{21} \cdot B_{13}) + (A_{22} \cdot B_{23}) + (A_{23} \cdot B_{33})$ Substituting the values: $C_{23} = (2 \cdot 2) + (1 \cdot \alpha) + ((-3) \cdot 3)$ $C_{23} = 4 + \alpha - 9$ $C_{23} = \alpha - 5$ Since $AB = I$, the element $C_{23}$ must be equal to $I_{23}$, which is $0$. So, we set: $\alpha - 5 = 0$ $\alpha = 5$ Both calculations yield the same value for $\alpha$, which increases our confidence in the result.

Tips for Success & Common Pitfalls:

• Careful with Matrix Multiplication: Matrix multiplication is not commutative ($AB \neq BA$ in general) and requires precise row-by-column dot products. Pay close attention to signs.
• Strategic Calculation: You don't always need to compute every element of a product matrix. Identify which elements contain the unknown variable and only calculate those. This saves time and reduces potential errors.
• Understanding the Identity Matrix: Remember the structure of the identity matrix for the given dimension – ones on the main diagonal, zeros elsewhere.
• Alternative: You could also use $BA = I$. However, $AB=I$ is sufficient and usually yields the same complexity.

Conclusion

By utilizing the fundamental property that the product of a matrix and its inverse is the identity matrix ($AB=I$), we strategically performed matrix multiplication for a specific element involving $\alpha$. Equating this element to its corresponding value in the identity matrix allowed us to directly solve for $\alpha$. The value of $\alpha$ is $5$.

The final answer is $\boxed{5}$.