This problem can be efficiently solved by combining the given individual linear equations into a single matrix equation and then utilizing fundamental properties of determinants.

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1. Key Concept: Combining Multiple Linear Systems and Determinant Properties

When we have several systems of linear equations of the form $Ax_i = b_i$ (where $A$ is the same coefficient matrix for all systems, but $x_i$ and $b_i$ vary), we can combine them into a single, more compact matrix equation.

Suppose we have $k$ such systems: $Ax_1 = b_1$ $Ax_2 = b_2$ $\vdots$ $Ax_k = b_k$

We can form two new matrices, $X$ and $B$, by using the vectors $x_i$ and $b_i$ as their respective columns: $X = \begin{bmatrix} x_1 & x_2 & \dots & x_k \end{bmatrix}$ $B = \begin{bmatrix} b_1 & b_2 & \dots & b_k \end{bmatrix}$

With this construction, the $k$ individual equations can be elegantly represented as a single matrix equation: $AX = B$

Now, to find the determinant of $A$, we can take the determinant of both sides of this matrix equation: $\det(AX) = \det(B)$

A crucial property of determinants states that for any two square matrices $P$ and $Q$ of the same order, the determinant of their product is the product of their individual determinants: $\det(PQ) = \det(P)\det(Q)$

Applying this property to $AX = B$, we get: $\det(A)\det(X) = \det(B)$

If $\det(X) \neq 0$, we can solve for $\det(A)$: $\det(A) = \frac{\det(B)}{\det(X)}$

This approach allows us to determine $\det(A)$ without needing to explicitly find the matrix $A$ itself, which often simplifies complex problems.

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2. Step-by-Step Working

Step 1: Construct the matrices $X$ and $B$ from the given vectors.

We are given three solution vectors $x_1, x_2, x_3$ and their corresponding right-hand side vectors $b_1, b_2, b_3$. These define the three linear systems:

1. $Ax_1 = b_1$
2. $Ax_2 = b_2$
3. $Ax_3 = b_3$

To apply the key concept, we form matrix $X$ using $x_1, x_2, x_3$ as its columns, and matrix $B$ using $b_1, b_2, b_3$ as its columns.

Given vectors: $x_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad x_2 = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}, \quad x_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ $b_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad b_2 = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}, \quad b_3 = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}$

Constructing matrix $X$: $X = \begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 1 \end{bmatrix}$

Constructing matrix $B$: $B = \begin{bmatrix} b_1 & b_2 & b_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

Now, the three individual equations $Ax_i=b_i$ are compactly represented by the single matrix equation $AX=B$.

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

We need $\det(X)$ for our formula. $X = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 1 \end{bmatrix}$ Notice that $X$ is a lower triangular matrix (all entries above the main diagonal are zero). For any triangular matrix (upper or lower), its determinant is simply the product of its diagonal elements. $\det(X) = (1)(2)(1) = 2$

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

We also need $\det(B)$ for our formula. $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$ Notice that $B$ is a diagonal matrix (all non-diagonal entries are zero). A diagonal matrix is a special case of a triangular matrix. Therefore, its determinant is also the product of its diagonal elements. $\det(B) = (1)(2)(2) = 4$

Step 4: Calculate $\det(A)$ using the formula.

Now we substitute the calculated values of $\det(X)$ and $\det(B)$ into the derived formula $\det(A) = \frac{\det(B)}{\det(X)}$. $\det(A) = \frac{4}{2}$ $\det(A) = 2$

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3. Tips and Common Mistakes

• Tip 1: Recognize Special Matrix Forms: Always look for special matrix structures like diagonal or triangular matrices. Their determinants are simply the product of their diagonal elements, which saves significant calculation time compared to cofactor expansion.
• Tip 2: Verify $\det(X) \neq 0$: Before using the formula $\det(A) = \frac{\det(B)}{\det(X)}$, ensure that $\det(X)$ is not zero. If $\det(X) = 0$, then $X$ is singular, and $A$ might not be uniquely determined or might not exist in a way that satisfies the given conditions for a non-singular $A$. In this problem, $\det(X) = 2 \neq 0$, so the formula is valid.
• Common Mistake: Incorrect Matrix Construction: Ensure that the vectors $x_i$ and $b_i$ are correctly placed as columns (not rows) when forming matrices $X$ and $B$. Reversing this would lead to incorrect determinants.
• Common Mistake: Forgetting Determinant Property: A common error is to assume $\det(A+B) = \det(A) + \det(B)$ (which is generally false) or to incorrectly apply $\det(AX) = \det(A)\det(X)$. Always remember the product rule for determinants.

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4. Summary and Key Takeaway

This problem beautifully illustrates how combining multiple linear systems into a single matrix equation ($AX=B$) can simplify finding properties of the unknown matrix $A$. By leveraging the determinant property $\det(PQ) = \det(P)\det(Q)$, we could efficiently calculate $\det(A) = \frac{\det(B)}{\det(X)}$ without ever needing to determine the matrix $A$ itself. This technique is a powerful tool in linear algebra for problems involving systems of equations.

The final answer is $\boxed{2}$.