Key Concepts and Formulas

• Properties of Determinants: The value of a determinant remains unchanged under elementary row or column operations of the form $R_i \to R_i + kR_j$ or $C_i \to C_i + kC_j$.
• Determinant Expansion: A $3 \times 3$ determinant can be expanded along any row or column. Expansion along a row/column with many zeros is significantly easier.
• Algebraic Simplification: Expanding and simplifying polynomial expressions.

Step-by-Step Solution

Step 1: Analyze the Determinant and the Given Condition We are given the determinant: f(x)=\left| {\matrix{ {x + a} & {x + 2} & {x + 1} \cr {x + b} & {x + 3} & {x + 2} \cr {x + c} & {x + 4} & {x + 3} \cr } } \right| and the condition $a - 2b + c = 1$. Observe that the elements in the first column are $x+a$, $x+b$, $x+c$. The second and third columns have constant terms that are consecutive integers. The given condition $a - 2b + c = 1$ suggests a linear combination of the rows that might simplify the determinant. A common strategy for determinants where elements involve 'x' and a linear relationship between parameters is to use row or column operations to eliminate 'x' or simplify the structure.

Step 2: Apply a Strategic Row Operation The condition $a - 2b + c = 1$ strongly suggests performing the row operation $R_1 \to R_1 + R_3 - 2R_2$. This operation is chosen because the coefficients $(1, -2, 1)$ directly correspond to the given condition. Let's apply this operation to each element of the first row:

• First Column Element: $(x+a) + (x+c) - 2(x+b) = x+a+x+c-2x-2b = (a+c-2b) + (x+x-2x) = (a+c-2b) + 0$. Given $a - 2b + c = 1$, so this element becomes $1$.

• Second Column Element: $(x+2) + (x+4) - 2(x+3) = x+2+x+4-2x-6 = (2x+6) - (2x+6) = 0$.

• Third Column Element: $(x+1) + (x+3) - 2(x+2) = x+1+x+3-2x-4 = (2x+4) - (2x+4) = 0$.

After applying the operation $R_1 \to R_1 + R_3 - 2R_2$, the determinant becomes: f(x)=\left| {\matrix{ 1 & 0 & 0 \cr {x + b} & {x + 3} & {x + 2} \cr {x + c} & {x + 4} & {x + 3} \cr } } \right|

Step 3: Expand the Determinant Now that the first row has two zeros, we can expand the determinant along the first row. The determinant of a $3 \times 3$ matrix $\begin{vmatrix} p & q & r \\ s & t & u \\ v & w & z \end{vmatrix}$ expanded along the first row is $p(tz-uw) - q(sz-uv) + r(sw-tv)$. In our case, $p=1$, $q=0$, and $r=0$. So, the expansion of $f(x)$ is: f(x) = 1 \cdot \left| {\matrix{ {x + 3} & {x + 2} \cr {x + 4} & {x + 3} \cr } } \right| - 0 \cdot (\text{minor}) + 0 \cdot (\text{minor}) f(x) = \left| {\matrix{ {x + 3} & {x + 2} \cr {x + 4} & {x + 3} \cr } } \right|

Step 4: Calculate the $2 \times 2$ Determinant Now we compute the value of the $2 \times 2$ determinant: $f(x) = (x+3)(x+3) - (x+2)(x+4)$ Expand the terms: $f(x) = (x^2 + 6x + 9) - (x^2 + 4x + 2x + 8)$ $f(x) = (x^2 + 6x + 9) - (x^2 + 6x + 8)$ Distribute the negative sign: $f(x) = x^2 + 6x + 9 - x^2 - 6x - 8$ Combine like terms: $f(x) = (x^2 - x^2) + (6x - 6x) + (9 - 8)$ $f(x) = 0 + 0 + 1$ $f(x) = 1$

Step 5: Evaluate $f(50)$ We have found that $f(x) = 1$ for all values of $x$. This means $f(x)$ is a constant function. Therefore, to find $f(50)$, we simply substitute $x=50$ into the constant value: $f(50) = 1$

This result corresponds to option (A).

Common Mistakes & Tips

• Incorrect Row/Column Operations: Ensure the operation applied ($R_i \to R_i + kR_j$ or $C_i \to C_i + kC_j$) is correctly executed and that the coefficients match the given condition.
• Algebraic Errors: Be meticulous when expanding and simplifying algebraic expressions, especially when dealing with signs.
• Ignoring the Condition: The condition $a-2b+c=1$ is crucial. Failing to use it will make the problem significantly harder or impossible to solve efficiently.

Summary

The problem involves evaluating a determinant function $f(x)$ with a given condition $a - 2b + c = 1$. By recognizing that the condition suggests a specific row operation ($R_1 \to R_1 + R_3 - 2R_2$), we strategically transformed the determinant. This operation, when applied, resulted in the first row having elements $1, 0, 0$. Expanding the determinant along this row simplified the problem to evaluating a $2 \times 2$ determinant. Upon calculation, the $2 \times 2$ determinant simplified to a constant value of $1$. Thus, $f(x) = 1$ for all $x$. Consequently, $f(50) = 1$.

The final answer is $\boxed{1}$.