Question
If x, y, z are in arithmetic progression with common difference d, x 3d, and the determinant of the matrix \left[ {\matrix{ 3 & {4\sqrt 2 } & x \cr 4 & {5\sqrt 2 } & y \cr 5 & k & z \cr } } \right] is zero, then the value of k 2 is :
Options
Solution
Key Concepts and Formulas
- Arithmetic Progression (AP): If three numbers are in AP, then the middle term is the arithmetic mean of the other two, meaning . This fundamental relationship is key to simplifying the determinant.
- Determinant Properties:
- The value of a determinant remains unchanged if we apply elementary row operations of the type (or more generally, ) to any row . Similarly, this holds for column operations.
- If a row or column of a determinant contains only zeros, the determinant's value is zero.
- To evaluate a determinant: Expanding along a row or column that contains more zeros significantly simplifies the calculation.
Step-by-Step Solution
Step 1: Write Down the Given Determinant Equation
We are provided with a matrix whose determinant is zero:
Step 2: Utilize the Arithmetic Progression Property
We are given that are in an Arithmetic Progression. Reasoning: For any three terms in an AP, the sum of the first and third terms is equal to twice the middle term. Therefore, we have the relation: This equation can be rearranged as . This specific relationship will be instrumental in simplifying the determinant.
Step 3: Perform a Strategic Row Operation
Reasoning: Our goal is to simplify the determinant by creating zeros. Observing the relation from the AP property, we can apply a row operation to the first row () that combines elements from all three rows to make the third element of the new first row zero. We apply the row operation . Let's calculate the new elements for the first row:
- New element at (1,1) position:
- New element at (1,2) position:
- New element at (1,3) position: . Using the AP property , we substitute with : .
Now, the determinant transforms into:
Step 4: Expand the Simplified Determinant
Reasoning: With two zeros in the first row, expanding the determinant along the first row is the most efficient method. The expansion formula along the first row is , where is the cofactor of element . Given , , and , the expansion simplifies to: This means we only need to calculate the cofactor . The cofactor . Substituting this back into the determinant equation: Multiplying by on both sides to simplify:
Step 5: Identify Potential Solutions for k
From the product of two terms being zero, we have two possible scenarios:
Step 6: Validate Solutions Using Problem Constraints
Reasoning: We must check if the second possibility, , is consistent with all the given conditions, especially . Let be the common difference of the arithmetic progression . Then we can express and in terms of and : Substitute these expressions into the condition : Rearranging the terms to solve for : Reasoning: The problem statement explicitly provides the constraint . Our derived condition directly contradicts this given information. Therefore, the possibility is not a valid solution under the problem's constraints.
Step 7: Calculate the Final Value of
Since the condition is ruled out by the problem's constraints, the only remaining valid possibility is . The question asks for the value of :
Common Mistakes & Tips
- Overlooking Constraints: Always pay close attention to all conditions given in the problem statement (e.g., ). These constraints are often crucial for eliminating extraneous solutions.
- Incorrect Row Operations: Ensure that the chosen row operation does not change the determinant's value (e.g., changes the determinant by a factor of ). Operations of the type are safe.
- Sign Errors in Cofactors: Be careful with the alternating signs when expanding a determinant. For , the sign is .
- Recognizing Patterns: For problems involving AP, GP, or HP, look for combinations of terms that can be formed using row/column operations to create zeros or simplify expressions. The operation is a classic trick for AP.
Summary
This problem effectively combines the properties of Arithmetic Progression with determinant evaluation. The key insight was to leverage the AP property () to perform a strategic row operation () that simplified the determinant by introducing zeros. After expanding the simplified determinant, two potential values for emerged. The critical constraint provided in the problem statement was then used to eliminate one of these possibilities, leading to a unique value for and subsequently .
The final answer is , which corresponds to option (A).