Skip to main content
Back to Matrices & Determinants
JEE Main 2023
Matrices & Determinants
Matrices and Determinants
Easy

Question

If x, y, z are in arithmetic progression with common difference d, x \ne 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

  1. Arithmetic Progression (AP): If three numbers x,y,zx, y, z are in AP, then the middle term yy is the arithmetic mean of the other two, meaning 2y=x+z2y = x+z. This fundamental relationship is key to simplifying the determinant.
  2. Determinant Properties:
    • The value of a determinant remains unchanged if we apply elementary row operations of the type RiRi+cRjR_i \to R_i + cR_j (or more generally, RiRi+c1Rj+c2RkR_i \to R_i + c_1R_j + c_2R_k) to any row RiR_i. 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 3×33 \times 3 determinant: abcdefghi=a(eifh)b(difg)+c(dheg)\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) 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 3×33 \times 3 matrix whose determinant is zero: 342x452y5kz=0\begin{vmatrix} 3 & {4\sqrt 2 } & x \\ 4 & {5\sqrt 2 } & y \\ 5 & k & z \end{vmatrix} = 0

Step 2: Utilize the Arithmetic Progression Property

We are given that x,y,zx, y, z 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: x+z=2yx+z = 2y This equation can be rearranged as x2y+z=0x - 2y + z = 0. 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 x2y+z=0x - 2y + z = 0 from the AP property, we can apply a row operation to the first row (R1R_1) that combines elements from all three rows to make the third element of the new first row zero. We apply the row operation R1R12R2+R3R_1 \to R_1 - 2R_2 + R_3. Let's calculate the new elements for the first row:

  • New element at (1,1) position: 32(4)+5=38+5=03 - 2(4) + 5 = 3 - 8 + 5 = 0
  • New element at (1,2) position: 422(52)+k=42102+k=k624\sqrt{2} - 2(5\sqrt{2}) + k = 4\sqrt{2} - 10\sqrt{2} + k = k - 6\sqrt{2}
  • New element at (1,3) position: x2y+zx - 2y + z. Using the AP property x+z=2yx+z=2y, we substitute x+zx+z with 2y2y: 2y2y=02y - 2y = 0.

Now, the determinant transforms into: 0k620452y5kz=0\begin{vmatrix} 0 & {k - 6\sqrt 2 } & 0 \\ 4 & {5\sqrt 2 } & y \\ 5 & k & z \end{vmatrix} = 0

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 a11C11+a12C12+a13C13a_{11} \cdot C_{11} + a_{12} \cdot C_{12} + a_{13} \cdot C_{13}, where CijC_{ij} is the cofactor of element aija_{ij}. Given a11=0a_{11}=0, a12=k62a_{12}=k-6\sqrt{2}, and a13=0a_{13}=0, the expansion simplifies to: 0C11+(k62)C12+0C13=00 \cdot C_{11} + (k - 6\sqrt{2}) \cdot C_{12} + 0 \cdot C_{13} = 0 This means we only need to calculate the cofactor C12C_{12}. The cofactor C12=(1)1+24y5z=(4z5y)C_{12} = (-1)^{1+2} \cdot \begin{vmatrix} 4 & y \\ 5 & z \end{vmatrix} = - (4z - 5y). Substituting this back into the determinant equation: (k62)((4z5y))=0(k - 6\sqrt{2}) \cdot (-(4z - 5y)) = 0 Multiplying by 1-1 on both sides to simplify: (k62)(4z5y)=0(k - 6\sqrt{2})(4z - 5y) = 0

Step 5: Identify Potential Solutions for k

From the product of two terms being zero, we have two possible scenarios:

  1. k62=0    k=62k - 6\sqrt{2} = 0 \implies k = 6\sqrt{2}
  2. 4z5y=0    4z=5y4z - 5y = 0 \implies 4z = 5y

Step 6: Validate Solutions Using Problem Constraints

Reasoning: We must check if the second possibility, 4z=5y4z = 5y, is consistent with all the given conditions, especially x3dx \ne 3d. Let dd be the common difference of the arithmetic progression x,y,zx, y, z. Then we can express yy and zz in terms of xx and dd: y=x+dy = x+d z=x+2dz = x+2d Substitute these expressions into the condition 4z=5y4z = 5y: 4(x+2d)=5(x+d)4(x+2d) = 5(x+d) 4x+8d=5x+5d4x + 8d = 5x + 5d Rearranging the terms to solve for xx: 8d5d=5x4x8d - 5d = 5x - 4x 3d=x3d = x Reasoning: The problem statement explicitly provides the constraint x3dx \ne 3d. Our derived condition x=3dx=3d directly contradicts this given information. Therefore, the possibility 4z5y=04z - 5y = 0 is not a valid solution under the problem's constraints.

Step 7: Calculate the Final Value of k2k^2

Since the condition 4z5y=04z - 5y = 0 is ruled out by the problem's constraints, the only remaining valid possibility is k62=0k - 6\sqrt{2} = 0. k=62k = 6\sqrt{2} The question asks for the value of k2k^2: k2=(62)2=62(2)2=362=72k^2 = (6\sqrt{2})^2 = 6^2 \cdot (\sqrt{2})^2 = 36 \cdot 2 = 72


Common Mistakes & Tips

  • Overlooking Constraints: Always pay close attention to all conditions given in the problem statement (e.g., x3dx \ne 3d). 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., RicRiR_i \to cR_i changes the determinant by a factor of cc). Operations of the type RiRi+cRjR_i \to R_i + cR_j are safe.
  • Sign Errors in Cofactors: Be careful with the alternating signs when expanding a determinant. For C12C_{12}, the sign is (1)1+2=1(-1)^{1+2} = -1.
  • 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 R1R12R2+R3R_1 \to R_1 - 2R_2 + R_3 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 (x+z=2yx+z=2y) to perform a strategic row operation (R1R12R2+R3R_1 \to R_1 - 2R_2 + R_3) that simplified the determinant by introducing zeros. After expanding the simplified determinant, two potential values for kk emerged. The critical constraint x3dx \ne 3d provided in the problem statement was then used to eliminate one of these possibilities, leading to a unique value for kk and subsequently k2k^2.

The final answer is 72\boxed{72}, which corresponds to option (A).

Practice More Matrices & Determinants Questions

View All Questions