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Matrices and Determinants question

2010 · Shift 2 · Q32
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Matrices and Determinants question

2010 · Shift 2 · Q32

JEE AdvancedMathematicsMatrices and DeterminantsNumerical+4 / −1
Let kkk be a positive real number and let A=[2k−12k2k2k1−2k−2k2k−1] and B=[02k−1k1−2k02k−k−2k0].\begin{aligned} A & =\left[\begin{array}{ccc} 2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \\ 2 \sqrt{k} & 1 & -2 k \\ -2 \sqrt{k} & 2 k & -1 \end{array}\right] \text { and } \\\\ \mathbf{B} & =\left[\begin{array}{ccc} 0 & 2 k-1 & \sqrt{k} \\ 1-2 k & 0 & 2 \sqrt{k} \\ -\sqrt{k} & -2 \sqrt{k} & 0 \end{array}\right] . \end{aligned}AB​=​2k−12k​−2k​​2k​12k​2k​−2k−1​​ and =​01−2k−k​​2k−10−2k​​k​2k​0​​.​ If det⁡(adj⁡A)+det⁡(adj⁡B)=106\operatorname{det}(\operatorname{adj} A)+\operatorname{det}(\operatorname{adj} B)=10^6det(adjA)+det(adjB)=106, then [k][k][k] is equal to ‾\underline{\hspace{2cm}}​. [ Note : adj M denotes the adjoint of a square matrix M and [k][k][k] denotes the largest integer less than or equal to kkk ].
Numerical answer
View written solutionFree

Correct answer: 4

Step-by-step Derivations

  1. Understand the relationship between the determinant of a matrix and its adjoint. For any square matrix M of order n, the determinant of its adjoint is given by the formula: det⁡(adj M)=(det⁡M)n−1\det(\text{adj } M) = (\det M)^{n-1}det(adj M)=(detM)n−1 In this problem, matrices A and B are of order n=3. Therefore, the formula becomes: det⁡(adj A)=(det⁡A)3−1=(det⁡A)2\det(\text{adj } A) = (\det A)^{3-1} = (\det A)^2det(adj A)=(detA)3−1=(detA)2 det⁡(adj B)=(det⁡B)3−1=(det⁡B)2\det(\text{adj } B) = (\det B)^{3-1} = (\det B)^2det(adj B)=(detB)3−1=(detB)2

  2. Substitute these into the given equation. The given equation is det⁡(adj A)+det⁡(adj B)=106\det(\text{adj } A) + \det(\text{adj } B) = 10^6det(adj A)+det(adj B)=106. Substituting the expressions from Step 1, we get: (det⁡A)2+(det⁡B)2=106(\det A)^2 + (\det B)^2 = 10^6(detA)2+(detB)2=106

  3. Calculate the determinant of matrix A. The matrix A is given by: A=[2k−12k2k2k1−2k−2k2k−1]A = \begin{bmatrix} 2k-1 & 2\sqrt{k} & 2\sqrt{k} \\ 2\sqrt{k} & 1 & -2k \\ -2\sqrt{k} & 2k & -1 \end{bmatrix}A=​2k−12k​−2k​​2k​12k​2k​−2k−1​​ We compute its determinant: det⁡A=(2k−1)∣1−2k2k−1∣−2k∣2k−2k−2k−1∣+2k∣2k1−2k2k∣\det A = (2k-1) \begin{vmatrix} 1 & -2k \\ 2k & -1 \end{vmatrix} - 2\sqrt{k} \begin{vmatrix} 2\sqrt{k} & -2k \\ -2\sqrt{k} & -1 \end{vmatrix} + 2\sqrt{k} \begin{vmatrix} 2\sqrt{k} & 1 \\ -2\sqrt{k} & 2k \end{vmatrix}detA=(2k−1)​12k​−2k−1​​−2k​​2k​−2k​​−2k−1​​+2k​​2k​−2k​​12k​​ det⁡A=(2k−1)(−1−(−4k2))−2k(−2k−4kk)+2k(4kk−(−2k))\det A = (2k-1)(-1 - (-4k^2)) - 2\sqrt{k}(-2\sqrt{k} - 4k\sqrt{k}) + 2\sqrt{k}(4k\sqrt{k} - (-2\sqrt{k}))detA=(2k−1)(−1−(−4k2))−2k​(−2k​−4kk​)+2k​(4kk​−(−2k​)) det⁡A=(2k−1)(4k2−1)−2k(−2k(1+2k))+2k(2k(2k+1))\det A = (2k-1)(4k^2 - 1) - 2\sqrt{k}(-2\sqrt{k}(1+2k)) + 2\sqrt{k}(2\sqrt{k}(2k+1))detA=(2k−1)(4k2−1)−2k​(−2k​(1+2k))+2k​(2k​(2k+1)) det⁡A=(2k−1)(2k−1)(2k+1)+4k(1+2k)+4k(2k+1)\det A = (2k-1)(2k-1)(2k+1) + 4k(1+2k) + 4k(2k+1)detA=(2k−1)(2k−1)(2k+1)+4k(1+2k)+4k(2k+1) det⁡A=(2k−1)2(2k+1)+8k(2k+1)\det A = (2k-1)^2(2k+1) + 8k(2k+1)detA=(2k−1)2(2k+1)+8k(2k+1) Factor out (2k+1)(2k+1)(2k+1): det⁡A=(2k+1)[(2k−1)2+8k]\det A = (2k+1)[(2k-1)^2 + 8k]detA=(2k+1)[(2k−1)2+8k] det⁡A=(2k+1)[(4k2−4k+1)+8k]\det A = (2k+1)[(4k^2 - 4k + 1) + 8k]detA=(2k+1)[(4k2−4k+1)+8k] det⁡A=(2k+1)[4k2+4k+1]\det A = (2k+1)[4k^2 + 4k + 1]detA=(2k+1)[4k2+4k+1] det⁡A=(2k+1)(2k+1)2\det A = (2k+1)(2k+1)^2detA=(2k+1)(2k+1)2 det⁡A=(2k+1)3\det A = (2k+1)^3detA=(2k+1)3

  4. Calculate the determinant of matrix B. The matrix B is given by: B=[02k−1k1−2k02k−k−2k0]B = \begin{bmatrix} 0 & 2k-1 & \sqrt{k} \\ 1-2k & 0 & 2\sqrt{k} \\ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix}B=​01−2k−k​​2k−10−2k​​k​2k​0​​ We can observe that B is a skew-symmetric matrix because BT=−BB^T = -BBT=−B. BT=[01−2k−k2k−10−2kk2k0]=−[02k−1k1−2k02k−k−2k0]=−BB^T = \begin{bmatrix} 0 & 1-2k & -\sqrt{k} \\ 2k-1 & 0 & -2\sqrt{k} \\ \sqrt{k} & 2\sqrt{k} & 0 \end{bmatrix} = -\begin{bmatrix} 0 & 2k-1 & \sqrt{k} \\ 1-2k & 0 & 2\sqrt{k} \\ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix} = -BBT=​02k−1k​​1−2k02k​​−k​−2k​0​​=−​01−2k−k​​2k−10−2k​​k​2k​0​​=−B A property of skew-symmetric matrices is that their determinant is zero if the order is odd. Since B is a 3x3 matrix (odd order), its determinant is zero. det⁡B=0\det B = 0detB=0 Alternatively, we can calculate it directly: det⁡B=0⋅(...)−(2k−1)((1−2k)(0)−(2k)(−k))+k((1−2k)(−2k)−0)\det B = 0 \cdot (...) - (2k-1)((1-2k)(0) - (2\sqrt{k})(-\sqrt{k})) + \sqrt{k}((1-2k)(-2\sqrt{k}) - 0)detB=0⋅(...)−(2k−1)((1−2k)(0)−(2k​)(−k​))+k​((1−2k)(−2k​)−0) det⁡B=−(2k−1)(2k)+k(−2k1/2(1−2k))\det B = -(2k-1)(2k) + \sqrt{k}(-2k^{1/2}(1-2k))detB=−(2k−1)(2k)+k​(−2k1/2(1−2k)) det⁡B=−2k(2k−1)−2k(1−2k)\det B = -2k(2k-1) - 2k(1-2k)detB=−2k(2k−1)−2k(1−2k) det⁡B=−4k2+2k−2k+4k2=0\det B = -4k^2 + 2k - 2k + 4k^2 = 0detB=−4k2+2k−2k+4k2=0

  5. Solve the equation for k. Substitute the values of det⁡A\det AdetA and det⁡B\det BdetB into the equation from Step 2: ((2k+1)3)2+(0)2=106( (2k+1)^3 )^2 + (0)^2 = 10^6((2k+1)3)2+(0)2=106 (2k+1)6=106(2k+1)^6 = 10^6(2k+1)6=106 Taking the 6th root of both sides, we get: ∣2k+1∣=10|2k+1| = 10∣2k+1∣=10 Since k is given as a positive real number, 2k+12k+12k+1 must be positive. Thus: 2k+1=102k+1 = 102k+1=10 2k=92k = 92k=9 k=92=4.5k = \frac{9}{2} = 4.5k=29​=4.5

  6. Find the required value. The question asks for [k][k][k], which denotes the greatest integer less than or equal to k. [k]=[4.5]=4[k] = [4.5] = 4[k]=[4.5]=4

Final Answer

The value of [k][k][k] is 4.

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