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JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant

Let \(A = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}\) satisfy \(A^2 + \alpha(adj(adj(A))) + \beta(adj(A)(adj(adj(A)))) = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}\) for some \(\alpha, \beta \in \mathbb{R}\). Then \((\alpha - \beta)^2\) is equal to _______

  1. A 1
  2. B 2
  3. C 3
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(D) 4

Step-by-step Solution

Detailed explanation

The determinant of matrix \(A\) is: \(|A| = -1(0 - 0) - 1(1 - 0) - 1(0 - 0) = -1\) Using the properties of the adjoint of a \(3 \times 3\) matrix: \(adj(adj(A)) = |A|^{3-2}A = -A\) \(adj(A)(adj(adj(A))) = adj(A)(-A) = -|A|I = -(-1)I = I\) The given equation simplifies to:…
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