Matrices & Determinants
Inverse of a matrix
Grade Class 12

Question:

An invertible matrix A of order 3 satisfies the relation A = A<sup>-1</sup> + 2I, (where I denotes identity matrix). The value of |A - I|.|A + I|.|A - 2I| is
4
6
8
10

Step-by-Step Solution

Key Concept: Given A = A^-1 + 2I, multiply by A to get A^2 - 2A - I = 0. This implies A^2 - 2A = I. The expression |A - I|.|A + I|.|A - 2I| can be evaluated by relating it to the characteristic polynomial or by manipulating the matrix equation.
Given A = A<sup>-1</sup> + 2I. Multiplying by A, we get A<sup>2</sup> = I + 2A, so A<sup>2</sup> - 2A - I = 0. Let f(x) = x<sup>2</sup> - 2x - 1 = 0. The roots are x = (2 \pm sqrt(4 + 4))/2 = 1 \pm sqrt(2). Let the eigenvalues of A be \lambda<sub>1</sub>, \lambda<sub>2</sub>, \lambda<sub>3</sub>. Then \lambda<sub>i</sub><sup>2</sup> - 2\lambda<sub>i</sub> - 1 = 0. We need to find |A - I|.|A + I|.|A - 2I| = Π(\lambda<sub>i</sub> - 1)Π(\lambda<sub>i</sub> + 1)Π(\lambda<sub>i</sub> - 2). Note that Π(\lambda<sub>i</sub> - 1)Π(\lambda<sub>i</sub> + 1) = Π(\lambda<sub>i</sub><sup>2</sup> - 1) = Π(2\lambda<sub>i</sub>). Also Π(\lambda<sub>i</sub> - 2) = -Π(2 - \lambda<sub>i</sub>). Using the characteristic equation, the product evaluates to 8.
Correct Answer: 3

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