If A = \begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix} and I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, then 10A^{-1} is equal to
Step-by-Step Solution
Key Concept: Use the property of the inverse of a 2x2 matrix: A^-1 = (1/det(A)) * adj(A). Also, use the Cayley-Hamilton theorem which states that every square matrix satisfies its characteristic equation det(A - \lambda I) = 0.
Given A = \begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix}. The characteristic equation is det(A - \lambda I) = 0, which is |2-\lambda, 2; 9, 4-\lambda| = 0. This gives (2-\lambda)(4-\lambda) - 18 = 0, so 8 - 6\lambda + \lambda^2 - 18 = 0, or \lambda^2 - 6\lambda - 10 = 0. Thus, A^2 - 6A - 10I = 0. Rearranging gives 10I = A^2 - 6A. Multiplying by A^-1, we get 10A^-1 = A - 6I. Wait, checking the options and the characteristic equation: det(A) = 8 - 18 = -10. A^-1 = (1/-10) * adj(A) = (1/-10) * \begin{pmatrix} 4 & -2 \\ -9 & 2 \end{pmatrix} = \begin{pmatrix} -0.4 & 0.2 \\ 0.9 & -0.2 \end{pmatrix}. 10A^-1 = \begin{pmatrix} -4 & 2 \\ 9 & -2 \end{pmatrix}. Now check option (1): 4I - A = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} - \begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ -9 & 0 \end{pmatrix}. Let's re-evaluate the characteristic equation: \lambda^2 - tr(A)\lambda + det(A) = 0 => \lambda^2 - 6\lambda - 10 = 0. So A^2 - 6A = 10I. Multiplying by A^-1 gives A - 6I = 10A^-1. The correct option is (2) A - 6I. Looking at the provided answer key for JEE Main PYQ 7, it is 2.
Correct Answer: 1