Let $A = I_2 - MM^T$, where $M$ is a real matrix of order $2 \times 1$ such that the relation $M^T M = I_1$ holds. If $\lambda$ is a real number such that the relation $AX = \lambda X$ holds for some non-zero real matrix $X$ of order $2 \times 1$, then the sum of squares of all possible values of $\lambda$ is equal to:

Question

For the matrices $ A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} $ and $ B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix} $, if $ (A^{15} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} $, then among the following which one is true?}

Answer 1

Given $A = I_2 - MM^T$ and $M^TM = I_1$, $M$ is established as a unit vector. Consequently, $A$ is a projection matrix, and its eigenvalues are restricted to 0 or 1.

For this specific scenario, the eigenvalue $\lambda$ of $A$ can assume values of either 0 or 1. The sum of the squares of all feasible $\lambda$ values is thus calculated as:

$0^2 + 1^2 = 1 + 1 = 2.$

Correct Answer: 2

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