Aspire Faculty ID #17142 · Topic: AMU MCA 2020 · Just now
AMU MCA 2020

If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ satisfies the matrix equation $A^2 - kA + 2I = 0$, then the value of $k$ is:

Solution

Trace of $A = 3 + (-2) = 1$

Determinant of $A = (3)(-2) - (-8) = -6 + 8 = 2$

Characteristic equation of $A$ is:

$\lambda^2 - (\text{trace})\lambda + \det = 0$

$\lambda^2 - \lambda + 2 = 0$

By Cayley–Hamilton theorem:

$A^2 - A + 2I = 0$

Comparing with $A^2 - kA + 2I = 0$

$k = 1$

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