GATE ECE 2012 | Question: 47

Question

Given that

$A=\begin{bmatrix} -5 &-3 \\ 2 &0\end{bmatrix}$ and $I=\begin{bmatrix} 1 & 0 \\ 0 &1\end{bmatrix}$, the value of $A^3$ is

• A. $15\:A+12\:I$
• B. $19\:A+30\:I$
• C. $17\:A+15\:I$
• D. $17\:A+21\:I$

Answer 1

• Characteristic Equation: $|A - \lambda I| = (-5-\lambda)(-\lambda) - (-6) = \lambda^2 + 5\lambda + 6 = 0$.

• Cayley-Hamilton Theorem: $A^2 + 5A + 6I = 0 \Rightarrow A^2 = -5A - 6I$.

• Find $A^3$: Multiply by $A$:

  $$A^3 = -5A^2 - 6A$$

  $$A^3 = -5(-5A - 6I) - 6A = 25A + 30I - 6A = 19A + 30I$$

• Final Answer: B. $19A + 30I$.