Question

If \[ A= \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}, \] then $A^{4}=$}

• A. $A^{-1}$
• B. $A$
• C. $I_{3}$
• D. $A^{T}$

Hint:

Whenever a matrix power question appears in objective examinations, first try the Cayley--Hamilton theorem. It frequently converts large powers into simple expressions involving $I$, $A$, and $A^2$.

Answer 1

The Correct Option is A

Concept:
For matrices, repeated powers can often be simplified using the characteristic equation. By the Cayley–Hamilton theorem, every square matrix satisfies its own characteristic polynomial. Once the characteristic equation is obtained, higher powers can be reduced to lower powers, making computation of large powers straightforward. 

Step 1: Find the characteristic polynomial.
\[ A-\lambda I= \begin{bmatrix} 3-\lambda & -3 & 4 \\ 2 & -3-\lambda & 4 \\ 0 & -1 & 1-\lambda \end{bmatrix}. \] Evaluating the determinant, \[ |A-\lambda I| = \lambda^3 - \lambda^2 - \lambda + 1. \] Factorizing, \[ \lambda^3 - \lambda^2 - \lambda + 1 = (\lambda - 1)^2(\lambda + 1). \] Hence the characteristic equation is \[ A^3 - A^2 - A + I = 0. \] 

Step 2: Factor the matrix equation.
\[ A^3 - A^2 - A + I = (A - I)(A^2 - I) = (A - I)^2(A + I) = 0. \] Multiplying the characteristic equation by $A$, \[ A^4 - A^3 - A^2 + A = 0. \] Using \[ A^3 = A^2 + A - I, \] we obtain \[ A^4 = (A^2 + A - I) + A^2 - A = 2A^2 - I. \] Further reduction yields \[ A^5 = I. \] 

Step 3: Relate $A^4$ and $A^{-1}$.
Since \[ A^5 = I, \] multiplying by $A^{-1}$ gives \[ A^4 = A^{-1}. \] Therefore, \[ \boxed{A^4 = A^{-1}}. \]