Question

If \(A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]_{3 \times 3}\), then \(A^{-1}=\)

• A. A
• B. \(\mathrm{A}^{2}\)
• C. \(\mathrm{A}^{3}\)
• D. \(\mathrm{A}^{4}\)

Hint:

Use the Cayley-Hamilton theorem or compute the powers of the matrix to find which power equals the inverse.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
By the Cayley-Hamilton theorem a matrix satisfies its own characteristic equation \(A^3-(\operatorname{tr}A)A^2+cA-|A|I=0\).

Step 2: Coefficients
\(\operatorname{tr}A=3-3+1=1\), the sum of principal minors is \(-3+3+1=1\), and \(|A|=1\).

Step 3: Characteristic equation
So \(A^3-A^2+A-I=0\), which gives \(A\left(A^2-A+I\right)=I\).

Step 4: Inverse
Hence \(A^{-1}=A^2-A+I\). From \(A^3-A^2+A-I=0\) we also get \(A^3=A^2-A+I\), so \(A^{-1}=A^3\), option (C).

Final Answer:
The inverse of A equals \(A^3\). \[ \boxed{\text{(C) }A^{3}} \]