1. **State the problem:** Find the index of nilpotency for each given 3x3 matrix. The index of nilpotency is the smallest positive integer $k$ such that $A^k = 0$. If no such $k$ exists, the matrix is not nilpotent. 2. **Matrix (i):** $$A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$ Calculate powers: $$A^2 = A \times A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ $$A^3 = A^2 \times A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ Since $A^3 = 0$ but $A^2 \neq 0$, the index of nilpotency is $3$. 3. **Matrix (ii):** $$B = \begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix}$$ Check if $B$ is nilpotent by calculating $B^2$: $$B^2 = B \times B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ Since $B^2 = 0$ and $B \neq 0$, the index of nilpotency is $2$. 4. **Matrix (iii):** $$C = \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix}$$ Calculate $C^2$: $$C^2 = C \times C = \begin{bmatrix} 6 & 3 & 12 \\ 19 & 9 & 39 \\ -9 & -4 & -18 \end{bmatrix} \neq 0$$ Calculate $C^3$: $$C^3 = C^2 \times C = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ Since $C^3 = 0$ but $C^2 \neq 0$, the index of nilpotency is $3$. **Final answers:** - (i) Index = 3 - (ii) Index = 2 - (iii) Index = 3