1. **Problem statement:** Given the matrix $$M = \begin{pmatrix} 0 & 0 & a \\ b & 0 & 0 \\ 0 & c & 0 \end{pmatrix}$$ with $a \in \mathbb{Q}^*$ and $0 \leq b,c \leq 1$, find conditions on $a,b,c$ such that $M^3 = E$ (the identity matrix). Then show that for these matrices $M^4 = M$ and analyze the powers of $M$. 2. **Step 1: Compute $M^2$** $$M^2 = M \cdot M = \begin{pmatrix} 0 & 0 & a \\ b & 0 & 0 \\ 0 & c & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & a \\ b & 0 & 0 \\ 0 & c & 0 \end{pmatrix} = \begin{pmatrix} 0 \cdot 0 + 0 \cdot b + a \cdot 0 & 0 \cdot 0 + 0 \cdot 0 + a \cdot c & 0 \cdot a + 0 \cdot 0 + a \cdot 0 \\ b \cdot 0 + 0 \cdot b + 0 \cdot 0 & b \cdot 0 + 0 \cdot 0 + 0 \cdot c & b \cdot a + 0 \cdot 0 + 0 \cdot 0 \\ 0 \cdot 0 + c \cdot b + 0 \cdot 0 & 0 \cdot 0 + c \cdot 0 + 0 \cdot c & 0 \cdot a + c \cdot 0 + 0 \cdot 0 \end{pmatrix} = \begin{pmatrix} 0 & a c & 0 \\ 0 & 0 & a b \\ c b & 0 & 0 \end{pmatrix}$$ 3. **Step 2: Compute $M^3 = M \cdot M^2$** $$M^3 = \begin{pmatrix} 0 & 0 & a \\ b & 0 & 0 \\ 0 & c & 0 \end{pmatrix} \begin{pmatrix} 0 & a c & 0 \\ 0 & 0 & a b \\ c b & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 \cdot 0 + 0 \cdot 0 + a \cdot c b & 0 \cdot a c + 0 \cdot 0 + a \cdot 0 & 0 \cdot 0 + 0 \cdot a b + a \cdot 0 \\ b \cdot 0 + 0 \cdot 0 + 0 \cdot c b & b \cdot a c + 0 \cdot 0 + 0 \cdot 0 & b \cdot 0 + 0 \cdot a b + 0 \cdot 0 \\ 0 \cdot 0 + c \cdot 0 + 0 \cdot c b & 0 \cdot a c + c \cdot 0 + 0 \cdot 0 & 0 \cdot 0 + c \cdot a b + 0 \cdot 0 \end{pmatrix} = \begin{pmatrix} a c b & 0 & 0 \\ 0 & b a c & 0 \\ 0 & 0 & c a b \end{pmatrix}$$ 4. **Step 3: Condition for $M^3 = E$** Since $E$ is the identity matrix, $$M^3 = E = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ This implies $$a b c = 1$$ (since all diagonal entries must be 1 and off-diagonal 0). 5. **Step 4: Show $M^4 = M$ for these matrices** Calculate $$M^4 = M \cdot M^3 = M \cdot E = M$$ because multiplying by the identity matrix leaves $M$ unchanged. 6. **Step 5: Analyze powers of $M$** Since $M^3 = E$, powers of $M$ cycle every 3 steps: - $M^0 = E$ - $M^1 = M$ - $M^2 = M^2$ - $M^3 = E$ - $M^4 = M$ - $M^5 = M^2$ - $M^6 = E$ Thus, the powers of $M$ repeat in cycles of length 3. **Final answer:** - The condition for $M^3 = E$ is $$a b c = 1$$ - For these matrices, $$M^4 = M$$ holds. - The powers of $M$ cycle every 3 steps, repeating $E, M, M^2$.