Question:

Step 1: Determine all possible integer triplets $(a,b,c)$. Given $a, b, c \in \mathbb{Z}$ and $a^2 + b^2 + c^2 = 2$. The only way for three integer squares to sum to 2 is if two of the squares are 1 and one is 0. This means two of the integers must be $\pm 1$ and one must be 0. The possible ordered triplets $(a,b,c)$ are permutations of $(\pm 1, \pm 1, 0)$. These 12 triplets are: $$ (1,1,0), (1,0,1), (0,1,1) $$ $$ (-1,-1,0), (-1,0,-1), (0,-1,-1) $$ $$ (1,-1,0), (1,0,-1), (0,1,-1) $$ $$ (-1,1,0), (-1,0,1), (0,-1,1) $$ Step 2: Compute the determinant of $A$ for each triplet. The matrix is $A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$. The determinant of $A$ is given by: $$ \det(A) = a(cb - a^2) - b(b^2 - ac) + c(ab - c^2) = 3abc - (a^3 + b^3 + c^3) $$ We calculate $\det(A)$ for each of the 12 triplets: 1. For $(1,1,0)$: $\det(A) = 3(1)(1)(0) - (1^3+1^3+0^3) = 0 - (1+1+0) = -2$. 2. For $(1,0,1)$: $\det(A) = 3(1)(0)(1) - (1^3+0^3+1^3) = 0 - (1+0+1) = -2$. 3. For $(0,1,1)$: $\det(A) = 3(0)(1)(1) - (0^3+1^3+1^3) = 0 - (0+1+1) = -2$. 4. For $(-1,-1,0)$: $\det(A) = 3(-1)(-1)(0) - ((-1)^3+(-1)^3+0^3) = 0 - (-1-1+0) = 2$. 5. For $(-1,0,-1)$: $\det(A) = 3(-1)(0)(-1) - ((-1)^3+0^3+(-1)^3) = 0 - (-1+0-1) = 2$. 6. For $(0,-1,-1)$: $\det(A) = 3(0)(-1)(-1) - (0^3+(-1)^3+(-1)^3) = 0 - (0-1-1) = 2$. 7. For $(1,-1,0)$: $\det(A) = 3(1)(-1)(0) - (1^3+(-1)^3+0^3) = 0 - (1-1+0) = 0$. 8. For $(1,0,-1)$: $\det(A) = 3(1)(0)(-1) - (1^3+0^3+(-1)^3) = 0 - (1+0-1) = 0$. 9. For $(0,1,-1)$: $\det(A) = 3(0)(1)(-1) - (0^3+1^3+(-1)^3) = 0 - (0+1-1) = 0$. 10. For $(-1,1,0)$: $\det(A) = 3(-1)(1)(0) - ((-1)^3+1^3+0^3) = 0 - (-1+1+0) = 0$. 11. For $(-1,0,1)$: $\det(A) = 3(-1)(0)(1) - ((-1)^3+0^3+1^3) = 0 - (-1+0+1) = 0$. 12. For $(0,-1,1)$: $\det(A) = 3(0)(-1)(1) - (0^3+(-1)^3+1^3) = 0 - (0-1+1) = 0$. In summary: * $\det(A) = -2$ for 3 triplets. * $\det(A) = 2$ for 3 triplets. * $\det(A) = 0$ for 6 triplets. Step 3: Analyze the given statements. **Statement 1: The probability that the system of equations has a unique solution is 1/2.** A system of linear equations $A\mathbf{x} = \mathbf{b}$ has a unique solution if and only if $\det(A) \neq 0$. From Step 2, there are $3+3=6$ triplets for which $\det(A) \neq 0$. The total number of possible triplets is 12. The probability of a unique solution is $\frac{6}{12} = \frac{1}{2}$. Thus, Statement 1 is true. **Statement 2: The number of triplets $(a, b, c)$ for which the system of equations has infinitely many solutions is 6.** A system $A\mathbf{x} = \mathbf{b}$ has infinitely many solutions if $\det(A) = 0$ and $\text{rank}(A) = \text{rank}([A | \mathbf{b}])$. From Step 2, there are 6 triplets for which $\det(A) = 0$. Let's examine one such case, for example, $(a,b,c) = (1,-1,0)$. The matrix $A = \begin{bmatrix} 1 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$ and $\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$. The augmented matrix is $[A | \mathbf{e}_1] = \begin{bmatrix} 1 & -1 & 0 & 1 \\ -1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \end{bmatrix}$. Perform row operations: $R_2 \to R_2 + R_1$: $\begin{bmatrix} 1 & -1 & 0 & 1 \\ 0 & -1 & 1 & 1 \\ 0 & 1 & 1 & 0 \end{bmatrix}$ $R_3 \to R_3 + R_2$: $\begin{bmatrix} 1 & -1 & 0 & 1 \\ 0 & -1 & 1 & 1 \\ 0 & 0 & 2 & 1 \end{bmatrix}$ From the row-reduced form, $\text{rank}(A) = 2$ (since the first three columns have two non-zero rows and $\det(A)=0$). However, $\text{rank}([A | \mathbf{e}_1]) = 3$ (since the last row $(0,0,2,1)$ implies three leading non-zero entries). Since $\text{rank}(A) \neq \text{rank}([A | \mathbf{e}_1])$, the system has no solution for $(1,-1,0)$. This pattern holds for all 6 triplets where $\det(A)=0$. For these triplets, the system has no solution. Therefore, the number of triplets for which the system has infinitely many solutions is 0. Thus, Statement 2 is false. **Statement 3: If $a = 0$, the number of ordered pairs $(b, c)$ for which the system of equations has no solution is 2.** If $a=0$, then $b^2+c^2=2$. The possible integer pairs $(b,c)$ are: * $(1,1)$ (corresponds to $(0,1,1)$ triplet) * $(-1,-1)$ (corresponds to $(0,-1,-1)$ triplet) * $(1,-1)$ (corresponds to $(0,1,-1)$ triplet) * $(-1,1)$ (corresponds to $(0,-1,1)$ triplet) From Step 2, for $(0,1,1)$ and $(0,-1,-1)$, $\det(A) \neq 0$, so they have unique solutions. For $(0,1,-1)$ and $(0,-1,1)$, $\det(A) = 0$. These are the two pairs $(b,c)$ for which the system might have no solution or infinitely many solutions. As shown in the analysis for Statement 2, for any triplet where $\det(A)=0$, the system has no solution. Specifically, for $(0,1,-1)$ and $(0,-1,1)$, the system has no solution. Thus, there are 2 ordered pairs $(b,c)$ for which the system has no solution when $a=0$. Thus, Statement 3 is true. **Statement 4: The number of elements in the range of $ab + bc + ca$ is 2.** We calculate $ab+bc+ca$ for each of the 12 triplets: 1. For $(1,1,0)$: $1 \cdot 1 + 1 \cdot 0 + 0 \cdot 1 = 1$. 2. For $(1,0,1)$: $1 \cdot 0 + 0 \cdot 1 + 1 \cdot 1 = 1$. 3. For $(0,1,1)$: $0 \cdot 1 + 1 \cdot 1 + 1 \cdot 0 = 1$. 4. For $(-1,-1,0)$: $(-1)(-1) + (-1)(0) + (0)(-1) = 1$. 5. For $(-1,0,-1)$: $(-1)(0) + (0)(-1) + (-1)(-1) = 1$. 6. For $(0,-1,-1)$: $(0)(-1) + (-1)(-1) + (-1)(0) = 1$. 7. For $(1,-1,0)$: $(1)(-1) + (-1)(0) + (0)(1) = -1$. 8. For $(1,0,-1)$: $(1)(0) + (0)(-1) + (-1)(1) = -1$. 9. For $(0,1,-1)$: $(0)(1) + (1)(-1) + (-1)(0) = -1$. 10. For $(-1,1,0)$: $(-1)(1) + (1)(0) + (0)(-1) = -1$. 11. For $(-1,0,1)$: $(-1)(0) + (0)(1) + (1)(-1) = -1$. 12. For $(0,-1,1)$: $(0)(-1) + (-1)(1) + (1)(0) = -1$. The values of $ab+bc+ca$ are either $1$ or $-1$. The range of $ab+bc+ca$ is $\{-1, 1\}$. The number of elements in this range is 2. Thus, Statement 4 is true.