Question:

Step 1: Write down the given system of equations and the coefficient matrix. The given system of linear equations is: $$ \begin{aligned} \alpha x+2y+z &=1 \\ 2\alpha x+3y+z &=1 \\ 3x+\alpha y+2z &=\beta \end{aligned} $$ The coefficient matrix $A$ is: $$ A = \begin{pmatrix} \alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & 2 \end{pmatrix} $$ Step 2: Calculate the determinant of the coefficient matrix, $D$. The determinant $D$ is given by: $$ D = \begin{vmatrix} \alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & 2 \end{vmatrix} $$ Expanding along the first row: $$ D = \alpha(3 \times 2 - 1 \times \alpha) - 2(2\alpha \times 2 - 1 \times 3) + 1(2\alpha \times \alpha - 3 \times 3) $$ $$ D = \alpha(6 - \alpha) - 2(4\alpha - 3) + (2\alpha^2 - 9) $$ $$ D = 6\alpha - \alpha^2 - 8\alpha + 6 + 2\alpha^2 - 9 $$ $$ D = \alpha^2 - 2\alpha - 3 $$ Factoring the quadratic expression: $$ D = (\alpha - 3)(\alpha + 1) $$ Step 3: Calculate the determinants $D_x$, $D_y$, and $D_z$. These determinants are formed by replacing one column of $A$ with the constant terms vector $\begin{pmatrix} 1 \\ 1 \\ \beta \end{pmatrix}$. $$ D_x = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 3 & 1 \\ \beta & \alpha & 2 \end{vmatrix} $$ Applying $R_2 \to R_2 - R_1$: $$ D_x = \begin{vmatrix} 1 & 2 & 1 \\ 0 & 1 & 0 \\ \beta & \alpha & 2 \end{vmatrix} $$ Expanding along the second row: $$ D_x = 1 \cdot \begin{vmatrix} 1 & 1 \\ \beta & 2 \end{vmatrix} = 1(2 - \beta) = 2 - \beta $$ $$ D_y = \begin{vmatrix} \alpha & 1 & 1 \\ 2\alpha & 1 & 1 \\ 3 & \beta & 2 \end{vmatrix} $$ Applying $R_2 \to R_2 - R_1$: $$ D_y = \begin{vmatrix} \alpha & 1 & 1 \\ \alpha & 0 & 0 \\ 3 & \beta & 2 \end{vmatrix} $$ Expanding along the second row: $$ D_y = -\alpha \cdot \begin{vmatrix} 1 & 1 \\ \beta & 2 \end{vmatrix} = -\alpha(2 - \beta) = \alpha(\beta - 2) $$ $$ D_z = \begin{vmatrix} \alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & \beta \end{vmatrix} $$ Expanding along the first column: $$ D_z = \alpha(3\beta - \alpha) - 2\alpha(2\beta - \alpha) + 3(2 - 3) $$ $$ D_z = 3\alpha\beta - \alpha^2 - 4\alpha\beta + 2\alpha^2 - 3 $$ $$ D_z = \alpha^2 - \alpha\beta - 3 $$ Step 4: Analyze the system's consistency based on the determinants. The system has: * A unique solution if $D \neq 0$. This occurs when $\alpha \neq 3$ and $\alpha \neq -1$. * No solution (inconsistent) if $D = 0$ and at least one of $D_x, D_y, D_z$ is non-zero. * Infinitely many solutions (consistent) if $D = 0$ and $D_x = D_y = D_z = 0$. We examine the cases where $D=0$: Case A: $\alpha = -1$. Substituting $\alpha = -1$ into the determinants: $D = (-1-3)(-1+1) = (-4)(0) = 0$ $D_x = 2 - \beta$ $D_y = -1(\beta - 2) = 2 - \beta$ $D_z = (-1)^2 - (-1)\beta - 3 = 1 + \beta - 3 = \beta - 2$ * If $\beta \neq 2$: Then $D=0$ but $D_x \neq 0$, $D_y \neq 0$, $D_z \neq 0$. Thus, the system has **no solution**. * If $\beta = 2$: Then $D=0$, $D_x = 0$, $D_y = 0$, $D_z = 0$. Thus, the system has **infinitely many solutions**. Case B: $\alpha = 3$. Substituting $\alpha = 3$ into the determinants: $D = (3-3)(3+1) = (0)(4) = 0$ $D_x = 2 - \beta$ $D_y = 3(\beta - 2)$ $D_z = (3)^2 - 3\beta - 3 = 9 - 3\beta - 3 = 6 - 3\beta = 3(2 - \beta)$ * If $\beta \neq 2$: Then $D=0$ but $D_x \neq 0$, $D_y \neq 0$, $D_z \neq 0$. Thus, the system has **no solution**. * If $\beta = 2$: Then $D=0$, $D_x = 0$, $D_y = 0$, $D_z = 0$. Thus, the system has **infinitely many solutions**. Step 5: Evaluate the given options. * **Option 1: It has no solution if $\alpha=-1$ and $\beta\neq2$.** From Case A, if $\alpha=-1$ and $\beta\neq2$, $D=0$ and $D_x \neq 0$. This means there is no solution. So, Option 1 is **CORRECT**. * **Option 2: It has no solution for $\alpha=-1$ and for all $\beta\in\mathbb{R}$.** From Case A, if $\alpha=-1$ and $\beta=2$, the system has infinitely many solutions, not no solution. So, this statement is false. Option 2 is **NOT CORRECT**. * **Option 3: It has no solution for $\alpha=3$ and for all $\beta\neq2$.** From Case B, if $\alpha=3$ and $\beta\neq2$, $D=0$ and $D_x \neq 0$. This means there is no solution. So, Option 3 is **CORRECT**. * **Option 4: It has a solution for all $\alpha\neq-1$ and $\beta=2$.** If $\alpha \neq -1$: * If $\alpha \neq 3$ (and $\alpha \neq -1$), then $D \neq 0$, so there's a unique solution. * If $\alpha = 3$ and $\beta=2$ (which is included in "$\alpha \neq -1$ and $\beta=2$"), from Case B, $D=D_x=D_y=D_z=0$, so there are infinitely many solutions. In both scenarios, a solution exists. So, Option 4 is **CORRECT**. Step 6: Conclude the incorrect statement. The question asks which of the given options is NOT correct. Based on the analysis, Option 2 is the incorrect statement. The final answer is $\boxed{\text{Option 2}}$.