If the system of equations 3x + y + 4z = 3$3x + y + 4z = 3$ 2x + alpha y - z = -3$2x + \alpha y - z = -3$ x + 2y + z = 4$x + 2y + z = 4$ has no solution, then the value of alpha$\alpha$ is equal to:

### Related Formula textFor no solution in AX = B, text we must have |A| = 0 text (i.e. Delta = 0 text) text and at least one of Delta_x, Delta_y, Delta_z neq 0.$$\text{For no solution in } AX = B, \text{ we must have } |A| = 0 \text{ (i.e. } \Delta = 0 \text{)} \text{ and at least one of } \Delta_x, \Delta_y, \Delta_z \neq 0.$$ ### Core Logic Evaluate the main determinant Delta$\Delta$ to 0$0$ to find alpha$\alpha$. Then quickly verify Delta_x neq 0$\Delta_x \neq 0$ to confirm it produces no solution. ### Step 1: Calculate Delta Delta = beginvmatrix 3 & 1 & 4 \\ 2 & alpha & -1 \\ 1 & 2 & 1 endvmatrix = 0$$\Delta = \begin{vmatrix} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0$$ Expand along the first row: 3(alpha + 2) - 1(2 - (-1)) + 4(4 - alpha) = 0$$3(\alpha + 2) - 1(2 - (-1)) + 4(4 - \alpha) = 0$$ 3alpha + 6 - 3 + 16 - 4alpha = 0$$3\alpha + 6 - 3 + 16 - 4\alpha = 0$$ 19 - alpha = 0 implies alpha = 19$$19 - \alpha = 0 \implies \alpha = 19$$ ### Step 2: Verification of Delta_x For alpha = 19$\alpha = 19$: Delta_x = beginvmatrix 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 endvmatrix$$\Delta_x = \begin{vmatrix} 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 \end{vmatrix}$$ Delta_x = 3(19 + 2) - 1(-3 + 4) + 4(-6 - 76)$$\Delta_x = 3(19 + 2) - 1(-3 + 4) + 4(-6 - 76)$$ Delta_x = 3(21) - 1(1) + 4(-82) = 63 - 1 - 328 neq 0$$\Delta_x = 3(21) - 1(1) + 4(-82) = 63 - 1 - 328 \neq 0$$ Since Delta = 0$\Delta = 0$ and Delta_x neq 0$\Delta_x \neq 0$, the system has no solution for alpha = 19$\alpha = 19$. ### Pattern Recognition Unless the problem specifically asks to check between infinite and no solution, setting the coefficient determinant Delta = 0$\Delta = 0$ directly yields the unique required parameter value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants