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(D) For a system of linear equations $AX = B$ to have a unique solution,the determinant of the coefficient matrix $A$ must be non-zero,i.e.,$|A| 
eq

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$\alpha$ only

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(D) For a system of linear equations $AX = B$ to have a unique solution,the determinant of the coefficient matrix $A$ must be non-zero,i.e.,$|A| 
eq 0$.The coefficient matrix $A$ is given by:$A = \begin{bmatrix} 2 & 1 & 1 \ 10 & -1 & \alpha \ 4 & 3 & -1 \end{bmatrix}$Calculating the determinant $|A|$:$|A| = 2((-1)(-1) - (3)(\alpha)) - 1((10)(-1) - (4)(\alpha)) + 1((10)(3) - (4)(-1))$$|A| = 2(1 - 3\alpha) - 1(-10 - 4\alpha) + 1(30 + 4)$$|A| = 2 - 6\alpha + 10 + 4\alpha + 34$$|A| = 46 - 2\alpha$For a unique solution,we require $|A| 
eq 0$:$46 - 2\alpha 
eq 0$$2\alpha 
eq 46$$\alpha 
eq 23$Since the condition for a unique solution depends only on the value of $\alpha$ and is independent of $\beta$,the existence of the unique solution depends on $\alpha$ only.

The existence of the unique solution of the system of equations $2x + y + z = \beta$,$10x - y + \alpha z = 10$ and $4x + 3y - z = 6$ depends on

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