The existence of a unique solution for the sy | vedclass

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(A) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix,denoted by $\Delta$,must be non-zero $(\Delta

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$\alpha$ only

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(A) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix,denoted by $\Delta$,must be non-zero $(\Delta 
eq 0)$.The coefficient matrix is:$A = \begin{bmatrix} 1 & 1 & 1 \ 5 & -1 & \alpha \ 2 & 3 & -1 \end{bmatrix}$Calculating the determinant $\Delta$:$\Delta = 1((-1)(-1) - (3)(\alpha)) - 1((5)(-1) - (2)(\alpha)) + 1((5)(3) - (2)(-1))$$\Delta = 1(1 - 3\alpha) - 1(-5 - 2\alpha) + 1(15 + 2)$$\Delta = 1 - 3\alpha + 5 + 2\alpha + 17$$\Delta = 23 - \alpha$For a unique solution,$\Delta 
eq 0$,which implies $23 - \alpha 
eq 0$,or $\alpha 
eq 23$.Since the condition depends only on $\alpha$ and is independent of $\beta$,the existence of a unique solution depends on $\alpha$ only.

The existence of a unique solution for the system of equations $x+y+z=\beta$,$5x-y+\alpha z=10$,and $2x+3y-z=6$ depends on:

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