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(A) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix $D$ must be non-zero $(D 
eq 0)$.The coeffic

Vedclass · Accepted Answer

$\mu$ only

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(A) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix $D$ must be non-zero $(D 
eq 0)$.The coefficient matrix is given by:$D = \begin{vmatrix} 1 & 1 & 1 \ 5 & -1 & \mu \ 2 & 3 & -1 \end{vmatrix}$Expanding the determinant along the first row:$D = 1((-1)(-1) - (3)(\mu)) - 1((5)(-1) - (2)(\mu)) + 1((5)(3) - (2)(-1))$$D = 1(1 - 3\mu) - 1(-5 - 2\mu) + 1(15 + 2)$$D = 1 - 3\mu + 5 + 2\mu + 17$$D = 23 - \mu$For a unique solution,$D 
eq 0$,which implies $23 - \mu 
eq 0$,or $\mu 
eq 23$.Since the condition $D 
eq 0$ depends only on the value of $\mu$ and is independent of $\lambda$,the existence of a unique solution depends on $\mu$ only.

The existence of the unique solution of the system $x + y + z = \lambda$,$5x - y + \mu z = 10$,and $2x + 3y - z = 6$ depends on

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