If the system of simultaneous linear equation | vedclass

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Question

(C) system of linear equations $AX=B$ has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero,i.e.,$|A| 
eq 0$

Vedclass · Accepted Answer

$\mu 
eq 23$ and $\lambda \in R$

Vedclass · Answer

(C) system of linear equations $AX=B$ has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero,i.e.,$|A| 
eq 0$.The coefficient matrix $A$ is given by:$A = \begin{bmatrix} 1 & 1 & 1 \ 5 & -1 & \mu \ 2 & 3 & -1 \end{bmatrix}$We calculate the determinant $|A|$:$|A| = 1((-1)(-1) - (3)(\mu)) - 1((5)(-1) - (2)(\mu)) + 1((5)(3) - (2)(-1))$$|A| = 1(1 - 3\mu) - 1(-5 - 2\mu) + 1(15 + 2)$$|A| = 1 - 3\mu + 5 + 2\mu + 17$$|A| = 23 - \mu$For a unique solution,we require $|A| 
eq 0$.$23 - \mu 
eq 0 \implies \mu 
eq 23$.Since the determinant is independent of $\lambda$,the condition for a unique solution depends only on $\mu$. Thus,$\mu 
eq 23$ and $\lambda$ can be any real number $(\lambda \in R)$.

If the system of simultaneous linear equations $x+y+z=\lambda$,$5x-y+\mu z=10$,and $2x+3y-z=6$ has a unique solution,then:

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