The system of equations x+y+z=5, x+2y+3z=9 an | vedclass

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(D) 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

Vedclass · Accepted Answer

$\lambda 
eq 5, \mu \in R$

Vedclass · Answer

(D) 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: $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & \lambda \end{bmatrix}$ The determinant $|A|$ is calculated as: $|A| = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2)$ $|A| = 2\lambda - 9 - \lambda + 3 + 1$ $|A| = \lambda - 5$ For a unique solution, we require $|A| 
eq 0$, which implies $\lambda - 5 
eq 0$, or $\lambda 
eq 5$. The value of $\mu$ does not affect the existence of a unique solution, so $\mu$ can be any real number $(\mu \in R)$. Therefore, the condition for a unique solution is $\lambda 
eq 5$ and $\mu \in R$.

The system of equations $x+y+z=5$, $x+2y+3z=9$ and $x+3y+\lambda z=\mu$ has a unique solution if

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