All the real values of p, q so that the syste | vedclass

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(C) The system of equations is given by: $2x + py + 6z = 8$ $x + 2y + qz = 5$ $x + y + 3z = 4$ For the system to have no solution, the determinant of

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$p 
eq 2, q=3$

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(C) The system of equations is given by: $2x + py + 6z = 8$ $x + 2y + qz = 5$ $x + y + 3z = 4$ For the system to have no solution, the determinant of the coefficient matrix $A$ must be zero, i.e.,$|A| = 0$. $|A| = \begin{vmatrix} 2 & p & 6 \ 1 & 2 & q \ 1 & 1 & 3 \end{vmatrix} = 0$ Expanding along the first row: $2(6 - q) - p(3 - q) + 6(1 - 2) = 0$ $12 - 2q - 3p + pq - 6 = 0$ $pq - 3p - 2q + 6 = 0$ $p(q - 3) - 2(q - 3) = 0$ $(p - 2)(q - 3) = 0$ This implies $p = 2$ or $q = 3$. Case $1$: If $p = 2$, the equations become: $2x + 2y + 6z = 8 \Rightarrow x + y + 3z = 4$ $x + 2y + qz = 5$ $x + y + 3z = 4$ Here, the first and third equations are identical. For the system to have no solution, the planes must be parallel or inconsistent. If $p=2$, the system reduces to two equations $x+y+3z=4$ and $x+2y+qz=5$. This system will have infinitely many solutions for any $q$, so $p=2$ does not lead to 'no solution'. Case $2$: If $q = 3$ and $p 
eq 2$, the equations are: $2x + py + 6z = 8$ $x + 2y + 3z = 5$ $x + y + 3z = 4$ Subtracting the third from the second: $y = 1$. Substituting $y=1$ into the third: $x + 3z = 3$. Substituting $y=1$ into the first: $2x + p + 6z = 8 \Rightarrow 2x + 6z = 8 - p$. Since $x + 3z = 3$, then $2x + 6z = 6$. For no solution, $6 
eq 8 - p$, which means $p 
eq 2$. Thus, the condition for no solution is $p 
eq 2$ and $q = 3$.

All the real values of $p, q$ so that the system of equations $\begin{cases} 2x + py + 6z = 8 \\ x + 2y + qz = 5 \\ x + y + 3z = 4 \end{cases}$ may have no solution are

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