If the system of equations 2x + py + 6z = 8,x | vedclass

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(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the augmented mat

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$2$

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(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the augmented matrix must be consistent.Given equations:$2x + py + 6z = 8$$x + 2y + qz = 5$$x + y + 3z = 4$First,find the determinant $D$ of the coefficient matrix:$D = \begin{vmatrix} 2 & p & 6 \ 1 & 2 & q \ 1 & 1 & 3 \end{vmatrix} = 2(6 - q) - p(3 - q) + 6(1 - 2) = 12 - 2q - 3p + pq - 6 = pq - 3p - 2q + 6 = (p - 2)(q - 3) = 0$.This implies $p = 2$ or $q = 3$.If $p = 2$,the equations become:$2x + 2y + 6z = 8 \implies x + y + 3z = 4$$x + 2y + qz = 5$$x + y + 3z = 4$Since the first and third equations are identical,we have two independent equations for three variables,which leads to infinitely many solutions for any $q$.Checking the options,$p = 2$ is given as option $B$.

If the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$,and $x + y + 3z = 4$ has infinitely many solutions,then $p=$

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