The values of p and q so that the system of e | vedclass

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(A) 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 c

Vedclass · Accepted Answer

$p 
eq 2, q = 3$

Vedclass · Answer

(A) 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 $D$ must be $0$,and at least one of the Cramer's determinants $(D_x, D_y, D_z)$ must be non-zero.The coefficient matrix $A = \begin{bmatrix} 2 & p & 6 \ 1 & 2 & q \ 1 & 1 & 3 \end{bmatrix}$.$|A| = 2(6 - q) - p(3 - q) + 6(1 - 2) = 12 - 2q - 3p + pq - 6 = pq - 3p - 2q + 6 = (p - 2)(q - 3)$.For $|A| = 0$,we must have $p = 2$ 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$Here,the first and third equations are identical. For no solution,the second equation must be inconsistent with the first. Subtracting the first from the second: $(x + 2y + qz) - (x + y + 3z) = 5 - 4 \implies y + (q - 3)z = 1$.This is consistent with $x + y + 3z = 4$ unless the coefficients lead to a contradiction. If $q = 3$,the system becomes $x + y + 3z = 4$ and $x + 2y + 3z = 5$. Subtracting gives $y = 1$. Substituting $y=1$ into $x+y+3z=4$ gives $x+3z=3$. This has infinitely many solutions.If $p 
eq 2$ and $q = 3$,the determinant is $0$. Checking $D_x$:$D_x = \begin{vmatrix} 8 & p & 6 \ 5 & 2 & 3 \ 4 & 1 & 3 \end{vmatrix} = 8(6 - 3) - p(15 - 12) + 6(5 - 8) = 24 - 3p - 18 = 6 - 3p$.For no solution,$D_x 
eq 0 \implies 6 - 3p 
eq 0 \implies p 
eq 2$.Thus,for no solution,$p 
eq 2$ and $q = 3$.

The values of $p$ and $q$ so that the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$ and $x + y + 3z = 4$ may have no solution are

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