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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

Consider the system of equations:
$x - 2y + 3z = -1$; $-x + y - 2z = k$; $x - 3y + 4z = 1$
$STATEMENT-1$: The system of equations has no solution for $k \neq 3$.
$STATEMENT-2$: The determinant $\left|\begin{array}{ccc}1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4\end{array}\right| = 0$.

If $AX=D$ represents the system of linear equations $3x-4y+7z+6=0$, $5x+2y-4z+9=0$ and $8x-6y-z+5=0$, then

If a system of three linear equations in three unknowns,which is in the matrix equation form of $AX = D$,is inconsistent,then $\frac{\text{rank of } A}{\text{rank of } AD}$ is

The system of equations $3x + 2y + z = 6$, $3x + 4y + 3z = 14$ and $6x + 10y + 8z = a$ has an infinite number of solutions if $a$ is equal to

Examine the consistency of the system of equations: $x+2y=2$ and $2x+3y=3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module