Question

For what values of p and q, the system of equations
$2x+py+6z=8$
$x+2y+qz=5$
$x+y+3z=4$ has a unique solution

• A. $p\ne 2,q\ne 3$
• B. $p\ne 3,q\ne 2$
• C. $p=2,q=3$
• D. $p=3,q=2$

Answer 1

The correct option is A $p\ne 2,q\ne 3$

Given equations are
$2x+py+6z=8$
$x+2y+qz=5$
$x+y+3z=4$
Here $D=\left|\begin{array}{ccc}2& p& 6\\ 1& 2& q\\ 1& 1& 3\end{array}\right|$
Applying $R_1\to R_1-2R_3$
$=\left|\begin{array}{ccc}0& p-2& 0\\ 1& 2& q\\ 1& 1& 3\end{array}\right|$
$=-(p-2)\left|\begin{array}{cc}1& q\\ 1& 3\end{array}\right|$
$=-(p-2)(3-q)$
$=(p-2)(q-3)$
$\therefore D_1=\left|\begin{array}{ccc}8& p& 6\\ 5& 2& q\\ 4& 1& 3\end{array}\right|$
Applying $C_1\to C_1-4C_2$ & $C_3\to C_3-3C_2$
$D_1=\left|\begin{array}{ccc}8-4p& p& 6-3p\\ -3& 2& q-6\\ 0& 1& 0\end{array}\right|$
Expanding along $R_3$, then
$D_1=(-1)\left|\begin{array}{cc}8-4p& 6-3p\\ -3& q-6\end{array}\right|$
$=(-1)\left\{\right(8-4p\left)\right(q-6)+3(6-3p\left)\right\}$
$=4(p-2)(q-6)+3(p-2)3$
$=(p-2)(4q-15)$
and $D_2=\left|\begin{array}{ccc}2& 8& 6\\ 1& 5& q\\ 1& 4& 3\end{array}\right|$
Applying $R_1\to -2R_3$
$=\left|\begin{array}{ccc}0& 0& 0\\ 1& 5& q\\ 1& 4& 3\end{array}\right|$
$=0$
and $D_3=\left|\begin{array}{ccc}2& p& 8\\ 1& 2& 5\\ 1& 1& 4\end{array}\right|$
Applying $R_1\to R_1-2R_3$
$\therefore D_3=\left|\begin{array}{ccc}0& p-2& 0\\ 1& 2& 5\\ 1& 1& 4\end{array}\right|$
Expanding along $R_1$ then
$D_3=-(p-2)\left|\begin{array}{cc}1& 5\\ 1& 4\end{array}\right|$
$D_3=(2-p)$
By Cramer's rule :
$x=\frac{D_1}{D},y=\frac{D_2}{D},z=\frac{D_3}{D}$
For unique solutions:
$D\ne 0$ and at least one of $D_1,D_2,D_3\ne 0$
$\Delta \ne 0,p\ne 2,q\ne 3$