Search any question & find its solution
Question:
Answered & Verified by Expert
If the unique solution of the simultaneous linear equations $3 x-2 y+z=5 k, 2 x+3 y-2 z=-5 k, x+4 y+3 z=k$ is $x=$ $\alpha, y=\beta, z=3$, then $k=$
Options:
Solution:
2861 Upvotes
Verified Answer
The correct answer is:
$2$
$3 x-2 y+z=5 K$
$\begin{aligned} & 2 x+3 y-2 z=-5 K \\ & x+4 y+3 z=K \\ & D=\left|\begin{array}{ccc}3 & -2 & 1 \\ 2 & 3 & -2 \\ 1 & 4 & 3\end{array}\right|=63+16+5=84 \\ & D_1=\left|\begin{array}{ccc}5 K & -2 & 1 \\ -5 K & 3 & -2 \\ K & 4 & 3\end{array}\right|=K\left|\begin{array}{ccc}5 & -2 & 1 \\ -5 & 3 & -2 \\ 1 & 4 & 3\end{array}\right| \\ & =K(85+(-26)-23)=36 K \\ & D_2=\left|\begin{array}{ccc}3 & 5 K & 1 \\ 2 & -5 K & -2 \\ 1 & K & 3\end{array}\right|=K\left|\begin{array}{ccc}3 & 5 & 1 \\ 2 & -5 & -2 \\ 1 & 1 & 3\end{array}\right| \\ & =K(-39-40+7)=-72 K \\ & D_3=\left|\begin{array}{ccc}3 & -2 & 5 K \\ 2 & 3 & -5 K \\ 1 & 4 & K\end{array}\right|=K\left|\begin{array}{ccc}3 & -2 & 5 \\ 2 & 3 & -5 \\ 1 & 4 & 1\end{array}\right| \\ & =K(69+14+25)=108 K \\ & x=\frac{D_1}{D}=\frac{36 K}{84} ; y=\frac{D_2}{D}=\frac{-72 K}{84} \\ & z=\frac{D_3}{\bar{D}}=\frac{108 K}{72} \Rightarrow \frac{108 K}{72}=3 \\ & {[\because z=3]} \\ & \Rightarrow \quad K=\frac{72 \times 3}{108}=2 \\ & \end{aligned}$
$\begin{aligned} & 2 x+3 y-2 z=-5 K \\ & x+4 y+3 z=K \\ & D=\left|\begin{array}{ccc}3 & -2 & 1 \\ 2 & 3 & -2 \\ 1 & 4 & 3\end{array}\right|=63+16+5=84 \\ & D_1=\left|\begin{array}{ccc}5 K & -2 & 1 \\ -5 K & 3 & -2 \\ K & 4 & 3\end{array}\right|=K\left|\begin{array}{ccc}5 & -2 & 1 \\ -5 & 3 & -2 \\ 1 & 4 & 3\end{array}\right| \\ & =K(85+(-26)-23)=36 K \\ & D_2=\left|\begin{array}{ccc}3 & 5 K & 1 \\ 2 & -5 K & -2 \\ 1 & K & 3\end{array}\right|=K\left|\begin{array}{ccc}3 & 5 & 1 \\ 2 & -5 & -2 \\ 1 & 1 & 3\end{array}\right| \\ & =K(-39-40+7)=-72 K \\ & D_3=\left|\begin{array}{ccc}3 & -2 & 5 K \\ 2 & 3 & -5 K \\ 1 & 4 & K\end{array}\right|=K\left|\begin{array}{ccc}3 & -2 & 5 \\ 2 & 3 & -5 \\ 1 & 4 & 1\end{array}\right| \\ & =K(69+14+25)=108 K \\ & x=\frac{D_1}{D}=\frac{36 K}{84} ; y=\frac{D_2}{D}=\frac{-72 K}{84} \\ & z=\frac{D_3}{\bar{D}}=\frac{108 K}{72} \Rightarrow \frac{108 K}{72}=3 \\ & {[\because z=3]} \\ & \Rightarrow \quad K=\frac{72 \times 3}{108}=2 \\ & \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.