Question

Q.1 Solve using Cramer's rule: 6x+y-
3z=5; x+3y-2z=5; 2x+y+4z=8.​

Answer 1

$\textbf{Given:}$

$6x+y-3z=5$

$x+3y-2z=5$

$2x+y+4z=8$

$\textbf{To find:}$

$\text{Solution of the given system of equations}$

$\textbf{Solution:}$

$\text{The given system of equations can be written as}$

​

$\left(\begin{array}{ccc}6&1&-3\\1&3&-2\\2&1&4\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}5\\5\\8\end{array}\right)$

$\Delta=\left|\begin{array}{ccc}6&1&-3\\1&3&-2\\2&1&4\end{array}\right|$

$\Delta=6(12+2)-1(4+4)-3(1-6)$

$\Delta=84-8+15=91$

$\Delta_x=\left|\begin{array}{ccc}5&1&-3\\5&3&-2\\8&1&4\end{array}\right|$

$\Delta_x=5(12+2)-1(20+16)-3(5-24)$

$\Delta_x=70-36+57=91$

$\Delta_y=\left|\begin{array}{ccc}6&5&-3\\1&5&-2\\2&8&4\end{array}\right|$

$\Delta_y=6(20+16)-5(4+4)-3(8-16)$

$\Delta_y=216-40+6=182$

$\Delta_z=\left|\begin{array}{ccc}6&1&5\\1&3&5\\2&1&8\end{array}\right|$

$\Delta_z=6(24-5)-1(8-10)+5(1-6)$

$\Delta_z=104+2-25=91$

$\text{By Cramer's rule}$

$x=\dfrac{\Delta_x}{\Delta}$

$x=\dfrac{91}{91}=1$

$y=\dfrac{\Delta_y}{\Delta}$

$y=\dfrac{182}{91}=2$

$z=\dfrac{\Delta_z}{\Delta}$

$z=\dfrac{91}{91}=1$

$\therefore\textbf{The solution is x=1, y=2 and z=1}$

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x + y-Z=1; 8x + 3y - 6Z=1; 4x - y +3Z =1​

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