Question

$3x+4y+5z=18$, $2x-y+8z=13$, $5x-2y+7z=20$. Solve the above system of equations by Cramer's method.

Answer 1

Given equations are $3x+4y+5z=18,2x-y+8z=13,5x-2y+7z=20$

$D=\left|\begin{array}{ccc}3& 4& 5\\ 2& -1& 8\\ 5& -2& 7\end{array}\right|=3(-7+16)-4(14-40)+5(-4+5)$

$=3\left(9\right)-4(-26)+5\left(1\right)=27+104+5=136$

$D_1=\left|\begin{array}{ccc}18& 4& 5\\ 13& -1& 8\\ 20& -2& 7\end{array}\right|=18(-7+16)-4(91-160)+5(-26+20)$

$=18\left(9\right)-4(-69)+5(-6)=162+276-30=408$

$D_2=\left|\begin{array}{ccc}3& 18& 5\\ 2& 13& 8\\ 5& 20& 7\end{array}\right|=3(91-160)-18(14-40)+5(40-65)$

$=3(-69)-18(-26)+5(-25)=-207+468-125=136$

$D_3=\left|\begin{array}{ccc}3& 4& 18\\ 2& -1& 13\\ 5& -2& 20\end{array}\right|=3(-20+26)-4(40-65)+18(-4+5)$

$=3\left(6\right)-4(-25)+18\left(1\right)=18+100+18=136$

Now, $x=\frac{D_1}{D}=\frac{408}{136}=3$

$y=\frac{D_2}{D}=\frac{136}{136}=1$

$z=\frac{D_3}{D}=\frac{136}{136}=1$

Hence, the solution for the system is $(3,1,1)$.