Question

By using Cramer's rule, solve
$3x+4y+5z=18,2x-y+8z=13,5x-2y+7z=20$

Answer 1

Consider the system of equations

$3x+4y+5z=18$

$2x-y+8z=13$

$5x-2y+7z=20$

Cramer's rule is applicable when determinant$\ne 0$

$D=\left|\begin{array}{ccc}3& 4& 5\\ 2& -1& 8\\ 5& -2& 7\end{array}\right|$

$=3(-7+16)-4(14-40)+5(-4+5)$

$=27+104+5=136\ne 0$

$\therefore$ Cramer's rule is aplicable.

$D_x=\left|\begin{array}{ccc}18& 4& 5\\ 13& -1& 8\\ 20& -2& 7\end{array}\right|$ by replacing the coefficents of $C_1$ by the Constants.

$=18(-7+16)-4(91-160)+5(-26+20)$

$=81+276-30=321$

$D_y=\left|\begin{array}{ccc}3& 18& 5\\ 2& 13& 8\\ 5& 20& 7\end{array}\right|$by replacing the coefficents of $C_2$ by the Constants.

$=3(91-160)-18(35-40)+5(40-65)$

$=-207+90-125=-242$

$D_z=\left|\begin{array}{ccc}3& 4& 18\\ 2& -1& 13\\ 5& -2& 20\end{array}\right|$by replacing the coefficents of $C_3$ by the Constants.

$=3(-20+26)-4(40-65)+8(-10+5)$

$=18+100-40=78$

The values of $x,y,z$ are calculated as

$x=\frac{D_x}{D}=\frac{321}{136}$

$y=\frac{D_y}{D}=\frac{-242}{136}=\frac{-121}{68}$

$z=\frac{D_z}{D}=\frac{78}{136}=\frac{39}{68}$