Question

2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11

Answer 1

Given: 2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11

$$\begin{gathered} D=\left|\begin{array}{ccc}2& -3& -4\\ -2& 5& -1\\ 3& -1& 5\end{array}\right| \\ =2(25-1)+3(-10+3)-4(2-15) \\ =2\left(24\right)+3(-7)-4(-13) \\ =79 \\ {D}_1=\left|\begin{array}{ccc}29& -3& -4\\ -15& 5& -1\\ -11& -1& 5\end{array}\right| \\ =29(25-1)+3(-75-11)-4(15+55) \\ =29\left(24\right)+3(-86)-4\left(70\right) \\ =158 \\ {D}_2=\left|\begin{array}{ccc}2& 29& -4\\ -2& -15& -1\\ 3& -11& 5\end{array}\right| \\ =2(-75-11)-29(-10+3)-4(22+45) \\ =2(-86)-29(-7)-4\left(67\right) \\ =-237 \\ {D}_3=\left|\begin{array}{ccc}2& -3& 29\\ -2& 5& -15\\ 3& -1& -11\end{array}\right| \\ =2(-55-15)+3(22+45)+29(2-15) \\ =2(-70)+3\left(67\right)+29(-13) \\ =-316 \\ \mathrm{Now}, \\ x=\frac{D_1}{D}=\frac{158}{79}=2 \\ y=\frac{D_2}{D}=\frac{-237}{79}=-3 \\ z=\frac{D_3}{D}=\frac{-316}{79}=-4 \\ \therefore x=2,y=-3\mathrm{and}z=-4 \end{gathered}$$