Question

x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1

Answer 1

Given: x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1


$$\begin{gathered} D=\left|\begin{array}{ccc}1& -4& -1\\ 2& -5& 2\\ -3& 2& 1\end{array}\right| \\ =1\left(-5-4\right)-(-4)(2+6)+(-1)(4-15) \\ =1(-9)-(-4)\left(8\right)+(-1)(-11)=34 \\ {D}_1=\left|\begin{array}{ccc}11& -4& -1\\ 39& -5& 2\\ 1& 2& 1\end{array}\right| \\ =11(-5-4)-(-4)(39-2)+(-1)(78+5) \\ =11(-9)-(-4)\left(37\right)+(-1)\left(83\right)=-34 \\ {D}_2=\left|\begin{array}{ccc}1& 11& -1\\ 2& 39& 2\\ -3& 1& 1\end{array}\right| \\ =1(39-2)-11(2+6)+(-1)(2+117) \\ =1\left(37\right)-11\left(8\right)+(-1)\left(119\right)=-170 \\ {D}_3=\left|\begin{array}{ccc}1& -4& 11\\ 2& -5& 39\\ -3& 2& 1\end{array}\right| \\ =1(-5-78)-(-4)(2+117)+11(4-15) \\ =1(-83)-(-4)\left(119\right)+11(-11)=272 \\ \mathrm{Now}, \\ x=\frac{D_1}{D}=\frac{-34}{34}=-1 \\ y=\frac{D_2}{D}=\frac{-170}{34}=-5 \\ z=\frac{D_3}{D}=\frac{272}{34}=8 \\ \therefore \mathrm{x\; =\; -1,}y=\; -5\mathrm{and}\mathrm{z}=8 \end{gathered}$$