Solution:
Cramer's rule: Δ = ∣ 3 3 − 1 2 − 1 2 4 3 2 ∣ = 3 ( − 2 − 6 ) − 3 ( 4 − 8 ) − 1 ( 6 + 4 ) = 3 ( − 8 ) − 3 ( − 4 ) − 1 ( 10 ) = − 24 + 12 − 10 = − 22 ≠ 0 Δ X = ∣ 11 3 − 1 9 − 1 2 25 3 2 ∣ = 11 ( − 2 − 6 ) − 3 ( 18 − 50 ) − 1 ( 27 + 25 ) = 11 ( − 8 ) − 3 ( 32 ) − 1 ( 52 ) = − 88 + 96 − 52 = − 44 Δ y = ∣ 3 11 − 1 2 9 2 4 25 2 ∣ \begin{aligned}
&\Delta=\left|\begin{array}{ccc}
3 & 3 & -1 \\
2 & -1 & 2 \\
4 & 3 & 2
\end{array}\right| \\
&=3(-2-6)-3(4-8)-1(6+4) \\
&=3(-8)-3(-4)-1(10) \\
&=-24+12-10 \\
&=-22 \neq 0 \\
&\Delta_{\mathrm{X}}=\left|\begin{array}{ccc}
11 & 3 & -1 \\
9 & -1 & 2 \\
25 & 3 & 2
\end{array}\right| \\
&=11(-2-6)-3(18-50)-1(27+25) \\
&=11(-8)-3(32)-1(52) \\
&=-88+96-52 \\
&=-44 \\
\Delta_y=|&\begin{array}{lll}
3 & 11 & -1 \\
2 & 9 & 2 \\
4 & 25 & 2
\end{array} \mid
\end{aligned} Δ y = ∣ Δ = ∣ ∣ 3 2 4 3 − 1 3 − 1 2 2 ∣ ∣ = 3 ( − 2 − 6 ) − 3 ( 4 − 8 ) − 1 ( 6 + 4 ) = 3 ( − 8 ) − 3 ( − 4 ) − 1 ( 10 ) = − 24 + 12 − 10 = − 22 = 0 Δ X = ∣ ∣ 11 9 25 3 − 1 3 − 1 2 2 ∣ ∣ = 11 ( − 2 − 6 ) − 3 ( 18 − 50 ) − 1 ( 27 + 25 ) = 11 ( − 8 ) − 3 ( 32 ) − 1 ( 52 ) = − 88 + 96 − 52 = − 44 3 2 4 11 9 25 − 1 2 2 ∣
= 3 ( 18 − 50 ) − 11 ( 4 − 8 ) − 1 ( 50 − 36 ) = 3 ( 32 ) − 11 ( 4 ) − 1 ( 14 ) = − 96 + 44 − 14 = − 66 Δ z = ∣ 3 3 11 2 − 1 9 4 3 25 ∣ = 3 ( − 25 − 27 ) − 3 ( 50 − 36 ) + 11 ( 6 + 4 ) = 3 ( − 52 ) − 3 ( 14 ) + 11 ( 10 ) = − 156 − 42 + 110 = − 88 By Cramer’s rule x = Δ x Δ = − 44 − 22 = 2 y = Δ y Δ = − 66 − 22 = 3 = − 88 Δ = − 22 − 22 = 4 ∴ x = 2 , y = 3 , z = 4 \begin{aligned}
&=3(18-50)-11(4-8)-1(50-36) \\
&=3(32)-11(4)-1(14) \\
&=-96+44-14 \\
&=-66 \\
&\begin{aligned}
\Delta_{\mathrm{z}} &=\left|\begin{array}{lll}
3 & 3 & 11 \\
2 & -1 & 9 \\
4 & 3 & 25
\end{array}\right| \\
&=3(-25-27)-3(50-36)+11(6+4) \\
&=3(-52)-3(14)+11(10) \\
&=-156-42+110 \\
&=-88 \\
& \text { By Cramer's rule } \mathrm{x}=\frac{\Delta_{x}}{\Delta}=\frac{-44}{-22}=2 \\
& \quad \mathrm{y}=\frac{\Delta_{y}}{\Delta}=\frac{-66}{-22}=3 \\
&=\frac{-88}{\Delta}=\frac{-22}{-22}=4
\end{aligned} \\
&\therefore x=2,y=3, \mathrm{z}=4
\end{aligned} = 3 ( 18 − 50 ) − 11 ( 4 − 8 ) − 1 ( 50 − 36 ) = 3 ( 32 ) − 11 ( 4 ) − 1 ( 14 ) = − 96 + 44 − 14 = − 66 Δ z = ∣ ∣ 3 2 4 3 − 1 3 11 9 25 ∣ ∣ = 3 ( − 25 − 27 ) − 3 ( 50 − 36 ) + 11 ( 6 + 4 ) = 3 ( − 52 ) − 3 ( 14 ) + 11 ( 10 ) = − 156 − 42 + 110 = − 88 By Cramer’s rule x = Δ Δ x = − 22 − 44 = 2 y = Δ Δ y = − 22 − 66 = 3 = Δ − 88 = − 22 − 22 = 4 ∴ x = 2 , y = 3 , z = 4
2. Graphical method:
We plot all the 3 equations in a 3d-system, where they intersect, that is our solution.
We find that they intersect at (2,3,4).
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