Question #297359

{F} Work Out


1. Solve by crammer's rule, the following equations.

3X + 3Y - Z = 11

2X - Y + 2Z = 9

4X + 3Y - 27 = 25

2. solve the above equations using graphical method. 


1
Expert's answer
2022-02-22T00:56:32-0500

Solution:

  1. Cramer's rule:

Δ=331212432=3(26)3(48)1(6+4)=3(8)3(4)1(10)=24+1210=220ΔX=11319122532=11(26)3(1850)1(27+25)=11(8)3(32)1(52)=88+9652=44Δy=31112924252\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}



=3(1850)11(48)1(5036)=3(32)11(4)1(14)=96+4414=66Δz=33112194325=3(2527)3(5036)+11(6+4)=3(52)3(14)+11(10)=15642+110=88 By Cramer’s rule x=ΔxΔ=4422=2y=ΔyΔ=6622=3=88Δ=2222=4x=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}

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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