ক্রেমারের নিয়ম
Solve, x+2y−z=5,3x−y+3z=7,2x+3y+z=11x + 2y -z= 5,3x - y + 3z = 7, 2x + 3y + z =11 x+2y−z=5,3x−y+3z=7,2x+3y+z=11
(−1,−2,−1) (-1,-2,-1 )(−1,−2,−1)
(−2,−9,9) (-2,-9,9 )(−2,−9,9)
(−1,2,1) (-1,2,1 )(−1,2,1)`
(2,2,1) (2,2,1 )(2,2,1)
Solve: দেওয়া আছে,
[12−13−13231][xyz]=[5711] \left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & -1 & 3 \\ 2 & 3 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}5 \\ 7 \\ 11\end{array}\right] 1322−13−131xyz=5711
⇒[x+2y−z3x−y+3z2x+3y+z]=[5711]∴x+2y−z=53x−y+3z=72x+3y+z=11 \begin{aligned} \Rightarrow & {\left[\begin{array}{c} x+2 y-z \\ 3 x-y+3 z \\ 2 x+3 y+z \end{array}\right]=\left[\begin{array}{c} 5 \\ 7 \\ 11 \end{array}\right] } \\ \therefore \quad & x+2 y-z=5 \\ & 3 x-y+3 z=7 \\ & 2 x+3 y+z=11 \end{aligned} ⇒∴x+2y−z3x−y+3z2x+3y+z=5711x+2y−z=53x−y+3z=72x+3y+z=11
এখখন, ক্রেমারের নিয়ম ব্যবহার করে আামরা পাই ,
D=∣12−13−13231∣=1(−1−9)−2(3−6)−1(9+2)=−10+6−11=−15Dx=∣52−17−131131∣=5(−1−9)−2(7−33)−1(21+11)=−50+52−32=−30Dy=∣15−13732111∣=1(7−33)−5(3−6)−1(33−14) \begin{array}{l} D=\left|\begin{array}{ccc} 1 & 2 & -1 \\ 3 & -1 & 3 \\ 2 & 3 & 1 \end{array}\right| \\ =1(-1-9)-2(3-6)-1(9+2) \\ =-10+6-11=-15 \\ D_{x}=\left|\begin{array}{ccc} 5 & 2 & -1 \\ 7 & -1 & 3 \\ 11 & 3 & 1 \end{array}\right| \\ =5(-1-9)-2(7-33)-1(21+11) \\ =-50+52-32=-30 \\ D_{y}=\left|\begin{array}{lll} 1 & 5 & -1 \\ 3 & 7 & 3 \\ 2 & 11 & 1 \end{array}\right| \\ =1(7-33)-5(3-6)-1(33-14) \end{array} D=1322−13−131=1(−1−9)−2(3−6)−1(9+2)=−10+6−11=−15Dx=57112−13−131=5(−1−9)−2(7−33)−1(21+11)=−50+52−32=−30Dy=1325711−131=1(7−33)−5(3−6)−1(33−14)
=−26+15−19=−30Dz=∣1253−172311∣=1(−11−21)−2(33−14)+5(9+2)=−32−38+55=−15∴x=DxD=−30−15=2,y=DyD=−30−15=2z=DzD=−15−15=1 \begin{aligned} & =-26+15-19=-30 \\ & D_{z}=\left|\begin{array}{ccc} 1 & 2 & 5 \\ 3 & -1 & 7 \\ 2 & 3 & 11 \end{array}\right| \\ & =1(-11-21)-2(33-14)+5(9+2) \\ & =-32-38+55=-15 \\ \therefore \quad & x=\frac{D_{x}}{D}=\frac{-30}{-15}=2, y=\frac{D_{y}}{D}=\frac{-30}{-15}=2 \\ & z=\frac{D_{z}}{D}=\frac{-15}{-15}=1 \end{aligned} ∴=−26+15−19=−30Dz=1322−135711=1(−11−21)−2(33−14)+5(9+2)=−32−38+55=−15x=DDx=−15−30=2,y=DDy=−15−30=2z=DDz=−15−15=1
2x - 3y= 2, 2y - x= -1; সমীকরণ জোটের ম্যাট্রিক্স আকার কোনটি?
N=[1−22212−22−1]=X=[xyz]N=\left[\begin{array}{ccc}1 & -2 & 2 \\ 2 & 1 & 2 \\ -2 & 2 & -1\end{array}\right]=X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]N=12−2−21222−1=X=xyz এবং B=[354]B=\left[\begin{array}{l}3 \\ 5 \\ 4\end{array}\right]B=354
L=[213−13−1−12−5],M=[611],N=[xyz] এবং A=∣P2a2a2 b2Q2 b2c2c2R2∣ \mathrm{L}=\left[\begin{array}{rrr} 2 & 1 & 3 \\ -1 & 3 & -1 \\ -1 & 2 & -5 \end{array}\right], \mathrm{M}=\left[\begin{array}{l} 6 \\ 1 \\ 1 \end{array}\right], \mathrm{N}=\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ z \end{array}\right] \text { এবং } \mathrm{A}=\left|\begin{array}{lll} \mathrm{P}^{2} & \mathrm{a}^{2} & \mathrm{a}^{2} \\ \mathrm{~b}^{2} & \mathrm{Q}^{2} & \mathrm{~b}^{2} \\ \mathrm{c}^{2} & \mathrm{c}^{2} & \mathrm{R}^{2} \end{array}\right| L=2−1−11323−1−5,M=611,N=xyz এবং A=P2 b2c2a2Q2c2a2 b2R2
f(θ)=cosθf\left(\theta\right)=\cos\thetaf(θ)=cosθ
