1.use Cramer's rule to solve the equation below (10 marks)
y-z =2
3x+2y+z=4
5x+4y=1
Δ=∣01−1321540∣=0+(−1)3(−5)−(−1)4(12−10)=3\Delta=\begin{vmatrix} 0& 1&-1 \\ 3& 2&1\\ 5&4&0 \end{vmatrix}=0+(-1)^3(-5)-(-1)^4(12-10)=3Δ=∣∣035124−110∣∣=0+(−1)3(−5)−(−1)4(12−10)=3
Δx=∣21−1421140∣=(−1)22(−4)+(−1)3(−1)+(−1)4(−1)(16−2)=−8+1−14=−21\Delta_{x}=\begin{vmatrix} 2& 1&-1 \\ 4& 2&1\\ 1&4&0 \end{vmatrix}=(-1)^22(-4)+(-1)^3(-1)+(-1)^4(-1)(16-2)=-8+1-14=-21Δx=∣∣241124−110∣∣=(−1)22(−4)+(−1)3(−1)+(−1)4(−1)(16−2)=−8+1−14=−21
Δy=∣02−1341510∣=(−1)32(−5)+(−1)4(−1)(3−20)=27\Delta_{y}=\begin{vmatrix} 0& 2&-1 \\ 3& 4&1\\ 5&1&0 \end{vmatrix}=(-1)^32(-5)+(-1)^4(-1)(3-20)=27Δy=∣∣035241−110∣∣=(−1)32(−5)+(−1)4(−1)(3−20)=27
Δz=∣012324541∣=(−1)3(3−20)+(−1)42(12−10)=21\Delta_{z}=\begin{vmatrix} 0& 1&2 \\ 3& 2&4\\ 5&4&1 \end{vmatrix}=(-1)^3(3-20)+(-1)^42(12-10)=21Δz=∣∣035124241∣∣=(−1)3(3−20)+(−1)42(12−10)=21
x=-21/3=-7
y=27/3=9
z=21/3=7
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