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Question Number 72569 by otchereabdullai@gmail.com last updated on 30/Oct/19

2x−y+3z =1 4x+2y−z = −8 3x+y+2z = −1

$$\mathrm{2x}−\mathrm{y}+\mathrm{3z}\:=\mathrm{1} \\ $$$$\mathrm{4x}+\mathrm{2y}−\mathrm{z}\:=\:−\mathrm{8} \\ $$$$\mathrm{3x}+\mathrm{y}+\mathrm{2z}\:=\:−\mathrm{1} \\ $$

Answered by MJS last updated on 30/Oct/19

(2) z=4x+2y+8 (1) 14x+5y+23=0 (3) 11x+5y+17=0 (1)−(3) 3x+6=0 ⇒ x=−2 (1) 5y−5=0 ⇒ y=1 ⇒ z=2

$$\left(\mathrm{2}\right)\:\:{z}=\mathrm{4}{x}+\mathrm{2}{y}+\mathrm{8} \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{14}{x}+\mathrm{5}{y}+\mathrm{23}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:\mathrm{11}{x}+\mathrm{5}{y}+\mathrm{17}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{3}\right)\:\:\mathrm{3}{x}+\mathrm{6}=\mathrm{0}\:\Rightarrow\:{x}=−\mathrm{2} \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{5}{y}−\mathrm{5}=\mathrm{0}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$$\Rightarrow\:{z}=\mathrm{2} \\ $$

Commented by otchereabdullai@gmail.com last updated on 30/Oct/19

thanks prof mjs

$$\mathrm{thanks}\:\mathrm{prof}\:\mathrm{mjs} \\ $$

Answered by Tanmay chaudhury last updated on 30/Oct/19

△=∣2 −1 3∣ ∣4 2 −1∣ ∣3 1 2∣=2(4+1)+1(8+3)+3(4−6) △=15 x=(1/(15))× ∣1 −1 3∣ ∣−8 2 −1∣ ∣−1 1 2∣ x=(1/(15))×{1(4+1)+1(−16−1)+3(−8+2)} =(1/(15))×{5−17−18}=−2 y=(1/(15))×∣ 2 1 3 ∣ ∣4 −8 −1∣=2(−16−1)−1(8+3)+3(−4+24) ∣3 −1 2 ∣=−34−11+60=15 so y=((15)/(15))=1 2x−y+3z=1 2×−2−1+3z=1 3z=1+5→z=2 x=−2 y=1 z=2

$$\bigtriangleup=\mid\mathrm{2}\:−\mathrm{1}\:\:\:\:\:\mathrm{3}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\mid\mathrm{4}\:\:\:\:\mathrm{2}\:\:\:−\mathrm{1}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\mid\mathrm{3}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{2}\mid=\mathrm{2}\left(\mathrm{4}+\mathrm{1}\right)+\mathrm{1}\left(\mathrm{8}+\mathrm{3}\right)+\mathrm{3}\left(\mathrm{4}−\mathrm{6}\right) \\ $$$$\:\:\:\bigtriangleup=\mathrm{15} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{15}}×\:\:\mid\mathrm{1}\:\:\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{3}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid−\mathrm{8}\:\:\:\:\:\mathrm{2}\:\:\:\:\:−\mathrm{1}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid−\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}\mid \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{15}}×\left\{\mathrm{1}\left(\mathrm{4}+\mathrm{1}\right)+\mathrm{1}\left(−\mathrm{16}−\mathrm{1}\right)+\mathrm{3}\left(−\mathrm{8}+\mathrm{2}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{15}}×\left\{\mathrm{5}−\mathrm{17}−\mathrm{18}\right\}=−\mathrm{2} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{15}}×\mid\:\mathrm{2}\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:\:\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{4}\:\:\:\:−\mathrm{8}\:\:\:−\mathrm{1}\mid=\mathrm{2}\left(−\mathrm{16}−\mathrm{1}\right)−\mathrm{1}\left(\mathrm{8}+\mathrm{3}\right)+\mathrm{3}\left(−\mathrm{4}+\mathrm{24}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{3}\:\:\:\:−\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\:\mid=−\mathrm{34}−\mathrm{11}+\mathrm{60}=\mathrm{15} \\ $$$${so}\:{y}=\frac{\mathrm{15}}{\mathrm{15}}=\mathrm{1} \\ $$$$\mathrm{2}{x}−{y}+\mathrm{3}{z}=\mathrm{1} \\ $$$$\mathrm{2}×−\mathrm{2}−\mathrm{1}+\mathrm{3}{z}=\mathrm{1} \\ $$$$\mathrm{3}{z}=\mathrm{1}+\mathrm{5}\rightarrow{z}=\mathrm{2} \\ $$$$\boldsymbol{{x}}=−\mathrm{2}\:\:\:\boldsymbol{{y}}=\mathrm{1}\:\:\:\boldsymbol{{z}}=\mathrm{2} \\ $$

Commented by otchereabdullai@gmail.com last updated on 30/Oct/19

thanks prof tanmay

$$\mathrm{thanks}\:\mathrm{prof}\:\mathrm{tanmay} \\ $$

Answered by malwaan last updated on 30/Oct/19

(1)+(3) 5x + 5z = 0 ⇒x = −z ; z = −x (2) 5x+2y = −8→∗ (3) x + y = −1 ⇒ 2x+2y=−2 (∗) 3x = −6 ∴ x = −2 ; z=2 ; y=2−1=1

$$\left(\mathrm{1}\right)+\left(\mathrm{3}\right)\:\mathrm{5}\boldsymbol{{x}}\:+\:\mathrm{5}\boldsymbol{{z}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\boldsymbol{{x}}\:=\:−\boldsymbol{{z}}\:;\:\boldsymbol{{z}}\:=\:−\boldsymbol{{x}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{5}\boldsymbol{{x}}+\mathrm{2}\boldsymbol{{y}}\:=\:−\mathrm{8}\rightarrow\ast \\ $$$$\left(\mathrm{3}\right)\:\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:=\:−\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{2}\boldsymbol{{x}}+\mathrm{2}\boldsymbol{{y}}=−\mathrm{2} \\ $$$$\left(\ast\right)\:\mathrm{3}\boldsymbol{{x}}\:=\:−\mathrm{6}\:\therefore\:\boldsymbol{{x}}\:=\:−\mathrm{2}\:;\:\boldsymbol{{z}}=\mathrm{2} \\ $$$$;\:\boldsymbol{{y}}=\mathrm{2}−\mathrm{1}=\mathrm{1} \\ $$

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