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Question-74328




Question Number 74328 by arthur.kangdani@gmail.com last updated on 22/Nov/19
Answered by ajfour last updated on 22/Nov/19
x=( determinant (((10),2,4),(4,(10),7),(9,5,(16)))_ / determinant ((6,2,4),(9,(10),7),((12),5,(16)))^ )  ⇒  x=((10(160−35)−2(64−63)+4(20−90))/(6(160−35)−2(144−84)+4(45−120)))       x=((1250−2−280)/(750−120−300)) = ((968)/(330)) = ((88)/(30))  .....
$${x}=\frac{\begin{vmatrix}{\mathrm{10}}&{\mathrm{2}}&{\mathrm{4}}\\{\mathrm{4}}&{\mathrm{10}}&{\mathrm{7}}\\{\mathrm{9}}&{\mathrm{5}}&{\mathrm{16}}\end{vmatrix}_{} }{\begin{vmatrix}{\mathrm{6}}&{\mathrm{2}}&{\mathrm{4}}\\{\mathrm{9}}&{\mathrm{10}}&{\mathrm{7}}\\{\mathrm{12}}&{\mathrm{5}}&{\mathrm{16}}\end{vmatrix}^{} } \\ $$$$\Rightarrow\:\:{x}=\frac{\mathrm{10}\left(\mathrm{160}−\mathrm{35}\right)−\mathrm{2}\left(\mathrm{64}−\mathrm{63}\right)+\mathrm{4}\left(\mathrm{20}−\mathrm{90}\right)}{\mathrm{6}\left(\mathrm{160}−\mathrm{35}\right)−\mathrm{2}\left(\mathrm{144}−\mathrm{84}\right)+\mathrm{4}\left(\mathrm{45}−\mathrm{120}\right)} \\ $$$$\:\:\:\:\:{x}=\frac{\mathrm{1250}−\mathrm{2}−\mathrm{280}}{\mathrm{750}−\mathrm{120}−\mathrm{300}}\:=\:\frac{\mathrm{968}}{\mathrm{330}}\:=\:\frac{\mathrm{88}}{\mathrm{30}} \\ $$$$….. \\ $$

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