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Question Number 204041 by hardmath last updated on 04/Feb/24
Find:    determinant ((1,7,(−1)),(9,(−3),5),((−1),5,3))= ?
$$\mathrm{Find}:\:\:\:\begin{vmatrix}{\mathrm{1}}&{\mathrm{7}}&{−\mathrm{1}}\\{\mathrm{9}}&{−\mathrm{3}}&{\mathrm{5}}\\{−\mathrm{1}}&{\mathrm{5}}&{\mathrm{3}}\end{vmatrix}=\:? \\ $$
Answered by AST last updated on 04/Feb/24
=1 determinant (((−3),5),(5,3))−7 determinant ((9,5),((−1),3))−1 determinant ((9,(−3)),((−1),5))  =−9−25−7(27+5)−1(45−3)=−300
$$=\mathrm{1}\begin{vmatrix}{−\mathrm{3}}&{\mathrm{5}}\\{\mathrm{5}}&{\mathrm{3}}\end{vmatrix}−\mathrm{7}\begin{vmatrix}{\mathrm{9}}&{\mathrm{5}}\\{−\mathrm{1}}&{\mathrm{3}}\end{vmatrix}−\mathrm{1}\begin{vmatrix}{\mathrm{9}}&{−\mathrm{3}}\\{−\mathrm{1}}&{\mathrm{5}}\end{vmatrix} \\ $$$$=−\mathrm{9}−\mathrm{25}−\mathrm{7}\left(\mathrm{27}+\mathrm{5}\right)−\mathrm{1}\left(\mathrm{45}−\mathrm{3}\right)=−\mathrm{300} \\ $$
Answered by som(math1967) last updated on 05/Feb/24
C_2 →C_2 −7C_1   C_3 →C_3 +C_1    determinant ((1,0,0),(9,(−66),(14)),((−1),(12),2))  =−66×(2)−12×14  =−132−168=−300
$${C}_{\mathrm{2}} \rightarrow{C}_{\mathrm{2}} −\mathrm{7}{C}_{\mathrm{1}} \\ $$$${C}_{\mathrm{3}} \rightarrow{C}_{\mathrm{3}} +{C}_{\mathrm{1}} \\ $$$$\begin{vmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{9}}&{−\mathrm{66}}&{\mathrm{14}}\\{−\mathrm{1}}&{\mathrm{12}}&{\mathrm{2}}\end{vmatrix} \\ $$$$=−\mathrm{66}×\left(\mathrm{2}\right)−\mathrm{12}×\mathrm{14} \\ $$$$=−\mathrm{132}−\mathrm{168}=−\mathrm{300} \\ $$

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