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Question Number 70385 by ®Ëƒ ¬Ë°¹¾¨ ‰¦Í¦¿¨ ¸Ë¹Ç² last updated on 04/Oct/19
The value of determinant   determinant (((x+2),( x+3),(x+5)),((x+4),( x+6),(x+9)),((x+8),(x+11),(x+15))) is
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{determinant} \\ $$$$\begin{vmatrix}{{x}+\mathrm{2}}&{\:{x}+\mathrm{3}}&{{x}+\mathrm{5}}\\{{x}+\mathrm{4}}&{\:{x}+\mathrm{6}}&{{x}+\mathrm{9}}\\{{x}+\mathrm{8}}&{{x}+\mathrm{11}}&{{x}+\mathrm{15}}\end{vmatrix}\:\mathrm{is} \\ $$
Answered by $@ty@m123 last updated on 04/Oct/19
= determinant (((−1),( −2),(x+5)),((−2),( −3),(x+9)),((−3),(−4),(x+15))) { ((by C_1 →C_1 −C_2 )),((& C_2 →C_2 −C_3 )) :}     = determinant ((1,( 1),(−4)),(1,( 1),(−6)),((−3),(−4),(x+15))) { ((by R_1 →R_1 −R_2 )),((& R_2 →R_2 −R_3 )) :}       = determinant ((0,( 0),2),(1,( 1),(−6)),((−3),(−4),(x+15))) by R_1 →R_1 −R_2    =2 determinant ((1,1),((−3),(−4)))  =2(−4+3)  =−2
$$=\begin{vmatrix}{−\mathrm{1}}&{\:−\mathrm{2}}&{{x}+\mathrm{5}}\\{−\mathrm{2}}&{\:−\mathrm{3}}&{{x}+\mathrm{9}}\\{−\mathrm{3}}&{−\mathrm{4}}&{{x}+\mathrm{15}}\end{vmatrix}\begin{cases}{{by}\:{C}_{\mathrm{1}} \rightarrow{C}_{\mathrm{1}} −{C}_{\mathrm{2}} }\\{\&\:{C}_{\mathrm{2}} \rightarrow{C}_{\mathrm{2}} −{C}_{\mathrm{3}} }\end{cases}\:\:\: \\ $$$$=\begin{vmatrix}{\mathrm{1}}&{\:\mathrm{1}}&{−\mathrm{4}}\\{\mathrm{1}}&{\:\mathrm{1}}&{−\mathrm{6}}\\{−\mathrm{3}}&{−\mathrm{4}}&{{x}+\mathrm{15}}\end{vmatrix}\begin{cases}{{by}\:{R}_{\mathrm{1}} \rightarrow{R}_{\mathrm{1}} −{R}_{\mathrm{2}} }\\{\&\:{R}_{\mathrm{2}} \rightarrow{R}_{\mathrm{2}} −{R}_{\mathrm{3}} }\end{cases}\:\:\:\:\: \\ $$$$=\begin{vmatrix}{\mathrm{0}}&{\:\mathrm{0}}&{\mathrm{2}}\\{\mathrm{1}}&{\:\mathrm{1}}&{−\mathrm{6}}\\{−\mathrm{3}}&{−\mathrm{4}}&{{x}+\mathrm{15}}\end{vmatrix}\:{by}\:{R}_{\mathrm{1}} \rightarrow{R}_{\mathrm{1}} −{R}_{\mathrm{2}} \: \\ $$$$=\mathrm{2}\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}\\{−\mathrm{3}}&{−\mathrm{4}}\end{vmatrix} \\ $$$$=\mathrm{2}\left(−\mathrm{4}+\mathrm{3}\right) \\ $$$$=−\mathrm{2} \\ $$

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