The value of determinant determinant (((x+2),( x+3),(x+5)),((x+4),( x+6),(x+9)),((x+8),(x+11),(x+15))) is


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Question Number 70385 by ®Ëƒ ¬Ë°¹¾¨ ‰¦Í¦¿¨ ¸Ë¹Ç² last updated on 04/Oct/19

$The value of determinant$ $\| \begin{matrix} x + 2 & x + 3 & x + 5 \\ x + 4 & x + 6 & x + 9 \\ x + 8 & x + 11 & x + 15 \end{matrix} \| is$
	Answered by $@ty@m123 last updated on 04/Oct/19
	$= determinant (((−1),( −2),(x+5)),((−2),( −3),(x+9)),((−3),(−4),(x+15))) { ((by C_1 →C_1 −C_2 )),((& C_2 →C_2 −C_3 )) :} = determinant ((1,( 1),(−4)),(1,( 1),(−6)),((−3),(−4),(x+15))) { ((by R_1 →R_1 −R_2 )),((& R_2 →R_2 −R_3 )) :} = determinant ((0,( 0),2),(1,( 1),(−6)),((−3),(−4),(x+15))) by R_1 →R_1 −R_2 =2 determinant ((1,1),((−3),(−4))) =2(−4+3) =−2$
	$= \| \begin{matrix} - 1 & - 2 & x + 5 \\ - 2 & - 3 & x + 9 \\ - 3 & - 4 & x + 15 \end{matrix} \| {\begin{cases} b y C_{1} \to C_{1} - C_{2} \\ & C_{2} \to C_{2} - C_{3} \end{cases}$ $= \| \begin{matrix} 1 & 1 & - 4 \\ 1 & 1 & - 6 \\ - 3 & - 4 & x + 15 \end{matrix} \| {\begin{cases} b y R_{1} \to R_{1} - R_{2} \\ & R_{2} \to R_{2} - R_{3} \end{cases}$ $= \| \begin{matrix} 0 & 0 & 2 \\ 1 & 1 & - 6 \\ - 3 & - 4 & x + 15 \end{matrix} \| b y R_{1} \to R_{1} - R_{2}$ $= 2 \| \begin{matrix} 1 & 1 \\ - 3 & - 4 \end{matrix} \|$ $= 2 (- 4 + 3)$ $= - 2$
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