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3 and 4 .Determinants and Matrices
medium
यदि $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{y - z}&{z - x}&{y - x}\\{z - y}&{z - x}&{x + y}\end{array}\,} \right| = k\,xyz$,तो $ k$ का मान है
A
$2$
B
$4$
C
$6$
D
$8$
Solution
(d) $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x – z}&{x – y}\\{y – z}&{z + x}&{y – x}\\{z – y}&{z – x}&{x + y}\end{array}\,} \right|$ $ = \left| {\,\begin{array}{*{20}{c}}{y + z}&{x – z}&{x – y}\\{2y}&{2x}&0\\{2z}&0&{2x}\end{array}\,} \right|$
(${R_2} \to {R_2} + {R_1}$ और ${R_3} \to {R_3} + {R_1}$ से)
$ = 4\left| {\begin{array}{*{20}{c}}{y + z}&{x – z}&{x – y}\\y&x&0\\z&0&x\end{array}} \right|$
$ = 4[(y + z)({x^2}) – (x – z)(xy)$ $ + (x – y)( – zx)]$
$ = 4[{x^2}y + z{x^2} – {x^2}y + xyz – z{x^2} + xyz]$ $ = 8xyz$
Hence, $k = 8$.
Std 12
Mathematics