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यदि $\left| {\,\begin{array}{*{20}{c}}{y + z}&x&y\\{z + x}&z&x\\{x + y}&y&z\end{array}\,} \right| = k(x + y + z){(x - z)^2}$, तब $k = $
$2xyz$
$1$
$xyz$
${x^2}{y^2}{z^2}$
Solution
(b) $\left| {\,\begin{array}{*{20}{c}}{y + z}&x&y\\{z + x}&z&x\\{x + y}&y&z\end{array}\,} \right|$=$(x + y + z)\,\left| {\,\begin{array}{*{20}{c}}2&1&1\\{z + x}&z&x\\{x + y}&y&z\end{array}\,} \right|$
${R_1} \to {R_1} + {R_2} + {R_3}$ द्वारा $= (x + y + z)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\x&z&x\\x&y&z\end{array}\,} \right|$ ;
${C_1} \to {C_1} – {C_2}$
द्वारा $= (x + y + z)\,.\,\{ ({z^2} – xy) – (xz – {x^2}) + (xy – xz)\} $= $(x + y + z)\,{(x – z)^2}$ $ \Rightarrow $ $k = 1$.
ट्रिक: $x = 1,\,y = 2$, $z = 3$ रखने पर,
$\left| {\,\begin{array}{*{20}{c}}5&1&2\\4&3&1\\3&2&3\end{array}\,} \right|\, = \,5(7)\, – 1(12 – 3) + 2(8 – 9)$ = $35 – 9 – 2 = 24$ और $(x + y + z)\,{(x – z)^2}$ $ = (6)\,{( – 2)^2} = 24$,
$\therefore \,\,k = \frac{{24}}{{24}} = 1$.
