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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
$\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + y}&1\\1&1&{1 + z}\end{array}\,} \right| = $
  • ✓
    $xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$
  • B
    $xyz$
  • C
    $1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$
  • D
    $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$
✓

Answer

Correct option: A.
$xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$
$\Delta=xyz$ $\,\left| \,\begin{matrix} 1+\frac{1}{x} & \frac{1}{x} & \frac{1}{x}  \\ \frac{1}{y} & 1+\frac{1}{y} & \frac{1}{y}  \\ \frac{1}{z} & \frac{1}{z} & 1+\frac{1}{z}  \\ \end{matrix}\, \right|$
$= xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$
$\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\{\frac{1}{y}}&{1 + \frac{1}{y}}&{\frac{1}{y}}\\{\frac{1}{z}}&{\frac{1}{z}}&{1 + \frac{1}{z}}\end{array}\,} \right|$
by ${R_1} \to {R_1} + {R_2} + {R_3}$
$=xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$
$\,\left| {\,\begin{array}{*{20}{c}}1&0&0\\{1/y}&1&0\\{1/z}&0&1\end{array}\,} \right|$,
by $\begin{array}{l}{C_2} \to {C_2} - {C_1}\\{C_3} \to {C_3} - {C_1}\end{array}$
$= xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$
$\left| {\,\begin{array}{*{20}{c}}1&0\\0&1\end{array}\,} \right| = xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$.
Trick : Put $x = 1,\,y = 2$ and $z = 3,$ then
$\left| {\,\begin{array}{*{20}{c}}2&1&1\\1&3&1\\1&1&4\end{array}\,} \right| $
$= 2(11) - 1(3) + 1(1 - 3) = 17$
Option $ (a)$ gives, $1 \times 2 \times 3\,\left( {1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{3}} \right) = 17$.

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