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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
For each real number $x$ such that $ - 1 < x < 1,$let $A(x)$ be the matrix ${(1 - x)^{ - 1}}\left[ {\begin{array}{*{20}{c}}1&{ - x}\\{ - x}&1\end{array}} \right]$ and $z = \frac{{x + y}}{{1 + xy}}$ Then
  • A
    $A(z) = A(x) + A(y)$
  • B
    $A(z) = A(x){[A(y)]^{ - 1}}$
  • ✓
    $A(z) = A(x)\,A(y)$
  • D
    $A(z) = A(x) - A(y)$
✓

Answer

Correct option: C.
$A(z) = A(x)\,A(y)$
c
(c) $A(z) = A\,\left( {\frac{{x + y}}{{1 + xy}}} \right) = \left[ {\frac{{1 + xy}}{{(1 - x)\,(1 - y)}}} \right]$

$\left[ {\begin{array}{*{20}{c}}1&{ - \left( {\frac{{x + y}}{{1 + xy}}} \right)}\\{ - \left( {\frac{{x + y}}{{1 + xy}}} \right)}&1\end{array}} \right]$

$\therefore$ $A(x)\,.\,A(y)\, = \,A(z)$.

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