A second-order determinant is written down at random using the numbers 1, -1 as elements. The probability that the value of the determinant is nonzero is

\frac{1}{2}

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\frac{3}{8}

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\frac{5}{8}

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\frac{1}{3}

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Solution

The correct option is A $\frac{1}{2}$
$T h e f a v o u r a b l e m a t r i c e s a r e [\begin{matrix} 1 & 1 1 & - 1 \end{matrix}] ⇝ 4 c a s e s o f i t w h e r e - 1 t a k e s 4 d i f f e r e n t p o s i t i o n s [\begin{matrix} 1 & - 1 - 1 & - 1 \end{matrix}] ⇝ 4 c a s e s o f i t w h e r e 1 t a k e s 4 d i f f e r e n t p o s i t i o n s A p a r t f r o m t h e a b o v e 8 c a s e s o f f a v o u r a b l e c a s e t h e n o n f a v o u r a b l e m a t r i c e s a r e [\begin{matrix} 1 & 1 1 & 1 \end{matrix}], [\begin{matrix} - 1 & - 1 - 1 & - 1 \end{matrix}] a n d [\begin{matrix} 1 & 1 - 1 & - 1 \end{matrix}] ⇝ 6 c a s e s o f i t {\frac{4!}{2! \times 2!}} T o t a l p o s s i b l e o u t c o m e s = 8 + 6 + 2 = 16 T o t a l f a v o u r a b l e o u t c o m e s = 8 \Rightarrow P r o b a b i l i t y t h a t t h e v a l u e o f t h e d e t e r m i n a n t i s n o n z e r o = \frac{8}{16} = \frac{1}{2}$

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