A determinant is chosen at random from the se | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) determinant of order $2$ is of the form $\Delta = \begin{vmatrix} a & b \ c & d \end{vmatrix}$.It is equal to $ad - bc$.The total number of ways

Vedclass · Accepted Answer

$\frac{3}{8}$

Vedclass · Answer

(B) determinant of order $2$ is of the form $\Delta = \begin{vmatrix} a & b \ c & d \end{vmatrix}$.It is equal to $ad - bc$.The total number of ways of choosing $a, b, c,$ and $d$ is $2 	imes 2 	imes 2 	imes 2 = 16$.Now,$\Delta 
eq 0$ if and only if $ad - bc 
eq 0$,which means $ad 
eq bc$.Since $a, b, c, d \in \{0, 1\}$,the possible values for $ad$ and $bc$ are $0$ or $1$.We have $\Delta 
eq 0$ in the following cases:$1$. If $ad = 1$ and $bc = 0$: This happens when $a=1, d=1$ and $(b, c) \in \{(0, 0), (0, 1), (1, 0)\}$. ($3$ cases)$2$. If $ad = 0$ and $bc = 1$: This happens when $(a, d) \in \{(0, 0), (0, 1), (1, 0)\}$ and $b=1, c=1$. ($3$ cases)Total favorable cases = $3 + 3 = 6$.Therefore,the required probability = $\frac{6}{16} = \frac{3}{8}$.

$A$ determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only. The probability that the determinant chosen is non-zero is

Similar Questions

The correct order of decreasing third ionisation enthalpy of $Ti, V, Cr, Mn$ is

$A$ non-uniformly shaped conductor is charged. Then,at its sharpest point:

Which of the following is the correct order of solubility of alkaline earth metal sulfates in water?

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are given by $f(x)=|x|$ and $g(x)=[x]$ for each $x \in R$,then $\{x \in R: g(f(x)) \leq f(g(x))\}$ is equal to

Liquids $A$ and $B$ form an ideal solution. If $X_A$ and $Y_A$ are the mole fractions of $A$ in the solution and in vapour phase in equilibrium,respectively,then (Given $P_A^o > P_B^o$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module