If a 2^{nd} order determinant with elements 0 | vedclass

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

Question

(B) Let the determinant be $\Delta = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$. Each element $a, b, c, d$ can be either $0$ or $1$. Sinc

Vedclass · Accepted Answer

$3/8$

Vedclass · Answer

(B) Let the determinant be $\Delta = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$. Each element $a, b, c, d$ can be either $0$ or $1$. Since there are $4$ positions and each has $2$ choices,the total number of possible determinants is $2^4 = 16$. For the determinant to be non-zero,$\Delta 
eq 0$,which means $ad - bc 
eq 0$,or $ad 
eq bc$. Since $a, b, c, d \in \{0, 1\}$,the possible values for $ad$ and $bc$ are $0$ or $1$. Case $1$: $ad = 1$ and $bc = 0$. $ad = 1 \implies a = 1, d = 1$. $bc = 0 \implies (b=0, c=0), (b=0, c=1), (b=1, c=0)$. This gives $3$ possible determinants: $\begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix}, \begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix}, \begin{vmatrix} 1 & 1 \ 0 & 1 \end{vmatrix}$. Case $2$: $ad = 0$ and $bc = 1$. $bc = 1 \implies b = 1, c = 1$. $ad = 0 \implies (a=0, d=0), (a=0, d=1), (a=1, d=0)$. This gives $3$ possible determinants: $\begin{vmatrix} 0 & 1 \ 1 & 0 \end{vmatrix}, \begin{vmatrix} 0 & 1 \ 1 & 1 \end{vmatrix}, \begin{vmatrix} 1 & 1 \ 1 & 0 \end{vmatrix}$. Total favorable outcomes = $3 + 3 = 6$. Probability $P = \frac{6}{16} = \frac{3}{8}$.

If a $2^{nd}$ order determinant with elements $0$ or $1$ is chosen from the set of all such determinants,find the probability that the determinant chosen is non-zero.

Similar Questions

If $A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$ satisfies the equation $x^2 + 4x - p = 0$, then $p$ is equal to

Let $\alpha, \beta, \gamma, \delta$ be distinct imaginary roots of $z^5=1$. Find the value of the determinant: $\left| \begin{array}{ccc} e^{\alpha} & e^{2\alpha} & e^{3\alpha+1} \\ e^{\beta} & e^{2\beta} & e^{3\beta+1} \\ e^{\gamma} & e^{2\gamma} & e^{3\gamma+1} \end{array} \right|$.

The number of cubic polynomials $P(x)$ satisfying $P(1)=2, P(2)=4, P(3)=6, P(4)=8$ is

Let $\alpha$ and $\beta$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^2 = 3A + \alpha I$. If $A^4 = 21A + \beta I$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module