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

(A) determinant of order $2$ is given by $\begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$. Each element can be either $0$ or $1$,so there are

Vedclass · Accepted Answer

$3/16$

Vedclass · Answer

(A) determinant of order $2$ is given by $\begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$. Each element can be either $0$ or $1$,so there are $2^4 = 16$ possible determinants. The value of the determinant is $ad - bc$. For the value to be positive,$ad - bc > 0$,which implies $ad > bc$. Since $a, b, c, d \in \{0, 1\}$,the possible values for $ad$ and $bc$ are $0$ or $1$. $ad - bc = 1$ occurs when $ad = 1$ and $bc = 0$. $ad = 1$ implies $a=1$ and $d=1$. $bc = 0$ implies $(b, c) \in \{(0, 0), (0, 1), (1, 0)\}$. These correspond to the matrices: $\begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} = 1$,$\begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix} = 1$,$\begin{vmatrix} 1 & 1 \ 0 & 1 \end{vmatrix} = 1$. There are $3$ such cases. Thus,the probability is $\frac{3}{16}$.

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

Similar Questions

Suppose $A$ is any $3 \times 3$ non-singular matrix and $(A - 3I)(A - 5I) = O$,where $I = I_3$ and $O = O_3$. If $\alpha A + \beta A^{-1} = 4I$,then $\alpha + \beta$ is equal to

If $A$ and $B$ are square matrices of order $3$ such that $(A + B)(A - B) = A^2 - B^2$,then $(ABA^{-1})^2$ is equal to

The number of singular matrices of order $2 \times 2$,whose elements are from the set $\{2, 3, 6, 9\}$ is

For each real number $x$ such that $-1 < x < 1$,let $A(x)$ be the matrix $\frac{1}{1-x^2} \begin{bmatrix} 1 & -x \\ -x & 1 \end{bmatrix}$. If $z = \frac{x+y}{1+xy}$,then which of the following is true?

Let $S = \left\{ \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix} : a, b \in \{1, 2, 3, \ldots, 100\} \right\}$ and let $T_n = \{A \in S : A^{n(n+1)} = I\}$. Then the number of elements in $\bigcap_{n=1}^{100} T_n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module