MCQ
The number of positive integral solutions $\left| {\,\,\begin{array}{*{20}{c}}{1 - \lambda }&2&1\\{ - 3}&\lambda &{ - 2}\\2&{ - 2}&{1 + \lambda }\end{array}\,\,} \right|$ $= 0$ is
  • A
    $0$
  • B
    $2$
  • $3$
  • D
    $1$

Answer

Correct option: C.
$3$
c

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The number of commutative binary operation that can be defined on a set of 2 elements is:
  1. 8
  2. 6
  3. 4
  4. 2
$\int_{\, - 1}^{\,2} {|x|\,dx} =$
The degree of the differential equation $\left(\frac{d s}{d t}\right)^4+3 s \frac{d^2 s}{d t^2}=0$ is :
If X follows a binomial distribution with parameter $\text{n}=100$ and $\text{p}=\frac{1}{3},$ then P(X = r) is maximum when r = 
  1. 32
  2. 34
  3. 33
  4. 31
Let $y(x)$ be the solution of the differential equation $2 x^{2} d y+\left(e^{y}-2 x\right) d x=0, x>0$. If $y(e)=1$, then $\mathrm{y}(1)$ is equal to :
A function $f$ satisfies the relation

$f(x) = f''(x) + f'''(x) + .......\infty $ where $f(x)$ is a differentiable function indefinitely. If $f(1) = 5$ , then the value of $f'(1) + f''(1)$ is equal to

Let $A = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
2&2&{ - 1}\\
3&0&k
\end{array}} \right]$ and $f(x) = {x^3} - 2{x^2} - \alpha x + \beta  = 0$ . If $A$ satisfies $f(x)=0$ ,then
For non-zero vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ the relation $\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ holds good, if:
  1. $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=0$
  2. $\vec{\text{a}}.\vec{\text{b}}=0=\vec{\text{c}}.\vec{\text{a}}$
  3. $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
  4. $\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
The corner points of the feasible region determined by the system of linear inequalities are $(0,0),(4,0)$, $(2,4)$ and $(0,5)$. If the maximum value of $z=a x+b y$, where $a, b>0$ occurs at both $(2,4)$ and $(4,0)$, then
$\int_{}^{} {\frac{{{x^5}}}{{\sqrt {1 + {x^3}} }}dx = } $