MCQ
Matrix ${A_\lambda } = \left[ {\begin{array}{*{20}{c}}
  \lambda &{\lambda  - 1} \\ 
  {\lambda  - 1}&\lambda  
\end{array}} \right],\lambda  \in N$ then the value of $\left| {{A_1}} \right| + \left| {{A_2}} \right| + \left| {{A_3}} \right| + ....... + \left| {{A_{300}}} \right|$ is
  • A
    $(299)^2$
  • $(300)^2$
  • C
    $(150)^2$
  • D
    $(301)^2$

Answer

Correct option: B.
$(300)^2$
b
$\left| {{{\rm{A}}_\lambda }} \right| = \left| {\begin{array}{*{20}{c}}
\lambda &{\lambda  - 1}\\
{\lambda  - 1}&\lambda 
\end{array}} \right| = {\lambda ^2} - {[\lambda  - 1)^2}$

$ = 2\lambda  - 1$

$\therefore \left| {{{\rm{A}}_1}} \right| + \left| {{{\rm{A}}_2}} \right| + \left| {{{\rm{A}}_3}} \right| +  \ldots . + \left| {{{\rm{A}}_{300}}} \right|$

$=1+3^{2}+5+\ldots . .+599 $ 

$= \frac{300}{2}(1+599)=(300)^{2} $

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 value of $'a'$ $(a>0)$ for which the area bounded by the curves $y = \frac{x}{6} + \frac{1}{{{x^2}}},\,y = 0,\,x = a$ and $x = 2a$ has the least value, is
The feasible region for an $LPP$ is shown in the Figure. Let $z=3 x-4 y$ be the objective function. Maximum value of $z$ is $....$
The function $f$ defined by $f(x)=4 x^4-2 x+1$ is increasing for
The value of $\cos ({\tan ^{ - 1}}(\tan 2))$ is
A real valued function $f(x)$ satisfies the function equation $f(x - y) = f(x)f(y) - f(a - x)f(a + y)$ where a is a given constant and $f(0) = 1$, $f(2a - x)$ is equal to
$\tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right)} \right] = $
Let $\mathrm{Q}$ and $\mathrm{R}$ be the feet of perpendiculars from the point $\mathrm{P}(\mathrm{a}, \mathrm{a}, \mathrm{a})$ on the lines $\mathrm{x}=\mathrm{y}, \mathrm{z}=1$ and $\mathrm{x}=-\mathrm{y}$, $\mathrm{z}=-1$ respectively. If $\angle \mathrm{QPR}$ is a right angle, then $12 \mathrm{a}^2$ is equal to
If the set $A$ contains 5 elements and the set $B$ contains 6 elements, then the number of one-one and onto mappings from $A$ to $B$ is
Let $\vec{v}=\alpha \hat{i}+2 \hat{j}-3 \hat{k}, \vec{w}=2 \alpha \hat{i}+\hat{j}-\hat{k}$, and $\overrightarrow{ u }$ be a vector such that $|\vec{u}|=\alpha > 0$. If the minimum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is $-\alpha \sqrt{3401}$, and $|\vec{u} . \hat{i}|^2=\frac{m}{n}$ where $m$ and $n$ are coprime natural numbers, then $m + n$ is equal to $.........$.
$z=30 x-30 y+1800$ is a objective function. The corner points of the feasible region are $(15,0),(15,15),(10,20),(0,20)$ and $(0,15) $. $z$ has the minimum value at $\ldots \ldots \ldots .$ point.