Matrix A_ = [ * 20 c & - 1 - 1 & ], N then the value of | A_1 | +… - Vidyadip

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

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 3 and 4 . determinant and metrices→STD 12 - 3 and 4 . determinant and metrices — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If $\text{S}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix},$ then adj A is:

The area of the region bounded by $y^{2}=8 x$ and $y^{2}=16(3-x)$ is equal to

If $\text{x},\text{ y}\in\text{R},$ then the determinant $\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$ lies in the interval:

If $\alpha ,\beta \ne 0$ and $f\left( n \right) = {\alpha ^n} + {\beta ^n}$ and $\left| {\begin{array}{*{20}{c}}3&{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}\\{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}\\{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}&{1 + f\left( 4 \right)}\end{array}} \right|\; = K{\left( {1 - \alpha } \right)^2}$ ${\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2}$ ,then $K=$ . . . . . .

The equation to the normal to the curve $\text{y}=\sin\text{x}$ at $(0, 0)$ is:

$4\, {\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{70}} + {\tan ^{ - 1}}\frac{1}{{99}} = $

If $X$ follows a binomial distribution with parameters $n = 6$ and $p$. If $9P\,(X = 4) = P\,(X = 2),$ then $p = $

If $g$ $(x)$ is the inverse of $f$ $(x) $ and $f$ $(x)$ has domain $x \in [1, 5], $ where $f (1) = 2$ and $f(5) = 10$ then the values of $\int\limits_1^5 {f(x)} dx$ $+\int\limits_2^{10} {g(y)} dy$ equals

The sine and cosine curves intersects infinitely many times giving bounded regions of equal areas. The area of one of such region is

If $y = x(x - 3)^2$ decreases for the values of $x$ given by: