For α, β ∈ mathrm{R} and a natural number mathrm{n} , let Aᵣ=left|begin{array}{ccc}r & 1 &

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3 and 4 .Determinants and Matrices

hard

For $\alpha, \beta \in \mathrm{R}$ and a natural number $\mathrm{n}$, let

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$

$4 \alpha+2 \beta$

$2 \alpha+4 \beta$

$2 n$

$0$

(JEE MAIN-2024)

Solution

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$

$2 \mathrm{~A}_{10}-\mathrm{A}_8=\left|\begin{array}{ccc}20 & 1 & \frac{\mathrm{n}^2}{2}+\alpha \\ 40 & 2 & \mathrm{n}^2-\beta \\ 56 & 3 & \frac{\mathrm{n}(3 \mathrm{n}-1)}{2}\end{array}\right|-\left|\begin{array}{ccc}8 & 1 & \frac{\mathrm{n}^2}{2}+\alpha \\ 16 & 2 & \mathrm{n}^2-\beta \\ 22 & 3 & \frac{\mathrm{n}(3 \mathrm{n}-1)}{2}\end{array}\right|$

$\Rightarrow\left|\begin{array}{ccc}12 & 1 & \frac{n^2}{2}+\alpha \\ 24 & 2 & n^2-\beta \\ 34 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$

$\Rightarrow\left|\begin{array}{ccc}0 & 1 & \frac{\mathrm{n}^2}{2}+\alpha \\ 0 & 2 & \mathrm{n}^2-\beta \\ -2 & 3 & \frac{\mathrm{n}(3 \mathrm{n}-1)}{2}\end{array}\right|$

$\Rightarrow-2\left(\left(\mathrm{n}^2-\beta\right)-\left(\mathrm{n}^2+2 \alpha\right)\right)$

$\Rightarrow-2(-\beta-2 \alpha) \Rightarrow 4 \alpha+2 \beta$

Standard 12

Mathematics

Let $k_1$, $k_2$ be the maximum and minimum values of $k$ for which the system of equations given by

$x + ky = 1$ ; $kx + y = 2$; $x + y = k$ are consistent then $k_1^2 + k_2^2$ is equal to

normal

Let the area of the triangle with vertices $A (1, \alpha)$, $B (\alpha, 0)$ and $C (0, \alpha)$ be $4\, sq.$ units. If the point $(\alpha,-\alpha),(-\alpha, \alpha)$ and $\left(\alpha^{2}, \beta\right)$ are collinear, then $\beta$ is equal to

medium

(JEE MAIN-2022)

Let $A = \left[ {\begin{array}{*{20}{c}}5&{5\alpha }&\alpha \\0&\alpha &{5\alpha }\\0&0&5\end{array}} \right]$, If ${\left| A \right|^2} = 25$, then $\left| \alpha \right|$ equals

medium

(AIEEE-2007)

If the following system of linear equations

$2 x+y+z=5$

$x-y+z=3$

$x+y+a z=b$

has no solution, then :

hard

(JEE MAIN-2021)

Evaluate the determinants : $\left|\begin{array}{ll}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right|$

easy

For $\alpha, \beta \in \mathrm{R}$ and a natural number $\mathrm{n}$, let $A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$

Solution

Similar Questions

Let $k_1$, $k_2$ be the maximum and minimum values of $k$ for which the system of equations given by $x + ky = 1$ ; $kx + y = 2$; $x + y = k$ are consistent then $k_1^2 + k_2^2$ is equal to