alpha, beta in mathbb{R} અને એક પ્રાકૃતિક સંખ | vedclass

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Question

$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}\

Vedclass · Accepted Answer

$4 \alpha+2 \beta$

Vedclass · Answer

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \ 2 r & 2 & n^2-\beta \ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}ight|$

$2 \mathrm{~A}_{10}-\mathrm{A}_8=\left|\begin{array}{ccc}20 & 1 & \frac{\mathrm{n}^2}{2}+\alpha \ 40 & 2 & \mathrm{n}^2-\beta \ 56 & 3 & \frac{\mathrm{n}(3 \mathrm{n}-1)}{2}\end{array}ight|-\left|\begin{array}{ccc}8 & 1 & \frac{\mathrm{n}^2}{2}+\alpha \ 16 & 2 & \mathrm{n}^2-\beta \ 22 & 3 & \frac{\mathrm{n}(3 \mathrm{n}-1)}{2}\end{array}ight|$

$\Rightarrow\left|\begin{array}{ccc}12 & 1 & \frac{n^2}{2}+\alpha \ 24 & 2 & n^2-\beta \ 34 & 3 & \frac{n(3 n-1)}{2}\end{array}ight|$

$\Rightarrow\left|\begin{array}{ccc}0 & 1 & \frac{\mathrm{n}^2}{2}+\alpha \ 0 & 2 & \mathrm{n}^2-\beta \ -2 & 3 & \frac{\mathrm{n}(3 \mathrm{n}-1)}{2}\end{array}ight|$

$\Rightarrow-2\left(\left(\mathrm{n}^2-\betaight)-\left(\mathrm{n}^2+2 \alphaight)ight)$

$\Rightarrow-2(-\beta-2 \alpha) \Rightarrow 4 \alpha+2 \beta$

Similar Questions

જો $k_1$, $k_2$ એ $k$ ની મહતમ અને ન્યૂનતમ કિમતો છે કે જેથી સમીકરણોની સહંતિ $x + ky = 1$ ; $kx + y = 2$; $x + y = k$ એ સુસંગત થાય છે તો $k_1^2 + k_2^2$ મેળવો.

જો $\left| {\begin{array}{*{20}{c}}
{\cos 2x}&{{{\sin }^2}x}&{\cos 4x} \\
{{{\sin }^2}x}&{\cos 2x}&{{{\cos }^2}x} \\
{\cos 4x}&{{{\cos }^2}x}&{\cos 2x}
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જો સમીકરણ સંહતિ $x+4 y-z=\lambda, 7 x+9 y+\mu z=-3,5 x+y+2 z=-1$ ને અનંત ઉકેલો હોય, તો $(2 \mu+3 \lambda)=$..............

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સમીકરણ સંહતિ $2x + y - z = 7,\,\,x - 3y + 2z = 1,\,x + 4y - 3z = 5$ ના ઉકેલની સંખ્યા મેળવો.