- Home
- Standard 12
- Mathematics
3 and 4 .Determinants and Matrices
hard
माना $\mathrm{A}$ वास्तविक अवयवों का एक $2 \times 2$ आव्यूह है जिसके लिए $\mathrm{A}^{\prime}=\alpha \mathrm{A}+\mathrm{I}$ है $\alpha \in \mathbb{R}-\{-1,1\}$ है। यदि $\operatorname{det}\left(A^2-A\right)=4$ है, तो $\alpha$ के सभी संभव मानों का योग बराबर है
A$0$
B$\frac{3}{2}$
C$\frac{5}{2}$
D$2$
(JEE MAIN-2023)
Solution
$A ^{ T }=\alpha A + I$
$A =\alpha A ^{ T }+ I$
$A =\alpha(\alpha A + I )+ I$
$A =\alpha^2 A +(\alpha+1) I$
$A \left(1-\alpha^2\right)=(\alpha+1) I$
$A =\frac{ I }{1-\alpha}$
$| A |=\frac{1}{(1-\alpha)^2}$
$\left| A ^2- A \right|=| A || A – I |$
$A – I =\frac{ I }{1-\alpha}- I =\frac{\alpha}{1-\alpha} I$
$| A – I |=\left(\frac{\alpha}{1-\alpha}\right)^2$
$\text { Now }\left| A ^2- A \right|=4$
$| A || A – I |=4$
$\Rightarrow \frac{1}{(1-\alpha)^2} \frac{\alpha^2}{(1-\alpha)^2}=4$
$\Rightarrow \frac{\alpha}{(1-\alpha)^2}= \pm 2$
$\Rightarrow 2(1-\alpha)^2= \pm \alpha$
$(C_1) 2(1-\alpha)^2=\alpha$ $ \left(C_2\right) 2(1-\alpha)^3=-\alpha$
$ 2 \alpha^2-5 \alpha+2=0 < _{\alpha_2}^{\alpha_1} $ $2 \alpha^2-3 \alpha+2=0$
$ \alpha_1+\alpha_2=\frac{5}{2}$ $ \alpha \notin R$
Sum of value of $\alpha=\frac{5}{2}$
$A =\alpha A ^{ T }+ I$
$A =\alpha(\alpha A + I )+ I$
$A =\alpha^2 A +(\alpha+1) I$
$A \left(1-\alpha^2\right)=(\alpha+1) I$
$A =\frac{ I }{1-\alpha}$
$| A |=\frac{1}{(1-\alpha)^2}$
$\left| A ^2- A \right|=| A || A – I |$
$A – I =\frac{ I }{1-\alpha}- I =\frac{\alpha}{1-\alpha} I$
$| A – I |=\left(\frac{\alpha}{1-\alpha}\right)^2$
$\text { Now }\left| A ^2- A \right|=4$
$| A || A – I |=4$
$\Rightarrow \frac{1}{(1-\alpha)^2} \frac{\alpha^2}{(1-\alpha)^2}=4$
$\Rightarrow \frac{\alpha}{(1-\alpha)^2}= \pm 2$
$\Rightarrow 2(1-\alpha)^2= \pm \alpha$
$(C_1) 2(1-\alpha)^2=\alpha$ $ \left(C_2\right) 2(1-\alpha)^3=-\alpha$
$ 2 \alpha^2-5 \alpha+2=0 < _{\alpha_2}^{\alpha_1} $ $2 \alpha^2-3 \alpha+2=0$
$ \alpha_1+\alpha_2=\frac{5}{2}$ $ \alpha \notin R$
Sum of value of $\alpha=\frac{5}{2}$
Standard 12
Mathematics
