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Let A be a symmetric matrix such that |A|=2 and \left[\begin{array}{ll}2 & 1 \\ 3 & \frac{3}{2}\end{array}\right] A-\left[\begin{array}{cc}1 & 2 \\ \alpha & \beta\end{array}\right]. If the sum of the diagonal elements of A is s, then  \frac{\beta s}{\alpha^2}  is equal to

Option: 1

5


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

A be a symmetric matrix such that  |\mathrm{A}|=2 \text { and }\left[\begin{array}{ll} 2 & 1 \\ 3 & 3 / 2 \end{array}\right] \mathrm{A}=\left[\begin{array}{ll} 1 & 2 \\ \alpha & \beta \end{array}\right]

A=\left[\begin{array}{ll} a & b \\ b & d \end{array}\right] \, \, \, \, \, \, \, \, \, \quad|A|=a d-b^2=2

\left[\begin{array}{cc} 2 & 1 \\ 3 & 3 / 2 \end{array}\right]\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{b} & \mathrm{d} \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ \alpha & \beta \end{array}\right]

\left[\begin{array}{cc} 2 a+b & 2 b+d \\ 3 a+3 / 2 b & 3 b+3 / 2 d \end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ \alpha & \beta \end{array}\right]

\begin{array}{ll} 2 \mathrm{a}+\mathrm{b}=1 & \, \, \, \, \, 2 \mathrm{~b}+\mathrm{d}=2 \\ \mathrm{~b}=1-2 \mathrm{a} & \, \, \, \, \mathrm{d}=2-2 \mathrm{~b} \\ & \, \, =2-2(1-2 a) \\ & \, \, =2-2+4 a \end{array}

a d-b^2=2

a. 4 a-(1-2 a)^2=2 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, Now \, \alpha=3 a+\frac{3}{2} b

\Rightarrow 4 \mathrm{a}^2-1-4 \mathrm{a}^2+4 \mathrm{a}=2 \, \, \, \, \, \, \, \, \quad=\frac{9}{4}+\frac{3}{2} \cdot\left(\frac{-1}{2}\right)

     4 \mathrm{a}=3 , \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{9-3}{4}=\frac{6}{4}=\frac{3}{2}

\mathrm{a}=\frac{3}{4}

b=1-2 \times \frac{3}{4} \quad \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \beta=3 b+\frac{3}{2} d

=\frac{-1}{2} \quad \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 3 \times\left(\frac{-1}{2}\right)+\frac{3}{2} \times 3

\mathrm{d}=4 \times \frac{3}{4}=3 \quad \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{-3+9}{2}=3                                     

   \begin{aligned} & \mathrm{A}=\left[\begin{array}{cc} 3 / 4 & -1 / 2 \\ -1 / 2 & 3 \end{array}\right] \mathrm{s}=\frac{3}{4}+3=\frac{15}{4} \\ & \frac{\mathrm{Bs}}{\alpha^2}=\frac{3 \times \frac{15}{4}}{\frac{9}{4}}=5 \end{aligned}

 

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seema garhwal

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