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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
Question
Let $A$ be a symmetric matrix such that $|A|=2$ and $\left[\begin{array}{ll}2 & 1 \\ 3 & \frac{3}{2}\end{array}\right] A =\left[\begin{array}{ll}1 & 2 \\ \alpha & \beta\end{array}\right]$. If the sum of the diagonal elements of $A$ is s, then $\frac{\beta s}{\alpha^2}$ is equal to $..........$.
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Answer

a
$\left[\begin{array}{ll}2 & 1 \\3 & \frac{3}{2}\end{array}\right]\left[\begin{array}{ll} a & b \\b & c\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\\alpha & \beta\end{array}\right]$

Now $a c-b^2=2$ and $2 a+b=1$ and $2 b + c =2$

solving all these above equations we get

$\frac{1-b}{2} \times\left(\frac{2-2 b}{1}\right)-b^2=2$

$\Rightarrow(1-b)^2-b^2=2$

$\Rightarrow 1-2 b=2$

$\Rightarrow b=-\frac{1}{2} \text { and } a=\frac{3}{4} \text { and } c=3$

Hence $\alpha=3 a+\frac{3 b}{2}=\frac{9}{4}-\frac{3}{4}=\frac{3}{2}$

and $\beta=3 b+\frac{3 c}{2}=-\frac{3}{2}+\frac{9}{2}=3$

also $s = a + c =\frac{15}{4}$

$\therefore \frac{\beta s}{\alpha^2}=\frac{3 \times 15}{4 \times \frac{9}{4}}=5$

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