Let A be a symmetric matrix such that |A|=2 and left[begin{array}{ll}2 & 1 3 & frac{3}{2}end

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3 and 4 .Determinants and Matrices

hard

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 $..........$.

$5$

$6$

$7$

$8$

(JEE MAIN-2023)

Solution

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

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

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

Standard 12

Mathematics

Let $\quad P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $Q=P Q P^{ T }$. If $P ^{ T } Q ^{2007} P =\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$, then $2 a+b-3 c-4 d$ equal to $……………….$.

hard

(JEE MAIN-2023)

If $A$ and $B$ are two square matrices of order $3$ such that $AB = A$ and $BA = B$ and matrices $X$,$Y$ and $Z$ are defined as $(X = A^4 + B^4)$, $Y$ = $A^{10}+ B^{10},$ then the matrix $X -Y$ is

normal

If $A = \left[ {\begin{array}{*{20}{c}}0&1&{ – 2}\\{ – 1}&0&5\\2&{ – 5}&0\end{array}} \right]$, then

easy

If $A=\left[\begin{array}{lll}3 & \sqrt{3} & 2 \\ 4 & 2 & 0\end{array}\right]$ and $B=\left[\begin{array}{rrr}2 & -1 & 2 \\ 1 & 2 & 4\end{array}\right],$ verify that $(k B)^{\prime}=k B^{\prime},$ where $k$ is any constant.

medium

If $A $ is a skew-symmetric matrix of order $ n$, and $ C$ is a column matrix of order $n \times 1$, then ${C^T}AC$ is