Question

plz give correct answers q 2​

Answer 1

Given,

$\displaystyle\longrightarrow A=\left [\begin{array}{cc}x&3\\y&3\end {array}\right]$

such that,

$\displaystyle\longrightarrow A^2=3I$

$\displaystyle\longrightarrow\left [\begin {array}{cc}x^2+3y&3(x+3)\\y(x+3)&3y+9\end{array}\right]=\left [\begin {array}{cc}3&0\\0&3\end{array}\right]$

Equating $a_{12}$ and $a_{22},$

$\displaystyle\longrightarrow 3x+9=0$

$\displaystyle\longrightarrow\underline {\underline {x=-3}}$

and,

$\displaystyle\longrightarrow 3y+9=3$

$\displaystyle\longrightarrow\underline {\underline {y=-2}}$

Answer 2

$\huge\underline{\underline{\bold{\pink{solution}}}}$

Given,

$\sf{\implies A=\left [\begin{array}{cc}x&3\\y&3\end {array}\right]}$

$\sf{ \implies A^2=3I}$

$\sf{ \implies \left [\begin {array}{cc}x^2+3y&3(x+3)\\y(x+3)&3y+9\end{array}\right]=\left [\begin {array}{cc}3&0\\0&3\end{array}\right]}$

Equating

$\\ a_{12} \: and \:\: a_{22},$

$\sf{\implies 3x+9=0}$

$\rm{\red{ \boxed{ \purple{\underline{x=-3}}}}}$

$\rm{3y+9=3}$

$\rm{\red{ \boxed{ \purple{y=-2}}}} \\ \\ \\ \\ \large{\purple{\mathfrak{x= -3\: , y= -2.}}}$

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