If A=xyyx and A2=abba , then a+b=

A=xyyxA2=xyyxxyyx=abba⇒x2+y22xy2xyx2+y2⇒x2+y22xy2xyx2+y2=abba⇒a=x2+y2,b=2xya+b=(x+y)2

If A=xyyx and A2=abba , then a+b=

A=xyyxA2=xyyxxyyx=abba⇒x2+y22xy2xyx2+y2⇒x2+y22xy2xyx2+y2=abba⇒a=x2+y2,b=2xya+b=(x+y)2

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If $A = [\begin{matrix} x & y \\ y & x \end{matrix}]$ and $A^{2} = [\begin{matrix} a & b \\ b & a \end{matrix}]$ , then $a + b =$

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${(x - y)}^{2}$

$x^{2} + y^{2} - x y$

${(x + y)}^{2}$

$x^{2} + y^{2} + x y$

answer is C.

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Detailed Solution

$\begin{array}{l} A = [\begin{matrix} x & y \\ y & x \end{matrix}] \\ A^{2} = [\begin{matrix} x & y \\ y & x \end{matrix}] [\begin{matrix} x & y \\ y & x \end{matrix}] = [\begin{matrix} a & b \\ b & a \end{matrix}] \\ \Rightarrow [\begin{matrix} x^{2} + y^{2} & 2 x y \\ 2 x y & x^{2} + y^{2} \end{matrix}] \\ \Rightarrow [\begin{matrix} x^{2} + y^{2} & 2 x y \\ 2 x y & x^{2} + y^{2} \end{matrix}] = [\begin{matrix} a & b \\ b & a \end{matrix}] \\ \Rightarrow a = x^{2} + y^{2}, b = 2 x y \\ a + b = {(x + y)}^{2} \end{array}$