Question

Heya frndzzz solve this question

Plz don't spam☺️

Tysm♥️✌️

Answer 1

From Attachment:

$(1+a^2+b^2)^2 \left[\begin{array}{ccc}1&0&-b\\0&1&a\\2b&-2a&1-a^2-b^2\end{array}\right]$

∴ Determinant of A equals a times d minus b times c.

|A| = ad - bc.

$=(1 + a^2 + b^2)^2[1(1 - a^2 - b^2) + 2a^2) - 0(0) - b(0 - 2b))]$

$=(1+a^2+b^2)^2[(1-a^2-b^2+2a^2)+(2b^2)]$

$=(1+a^2+b^2)^2[1-a^2-b^2+2a^2+2b^2]$

$=(1+a^2+b^2)^2(1+a^2+b^2)$

$=(1+a^2+b^2)^3$

Hope it helps!

Answer 2

$\huge\boxed{\texttt{\fcolorbox{purple}{pink} {hey friend}}}$

$<b><i>ANSWER!$

⬇⬇⬇

=> (1+a2+b2)2[1(1−a2−b2)+2a2)−0(0)−b(0−2b))]

=> (1+a^2+b^2)^2[(1-a^2-b^2+2a^2)+(2b^2)]=(1+a2+b2)2[(1−a2−b2+2a2)+(2b2)]

=> (1+a^2+b^2)^2[1-a^2-b^2+2a^2+2b^2]=(1+a2+b2)2[1−a2−b2+2a2+2b2]

=> (1+a^2+b^2)^2(1+a^2+b^2)=(1+a2+b2)2(1+a2+b2)

=> (1+a^2+b^2)^3=(1+a2+b2)3

hope it helps u :)
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