Question

Question for Brainly Teachers​

Answer 1

Step-by-step explanation:

To find:

$\left|\begin{array}{ccc}1+a^2-b^2&2ab&-2b\\2ab&1-a^2+b^2&2a\\</p><p>2b&-2a&1-a^2-b^2\end{array}\right|$

Is a perfect cube.

Apply C1 -> C1-bC3

C2 ->C2+aC3

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

Take common from C1 and C2

$(1+a^2+b^2)^2\left|\begin{array}{ccc}</p><p>1&0&-2b\\0&1&2a\\b&-a&1-a^2-b^2\end{array}\right|$

R3 -> R3-b R1

$(1+a^2+b^2)^2\left|\begin{array}{ccc}</p><p>1&0&-2b\\0&1&2a\\0&-a&1-a^2-b^2\end{array}\right|$

Expand the determinant along C1

$(1+a^2+b^2)^2 [1-a^2+b^2+2a^2]\\\\(1+a^2+b^2)^2(1+a^2+b^2)\\\\=(1+a^2+b^2)^3$

Hence

$Determinant=\left|\begin{array}{ccc}1+a^2-b^2&2ab&-2b\\2ab&1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2\end{array}\right|$

is a perfect cube.

Hope it helps you.

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https://brainly.in/question/13362759

Answer 2

$To find</p><p></p><p>\begin{gathered}|\begin{array}{ccc} 1+a^2-b^2&2ab&-2b\\ 2ab&1-a^2+b^2&2a\\ 2b&-2a&1-a^2-b^2\end{array} |\end{gathered}∣1+a2−b22ab2b2ab1−a2+b2−2a−2b2a1−a2−b2∣</p><p></p><p>Is a perfect cube.</p><p></p><p>Apply C1 -> C1-bC3</p><p></p><p>C2 ->C2+aC3</p><p></p><p>\begin{gathered}|\begin{array}{ccc} 1+a^2+b^2&0&-2b\\ 0&1+a^2+b^2&2a\\ b(1+a^2+b^2)&-a(1+a^2+b^2)&1-a^2-b^2\end{array} |\end{gathered}∣1+a2+b20b(1+a2+b2)01+a2+b2−a(1+a2+b2)−2b2a1−a2−b2∣</p><p></p><p>Take common from C1 and C2</p><p></p><p>\begin{gathered}(1+a^2+b^2)^2 |\begin{array}{ccc} 1&0&-2b\\ 0&1&2a\\ b&-a&1-a^2-b^2\end{array} |\end{gathered}(1+a2+b2)2∣10b01−a−2b2a1−a2−b2∣</p><p></p><p>R3 -> R3-b R1</p><p></p><p>\begin{gathered}(1+a^2+b^2)^2 |\begin{array}{ccc} 1&0&-2b\\ 0&1&2a\\ 0&-a&1-a^2+b^2\end{array} |\end{gathered}(1+a2+b2)2∣10001−a−2b2a1−a2+b2∣</p><p></p><p>Expand the determinant along C1</p><p></p><p>\begin{gathered}(1+a^2+b^2)^2 [1-a^2+b^2+2a^2]\\\\(1+a^2+b^2)^2(1+a^2+b^2)\\\\=(1+a^2+b^2)^3\\\end{gathered}(1+a2+b2)2[1−a2+b2+2a2](1+a2+b2)2(1+a2+b2)=(1+a2+b2)3</p><p></p><p>Hence</p><p></p><p>\begin{gathered}determinant |\begin{array}{ccc} 1+a^2-b^2&2ab&-2b\\ 2ab&1-a^2+b^2&2a\\ 2b&-2a&1-a^2-b^2\end{array} |\end{gathered}determinant∣1+a2−b22ab2b2ab1−a2+b2−2a−2b2a1−a2−b2∣</p><p></p><p>is a perfect cube$