Question

prove determinant | 1+a2-b2, 2ab, -2b, 2ab, 1-a2+b2 , 2a , 2b , -2a , 1-a2-b2 | = ( 1+ a2 + b2 )^3

Answer 1

it was easy and lengthy hope it helps you

Answer 2

We have to Prove that

$\left[\begin{array}{ccc}1+a^{2}-b^{2} &2ab&-2b\\2ab&1-a^{2}+b^{2} &2a\\2b&-2a&1-a^{2}-b^{2} \end{array}\right] = (1+a^{2}+b^{2} )^{3}$

Using properties of determinants

• Now, we have

$\left[\begin{array}{ccc}1+a^{2}-b^{2} &2ab&-2b\\2ab&1-a^{2}+b^{2} &2a\\2b&-2a&1-a^{2}-b^{2} \end{array}\right] = (1+a^{2}+b^{2} )^{3}$

• Solving LHS :

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

As we know that the value of determinants doesn't change by applying row transformations, therefore

Using row transformation

• R1→ R1 +bR3

$\left[\begin{array}{ccc}1+a^{2}-b^{2} +2b^{2} &2ab-2ab&-2b+b-a^{2}b-b^{3} \\2ab&1-a^{2}+b^{2} &2a\\2b&-2a&1-a^{2}-b^{2} \end{array}\right]$

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

Using row transformation

• R2→ R2 - aR3

$\left[\begin{array}{ccc}1+a^{2}+b^{2} &0&-b(1+a^{2}+b^{2}) \\2ab-2ab&1-a^{2}+b^{2}+2a^{2} &2a-a+a^{3}+b^{2}a \\2b&-2a&1-a^{2}-b^{2} \end{array}\right]$

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

Taking $1 + a^{2} +b^{2}$ common from R1 and R2, we get

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

• Now expanding the above determinant along R1, we get

$= [1 ( 1 - a^{2} -b^{2} +2a^{2}) - 0(-2ab) -b(-2b)] (1 + a^{2}+b^{2} ) ^{2} \\ = (1 + a^{2} - b^{2} + 2b^{2} ) (1 + a^{2}+b^{2} ) ^{2}\\= (1 + a^{2}+b^{2} )(1 + a^{2}+b^{2} ) ^{2}\\= (1 + a^{2}+b^{2} ) ^{3} \\= RHS$

$\left[\begin{array}{ccc}1+a^{2}-b^{2} &2ab&-2b\\2ab&1-a^{2}+b^{2} &2a\\2b&-2a&1-a^{2}-b^{2} \end{array}\right] = (1+a^{2}+b^{2} )^{3}$

hence proved.