question_answer

Using properties of determinants, prove the following

\[={{(1+{{a}^{2}}+{{b}^{2}})}^{3}}.\]

OR

If x, y, z are in GP, using properties of determinants, show that

where \[x\ne y\ne z\]and p is any real number.

Answer:

Applying \[{{C}_{1}}\to {{C}_{1}}-b{{C}_{3}},\] \[{{C}_{2}}\to {{C}_{2}}+a{{C}_{3}},\] we get

Taking \[(1+{{a}^{2}}+{{b}^{2}})\] common from \[{{C}_{1}}\] and \[{{C}_{2}},\] we get

Applying \[{{R}_{3}}\to {{R}_{3}}-b{{R}_{1}}+a{{R}_{2}},\] we get

Expanding along \[{{C}_{1}},\] we get

\[LHS={{(1+{{a}^{2}}+{{b}^{2}})}^{2}}[1(1+{{a}^{2}}+{{b}^{2}})]={{(1+{{a}^{2}}+{{b}^{2}})}^{3}}\]= RHS                                         Hence proved.

OR

Given x, y and z are in GP, then \[{{y}^{2}}=xz\]             ?(i)

To show

Consider, LHS

Applying \[{{C}_{1}}\to {{C}_{1}}-p{{C}_{2}}-{{C}_{3}},\] we get

On expanding along \[{{C}_{1}},\] we get

\[=(-{{p}^{2}}x-2py-z)(xz-{{y}^{2}})\]

\[=(-{{p}^{2}}x-2py-z)\times 0\]            [using eq. (i)]

= 0 = RHS                  Hence proved