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
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