If a,b,c and d are linearly independent set of vectors and ${{\text{k}}_{\text{1}}}{\text{a + }}{{\text{k}}_{\text{2}}}{\text{b + }}{{\text{k}}_{\text{3}}}{\text{c + }}{{\text{k}}_{\text{4}}}{\text{d = 0}}$=0, then
A. ${{\text{k}}_{\text{1}}}{\text{ + }}{{\text{k}}_{\text{2}}}{\text{ + }}{{\text{k}}_{\text{3}}}{\text{ + }}{{\text{k}}_{\text{4}}}{\text{ = 0}}$
B. ${{\text{k}}_{\text{1}}} + {{\text{k}}_3} = {{\text{k}}_2} + {{\text{k}}_4} = 1$
C. ${{\text{k}}_{\text{1}}} + {{\text{k}}_2} = {{\text{k}}_3} + {{\text{k}}_4}$
D. None of the above
Answer
640.8k+ views
Hint: To solve this question, we need to know the basic theory related to the linearly dependent vector. As we know the span of a set of vectors is the set of all linear combinations of the vectors. A set of vectors is linearly independent if the only solution to ${c_1}{v_1} + {c_2}{v_2} + {c_3}{v_3} + .... + {c_k}{v_k}$ = 0 is ${c_i}$ = 0 for all i.
Complete step-by-step answer:
Given,
${{\text{k}}_{\text{1}}}{\text{a + }}{{\text{k}}_{\text{2}}}{\text{b + }}{{\text{k}}_{\text{3}}}{\text{c + }}{{\text{k}}_{\text{4}}}{\text{d = 0}}$
'a', 'b', 'c' & 'd' are linearly independent vectors.
And also \[{{\text{k}}_{\text{1}}}{\text{,}}{{\text{k}}_{\text{2}}}{\text{,}}{{\text{k}}_{\text{3}}}{\text{,}}{{\text{k}}_{\text{4}}}\]are scalers.
\[\therefore {{\text{k}}_{\text{1}}}{\text{ = 0,}}{{\text{k}}_{\text{2}}}{\text{ = 0,}}{{\text{k}}_{\text{3}}}{\text{ = 0,}}{{\text{k}}_{\text{4}}}{\text{ = 0}}\]
$\therefore $${{\text{k}}_{\text{1}}}{\text{ + }}{{\text{k}}_{\text{2}}}{\text{ + }}{{\text{k}}_{\text{3}}}{\text{ + }}{{\text{k}}_{\text{4}}}{\text{ = 0}}$
$\therefore $ ${{\text{k}}_{\text{1}}} + {{\text{k}}_3} = {{\text{k}}_2} + {{\text{k}}_4} = 0$ and ${{\text{k}}_{\text{1}}} + {{\text{k}}_2} = {{\text{k}}_3} + {{\text{k}}_4}$
Therefore, option (A) and (C) are the correct answers.
Note: A set of two vectors is linearly independent if and only if neither of the vectors is a multiple of the other. A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, any set consisting of a single nonzero vector is linearly independent. Also remember that A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, any set consisting of a single nonzero vector is linearly independent.
Complete step-by-step answer:
Given,
${{\text{k}}_{\text{1}}}{\text{a + }}{{\text{k}}_{\text{2}}}{\text{b + }}{{\text{k}}_{\text{3}}}{\text{c + }}{{\text{k}}_{\text{4}}}{\text{d = 0}}$
'a', 'b', 'c' & 'd' are linearly independent vectors.
And also \[{{\text{k}}_{\text{1}}}{\text{,}}{{\text{k}}_{\text{2}}}{\text{,}}{{\text{k}}_{\text{3}}}{\text{,}}{{\text{k}}_{\text{4}}}\]are scalers.
\[\therefore {{\text{k}}_{\text{1}}}{\text{ = 0,}}{{\text{k}}_{\text{2}}}{\text{ = 0,}}{{\text{k}}_{\text{3}}}{\text{ = 0,}}{{\text{k}}_{\text{4}}}{\text{ = 0}}\]
$\therefore $${{\text{k}}_{\text{1}}}{\text{ + }}{{\text{k}}_{\text{2}}}{\text{ + }}{{\text{k}}_{\text{3}}}{\text{ + }}{{\text{k}}_{\text{4}}}{\text{ = 0}}$
$\therefore $ ${{\text{k}}_{\text{1}}} + {{\text{k}}_3} = {{\text{k}}_2} + {{\text{k}}_4} = 0$ and ${{\text{k}}_{\text{1}}} + {{\text{k}}_2} = {{\text{k}}_3} + {{\text{k}}_4}$
Therefore, option (A) and (C) are the correct answers.
Note: A set of two vectors is linearly independent if and only if neither of the vectors is a multiple of the other. A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, any set consisting of a single nonzero vector is linearly independent. Also remember that A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, any set consisting of a single nonzero vector is linearly independent.
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