# If $({a_1},{a_2},{a_3})$,$({b_1},{b_2},{b_3})$and $({c_1},{c_2},{c_3})$ are the linearly independent and $x({a_1},{a_2},{a_3}) + y({b_1},{b_2},{b_3}) + z({c_1},{c_2},{c_3})$$ = 0$, then show that $x = y = z = 0$

Answer

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**Hint:**First we have to define what the terms we need to solve the problem are.

Since the value of $a's,b's,c's$ are represented the matrix elements and which are linear independent

A linear combination is an expression constructed from a set of terms by multiplying each term by a constant and adding the results from the linear combination we derive linearly independent where there is a nontrivial linear combination of the vectors that equals the zero vector. If no such linear combination exists, then the vectors are said to be linearly independent.

**Complete step by step answer:**

Since the elements are linearly independent, we frame a $3 \times 3$matrix which is

$\left( {\begin{array}{*{20}{c}}

{{a_1}}&{{a_2}}&{{a_3}} \\

{{b_1}}&{{b_2}}&{{b_3}} \\

{{c_1}}&{{c_2}}&{{c_3}}

\end{array}} \right)$ thus, further proceeding to find the determinant of this matrix

Hence $\left| {\begin{array}{*{20}{c}}

{{a_1}}&{{a_2}}&{{a_3}} \\

{{b_1}}&{{b_2}}&{{b_3}} \\

{{c_1}}&{{c_2}}&{{c_3}}

\end{array}} \right| \ne 0$(since it is linearly independent hence its determinant is non-zero as nonsingular invertible elements)

Thus, to find the linear combinations of elements $ax + by + cz = 0$ will be a general formula where $a,b,c$are non-zero values.

Now approaching the same in given problem we get

${a_1}x + {b_1}y + c{}_1z = 0$

${a_2}x + {b_2}y + c{}_2z = 0$

${a_3}x + {b_3}y + c{}_3z = 0$ [Since the elements $a's,b's,c's$ are linearly independent it cannot able to zero at any point]

Thus for $x({a_1},{a_2},{a_3}) + y({b_1},{b_2},{b_3}) + z({c_1},{c_2},{c_3})$$ = 0$

Hence, we get $x = y = z = 0$(the elements $a's,b's,c's$ are linearly independent are determinant not zero, thus the values multiplied by it will be zero)

Hence for the given problem $x = y = z = 0$ only satisfied the linearly independent condition for$({a_1},{a_2},{a_3})$,$({b_1},{b_2},{b_3})$and $({c_1},{c_2},{c_3})$.

**Note:**Non singular means determinant non zero and thus a matrix is non singular if and only is it is invertible (inverse exists matrix)

Linearly dependent if there is a nontrivial linear combination of the vectors that equals the zero vector. If no such linear combination exists, then the vectors are said to be linearly independent

Also the determinant of the matrix is zero; it is linearly dependent.

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