Linear Combinations of Vectors

Definition of Linear Combinations of Vectors

Vector Letter V is a linear combinations of the vectors Vectors if it can be written in the form Linear Combination of Vectors where Scalars r1, r2, ... are scalars.


Examples with Solutions


Example 1
Express vector Vector v as a linear combination of the vectors Vectors u1 and u2

Solution to Example 1
For vector Letter V to be a linear combination of the vectors Vectors u1 and u2, we need to find a scalars Scalars r1 and r2 such that (see definition above)
Linear Combination of Vectors
Substitute Vectors v, u1 and u2 by their components and scalar multiply to obtain the equation
Linear Combination of Vectors
Equality of the components of the vectors above gives the system of equations
System of Equations
The solutions are
Solutions of r1 and r2
Hence
Final Form of Linear Combination



Example 2 \( \) \( \) \( \) Show that the vector \( \textbf {v} = \begin{bmatrix} 1 \\ 3 \\ \end{bmatrix} \) cannot be expressed as a linear combination of the vectors \( \textbf{u}_1 = \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix} 2 \\ 4 \\ \end{bmatrix} \)
Solution to Example 2
We need to show that we cannot find \( r_1 \) and \( r_2 \) such that
\( \begin{bmatrix} 1 \\ 3 \\ \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} + r_2 \begin{bmatrix} 2 \\ 4 \\ \end{bmatrix} \)
Scalar multiply and add the vectors on the right side of the above equation
\( \begin{bmatrix} 1 \\ 3 \\ \end{bmatrix} = \begin{bmatrix} r_1 + 2 r_2 \\ 2 r_1 + 4 r_2\\ \end{bmatrix} \)
Solve the above as a system of equations
\( \begin{array}{lcl} r_1 + 2 r_2 & = & 1 \\ 2 r_1 + 4 r_2 & = & 3 \end{array} \)
Solve the above system using any method. We use the method of elimination.
Multiply the top equation by \( - 2 \)
\( \begin{array}{lcl} -2 r_1 - 4 r_2 & = & -2 \\ 2 r_1 + 4 r_2 & = & 3 \end{array} \)
Add the two equations to obtain
\( 0 = 1 \)
The system has no solution; therefore vector \( \textbf {v} \) cannot be expressed as a linear combinations of the given vectors \( \textbf {u}_1 \) and \( \textbf {u}_2 \).



Example 3
For what values of \( k \) the vector \( \textbf {v} = \begin{bmatrix} -1 \\ 5 \\ \end{bmatrix} \) can be expressed as a linear combination of the vectors \( \textbf{u}_1 = \begin{bmatrix} - 3 \\ 2 \\ \end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix} k \\ 8 \\ \end{bmatrix} \)
Solution to Example 3
We need to find \( r_1 \) and \( r_2 \) such that
\( \begin{bmatrix} -1 \\ 5 \\ \end{bmatrix} = r_1 \begin{bmatrix} -3 \\ 2 \\ \end{bmatrix} + r_2 \begin{bmatrix} k \\ 8 \\ \end{bmatrix} \)
Scalar multiply and add the vectors on the right side of the above equation
\( \begin{bmatrix} -1 \\ 5 \\ \end{bmatrix} = \begin{bmatrix} -3 r_1 + k r_2 \\ 2 r_1 + 8 r_2\\ \end{bmatrix} \)
Rewrite the above as a system of equations
\( \begin{array}{lcl} -3 r_1 + k r_2 & = & -1 \\ 2 r_1 + 8 r_2 & = & 5 \end{array} \)
For the above system to have solutions \( r_1 \) and \( r_2 \), the determinant.
\( Det \begin{bmatrix} -3 & k \\ 2 & 8 \\ \end{bmatrix} = -24 - 8 k\)
must be different from zero. \)
Solve the inequality \( -24 + 2 k \ne 0\)
to obtain \( k \ne - 12 \)
Vector \( \textbf {v} \) can be expressed as a linear combinations of the given vectors \( \textbf {u}_1 \) and \( \textbf {u}_2 \) for any value of \( k \ne - 12 \).


More References and links

  1. Linear Algebra - Questions with Solutions
  2. Linear Algebra and its Applications - 5 th Edition - David C. Lay , Steven R. Lay , Judi J. McDonald
  3. Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres

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