is a linear combinations of the vectors
if it can be written in the form
where
are scalars.
Vector
is a linear combinations of the vectors
if it can be written in the form
where
are scalars.
Example 1
Express vector
as a linear combination of the vectors
Solution to Example 1
For vector to be a linear combination of the vectors
, we need to find a scalars
such that (see definition above)
Substitute 
by their components and scalar multiply to obtain the equation
Equality of the components of the vectors above gives the system of equations
The solutions are
Hence
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 \( k \ne - 12 \).
