Consider the vectors \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), \({{\bf{v}}_3}\), and \({\bf{b}}\) in \({\mathbb{R}^2}\), shown in the figure. Does the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} = {\bf{b}}\) have a solution? Is the solution unique? Use the figure to explain your answers.

From the figure, it is observed that vectors \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), \({{\bf{v}}_3}\), and b have the same initial point, i.e., 0 but different final points.

Moreover, it is known that the vectors whose initial points are the same but final points are different are considered non-collinear vectors.

Thus, vectors \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), \({{\bf{v}}_3}\), and b are non-collinear.

From the figure, the vectors are in \({\mathbb{R}^2}\). Vector b can be written as the linear combination of vectors \({{\bf{v}}_1}\), and \({{\bf{v}}_2}\) (as joining vectors \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), and b forms the vertices of the parallelogram).

Thus, it can be written as shown below:

\({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + \left( 0 \right){{\bf{v}}_3} = {\bf{b}}\)

From the figure, the vectors are in \({\mathbb{R}^2}\). Vector b can also be written as the linear combination of vectors \({{\bf{v}}_1}\), and \({{\bf{v}}_3}\) (as joining vectors \({{\bf{v}}_1}\), \({{\bf{v}}_3}\), and b forms the vertices of the parallelogram).

Thus, it can be written as shown below:

\({x_1}{{\bf{v}}_1} + \left( 0 \right){{\bf{v}}_2} + {x_3}{{\bf{v}}_3} = {\bf{b}}\)

On substituting \({x_1} = 0\), it is observed that the solution is \(\left( {0,{x_1},{x_2}} \right)\), or vice versa.

From the above observations, the equation has more than two solutions, or many solutions.

Thus, the solution cannot be unique, and it has infinitely many solutions.