# Express a Vector as a Linear Combination of Other Vectors

## Problem 115

Express the vector $\mathbf{b}=\begin{bmatrix} 2 \\ 13 \\ 6 \end{bmatrix}$ as a linear combination of the vectors
$\mathbf{v}_1=\begin{bmatrix} 1 \\ 5 \\ -1 \end{bmatrix}, \mathbf{v}_2= \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \mathbf{v}_3= \begin{bmatrix} 1 \\ 4 \\ 3 \end{bmatrix}.$

(The Ohio State University, Linear Algebra Exam)

## Solution.

We need to find numbers $x_1, x_2, x_3$ satisfying
$x_1 \mathbf{v}_1 + x_2\mathbf{v}_2+x_3\mathbf{v}_3=\mathbf{b}.$

This vector equation is equivalent to the following matrix equation.
$[\mathbf{v}_1 \mathbf{v}_2 \mathbf{v}_3]\mathbf{x}=\mathbf{b}$ or more explicitly we can write it as
$\begin{bmatrix} 1 & 1 & 1 \\ 5 &2 &4 \\ -1 & 1 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}= \begin{bmatrix} 2 \\ 13 \\ 6 \end{bmatrix}.$

Thus the problem is to find the solution of this matrix equation.
Let us consider the augmented matrix for this system to apply Gauss-Jordan elimination.

The augmented matrix is
$\left[\begin{array}{rrr|r} 1 & 1 & 1 & 2 \\ 5 &2 & 4 & 13 \\ -1 & 1 & 3 & 6 \end{array} \right].$ We apply elementary row operations and obtain a matrix in reduced row echelon form as follows.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
5 &2 & 4 & 13 \\
-1 & 1 & 3 & 6
\end{array} \right] \xrightarrow[R_3+R_1]{R_2-5R_1}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
0 &-3 & -1 & 3 \\
0 & 2 & 4 & 8
\end{array} \right]\6pt] \xrightarrow[-R_2]{\frac{1}{2}R_3} \left[\begin{array}{rrr|r} 1 & 1 & 1 & 2 \\ 0 &3 & 1 & -3 \\ 0 & 1 & 2 & 4 \end{array} \right] \xrightarrow{R_2 \leftrightarrow R_3} \left[\begin{array}{rrr|r} 1 & 1 & 1 & 2 \\ 0 & 1 & 2 & 4 \\ 0 &3 & 1 & -3 \end{array} \right]\\[6pt] \xrightarrow[R_3-3R_2]{R_1-R_2} \left[\begin{array}{rrr|r} 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 4 \\ 0 &0 & -5 & -15 \end{array} \right] \xrightarrow{\frac{-1}{5}R_3} \left[\begin{array}{rrr|r} 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 4 \\ 0 &0 & 1 & 3 \end{array} \right]\\[6pt] \xrightarrow[R_2-2R_3]{R_1+R_3} \left[\begin{array}{rrr|r} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -2 \\ 0 &0 & 1 & 3 \end{array} \right]. \end{align*} Therefore the solution for the system is \[x_1=1, x_2=-2, \text{ and } x_3=3 and we obtain the linear combination
$\mathbf{b}=\mathbf{v}_1-2\mathbf{v}_2+3\mathbf{v}_3.$

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### 1 Response

1. 07/19/2017

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