Express a Vector as a Linear Combination of Other Vectors (2024)

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)

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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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