# Express a Vector as a Linear Combination of Given Three Vectors

## Problem 298

Let

\[\mathbf{v}_1=\begin{bmatrix}

1 \\

5 \\

-1

\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}

1 \\

4 \\

3

\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}

1 \\

2 \\

1

\end{bmatrix}, \mathbf{b}=\begin{bmatrix}

2 \\

13 \\

6

\end{bmatrix}.\]
Express the vector $\mathbf{b}$ as a linear combination of the vector $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$.

(*The Ohio State University, Linear Algebra Midterm Exam Problem*)

Add to solve later

## Solution.

We want to find scalars $x_1, x_2, x_3$ such that

\[\mathbf{b}=x_1\mathbf{v}_1+x_2\mathbf{v}_2+x_3\mathbf{v}_3.\]
This is equivalent to solving the matrix equation $A\mathbf{x}=\mathbf{b}$, where

\[A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]=\begin{bmatrix}

1 & 1 & 1 \\

5 &4 &2 \\

-1 & 3 & 1

\end{bmatrix} \text{ and } \mathbf{x}=\begin{bmatrix}

x_1 \\

x_2 \\

x_3

\end{bmatrix}.\]

We solve this by Gauss-Jordan elimination.

The augmented matrix is

\[ \left[\begin{array}{rrr|r}

1 & 1 & 1 & 2 \\

5 &4 & 2 & 13 \\

-1 & 3 & 1 & 6

\end{array} \right].\]
We apply elementary row operations as follows.

\begin{align*}

\left[\begin{array}{rrr|r}

1 & 1 & 1 & 2 \\

5 &4 & 2 & 13 \\

-1 & 3 & 1 & 6

\end{array} \right]
\xrightarrow{\substack{R_2-5R_1\\R_3+R_1}}

\left[\begin{array}{rrr|r}

1 & 1 & 1 & 2 \\

0 &-1 & -3 & 3 \\

0 & 4 & 2 & 8

\end{array} \right]\\[10pt]
\xrightarrow{\substack{-R_2\\ \frac{1}{2}R_3}}

\left[\begin{array}{rrr|r}

1 & 1 & 1 & 2 \\

0 &1 & 3 & -3 \\

0 & 2 & 1 & 4

\end{array} \right]
\xrightarrow{\substack{R_1-R_2\\ R_3-2R_2}}

\left[\begin{array}{rrr|r}

1 & 0 & -2 & 5 \\

0 &1 & 3 & -3 \\

0 & 0 & -5 & 10

\end{array} \right]\\[10pt]
\xrightarrow{-\frac{1}{2}R_3}

\left[\begin{array}{rrr|r}

1 & 0 & -2 & 5 \\

0 &1 & 3 & -3 \\

0 & 0 & 1 & -2

\end{array} \right]
\xrightarrow{\substack{R_1+2R_3 \\ R_2-3R_3}}

\left[\begin{array}{rrr|r}

1 & 0 & 0 & 1 \\

0 &1 & 0 & 3 \\

0 & 0 & 1 & -2

\end{array} \right].

\end{align*}

The last matrix is in reduced row echelon form.

Therefore, the solution is

\[x_1=1, x_2=3, x_3=-2.\]
Thus the linear combination we are looking for is

\[\mathbf{b}=\mathbf{v}_1+3\mathbf{v}_2-2\mathbf{v}_3.\]

## Comment.

This is one of the midterm exam 1 problems of linear algebra (Math 2568) at the Ohio State University.

Many of students started their solutions by writing the augmented matrix.

But it is good to start with the linear combination (because that’s what we need to find) and proceed to the augmented matrix to find the coefficients.

### Double check

Once you obtained the linear combination

\[\mathbf{b}=\mathbf{v}_1+3\mathbf{v}_2-2\mathbf{v}_3,\]
you should check whether this equality really holds.

Since vectors $\mathbf{b}, \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are given explicitly, checking this won’t take time.

If the equality doesn’t hold, then you must have made a computational error.

## Midterm 1 problems and solutions

The following list is the problems and solutions/proofs of midterm exam 1 of linear algebra at the Ohio State University in Spring 2017.

- Problem 1 and its solution: Possibilities for the solution set of a system of linear equations
- Problem 2 and its solution: The vector form of the general solution of a system
- Problem 3 and its solution: Matrix operations (transpose and inverse matrices)
- Problem 4 and its solution (The current page): Linear combination
- Problem 5 and its solution: Inverse matrix
- Problem 6 and its solution: Nonsingular matrix satisfying a relation
- Problem 7 and its solution: Solve a system by the inverse matrix
- Problem 8 and its solution:A proof problem about nonsingular matrix

Add to solve later

## 6 Responses

[…] Problem 4 and its solution: Linear combination […]

[…] Problem 4 and its solution: Linear combination […]

[…] Problem 4 and its solution: Linear combination […]

[…] Problem 4 and its solution: Linear combination […]

[…] Problem 4 and its solution: Linear combination […]

[…] Problem 4 and its solution: Linear combination […]