# Find a Condition that a Vector be a Linear Combination

## Problem 312

Let
$\mathbf{v}=\begin{bmatrix} a \\ b \\ c \end{bmatrix}, \qquad \mathbf{v}_1=\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \qquad \mathbf{v}_2=\begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix}.$ Find the necessary and sufficient condition so that the vector $\mathbf{v}$ is a linear combination of the vectors $\mathbf{v}_1, \mathbf{v}_2$.

Sponsored Links

We give two solutions.

## Solution 1. (Use the range)

The question is equivalent to finding the condition so that the vector $\mathbf{v}$ is in the range of the matrix
$A=[\mathbf{v}_1, \mathbf{v}_2]=\begin{bmatrix} 1 & 2 \\ 2 & -1 \\ 0 &2 \end{bmatrix}.$ The vector $\mathbf{v}$ is in the range $\calR(A)$ if and only if the system $A\mathbf{x}=\mathbf{v}$ is consistent.

We reduce the augmented matrix of the system by elementary row operations as follows.
\begin{align*}
[A\mid \mathbf{v}]=\left[\begin{array}{rr|r}
1 & 2 & a \\
2 &-1 &b \\
0 & 2 & c
\end{array}\right] \xrightarrow{R_2-2R_1}
\left[\begin{array}{rr|r}
1 & 2 & a \\
0 &-5 &b-2a \\
0 & 2 & c
\end{array}\right]\6pt] \xrightarrow{R_2+3R_3} \left[\begin{array}{rr|r} 1 & 2 & a \\ 0 &1 &b-2a+3c \\ 0 & 2 & c \end{array}\right]\\ \xrightarrow{R_3-2R_2} \left[\begin{array}{rr|r} 1 & 2 & a \\ 0 &1 &b-2a+3c \\ 0 & 0 & 4a-2b-5c \end{array}\right]. \end{align*} The last matrix is in echelon form and the system is consistent if and only if 4a-2b-5c=0. Therefore, the condition that \mathbf{v} be a linear combination of \mathbf{v}_1, \mathbf{v}_2 is 4a-2b-5c=0. ## Solution 2. (Use the cross product) Note that the vectors \mathbf{v}_1, \mathbf{v}_2 spans a plane P in \R^3. Thus, the vector \mathbf{v} is a linear combination of \mathbf{v}_1, \mathbf{v}_2 if and only if \mathbf{v} lies on the plane P. The cross product \[\mathbf{v}_1\times \mathbf{v}_2=\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}\times \begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix}= \begin{bmatrix} \begin{vmatrix} 2 & -1\\ 0& 2 \end{vmatrix} \\[10pt] – \begin{vmatrix} 1 & 2\\ 0& 2 \end{vmatrix} \\[10pt] \begin{vmatrix} 1 & 2\\ 2& -1 \end{vmatrix} \end{bmatrix}=\begin{bmatrix} 4 \\ -2 \\ -5 \end{bmatrix} is perpendicular to the plane $P$.

Therefore, the vector $\mathbf{v}$ is on the plane if and only if the dot (inner) product
$\mathbf{v}\cdot (\mathbf{v}_1\times \mathbf{v}_2)=0.$ Namely,
$\begin{bmatrix} a \\ b \\ c \end{bmatrix}\cdot \begin{bmatrix} 4 \\ -2 \\ -5 \end{bmatrix}=4a-2b-5c=0,$ and we obtained the same condition as in Solution 1.

Sponsored Links

### More from my site

#### You may also like...

##### Intersection of Two Null Spaces is Contained in Null Space of Sum of Two Matrices

Let $A$ and $B$ be $n\times n$ matrices. Then prove that $\calN(A)\cap \calN(B) \subset \calN(A+B),$ where $\calN(A)$ is the null...

Close