Find a Condition that a Vector be a Linear Combination

Linear algebra problems and solutions

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 $\mathbf{v}_1, \mathbf{v}_2$.

 
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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.


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