Answer:
Step-by-step explanation:
[tex]w=(\frac{1}{7}).V_{1}+(\frac{4}{7}).V_{2}[/tex]
We have the following vectors :
[tex]w=\left[\begin{array}{c}3&-1&2\end{array}\right][/tex]
[tex]V_{1}=\left[\begin{array}{c}-3&1&2\end{array}\right][/tex]
[tex]V_{2}=\left[\begin{array}{c}6&-2&3\end{array}\right][/tex]
In order to express [tex]w[/tex] as a linear combination of the vectors [tex]V_{1}[/tex] and [tex]V_{2}[/tex], we will search for [tex]a,b[/tex] ∈ IR such that :
[tex]a.V_{1}+b.V_{2}=w[/tex] (I)
Now we are going to work matrixically with the equation (I) :
[tex]a\left[\begin{array}{c}-3&1&2\end{array}\right]+b\left[\begin{array}{c}6&-2&3\end{array}\right]=\left[\begin{array}{c}3&-1&2\end{array}\right][/tex]
Distributing mathematically and matching ''component to component'' we lead to the following equations :
[tex]-3a+6b=3\\a-2b=-1\\2a+3b=2[/tex]
Working with the system associated matrix :
[tex]\left[\begin{array}{ccc}-3&6&3\\1&-2&-1\\2&3&2\end{array}\right][/tex]
Applying matrix operations we lead to the following equivalent matrix :
[tex]\left[\begin{array}{ccc}1&0&\frac{1}{7}\\0&1&\frac{4}{7}\\0&0&0\end{array}\right][/tex]
In this matrix we obtain that :
[tex]a=\frac{1}{7}[/tex] and [tex]b=\frac{4}{7}[/tex]
We can verify this solution by replacing the values of [tex]a[/tex] and [tex]b[/tex] in the equation (I) :
[tex](\frac{1}{7}).V_{1}+(\frac{4}{7}).V_{2}=w[/tex] ⇒
[tex](\frac{1}{7}).\left[\begin{array}{c}-3&1&2\end{array}\right]+(\frac{4}{7}).\left[\begin{array}{c}6&-2&3\end{array}\right]=\left[\begin{array}{c}3&-1&2\end{array}\right][/tex] ⇒
[tex]\left[\begin{array}{c}3&-1&2\end{array}\right]=\left[\begin{array}{c}3&-1&2\end{array}\right][/tex]
