Two different ways to express begin{bmatrix}11 end{bmatrix} as a linear combination of v_{1}, c_{2} and v_{3}.

Two different ways to express $\left[\begin{array}{c}1\\ 1\end{array}\right]$
as a linear combination of
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Given vectors are
Calculation:
Let the vector w = $\left[\begin{array}{c}1\\ 1\end{array}\right]$.
The vector w can be represented in terms of as $w=a{v}_{1}+v{v}_{2}+c{v}_{3}.$
Determine the coefficients a, b, c.
Write the vectors
as the columns of a matrix and reduce the matrix in to row reduce echelon form.
$\left[\begin{array}{cccc}1& 2& -3& 1\\ -3& -8& 7& 1\end{array}\right]{R}_{2}\to 3{R}_{1}+{R}_{2}\left[\begin{array}{cccc}1& 2& -3& 1\\ 0& -2& -2& 4\end{array}\right]$
$\left[\begin{array}{cccc}1& 2& -3& 1\\ 0& -2& -2& 4\end{array}\right]{R}_{2}\to -\frac{1}{2}{R}_{2}\left[\begin{array}{cccc}1& 2& -3& 1\\ 0& 1& 1& -2\end{array}\right]$
$\left[\begin{array}{cccc}1& 2& -3& 1\\ 0& 1& 1& -2\end{array}\right]{R}_{1}\to {R}_{1}-2{R}_{2}\left[\begin{array}{cccc}1& 0& -5& 5\\ 0& 1& 1& 2\end{array}\right]$
This gives the following equations,
$a-5c=5$
$b+c=-2$
Here, c is a free variable.
Two find two different ways to write w as a linear combination of ${v}_{1},{v}_{2},{v}_{3},$ choose two different values of c, which will give two different values of corresponding a and b.
Let, $c=1$, this gives,
$a=10$
$b=-3$
Then, the vector w can be expressed as $w=10{v}_{1}-3{v}_{2}+{v}_{3}.$
Let $c=2,$ this gives,
$a=15$
$b=-4$
Then, the vector w can be expressed as $w=15{v}_{1}-4{v}_{2}+2{v}_{3}.$
Therefore, two different ways of expressing $\left[\begin{array}{c}1\\ 1\end{array}\right]$
as a linear combination of