# If possible, find scalars c_1, c_2, text{ and } c_3 so that c_1begin{bmatrix}1 2-3 end{bmatrix}+c_2begin{bmatrix}-1 11 end{bmatrix}+c_3begin{bmatrix}-1 4-1 end{bmatrix}=begin{bmatrix}2 -23 end{bmatrix}

If possible, find scalars so that
${c}_{1}\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]+{c}_{2}\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]+{c}_{3}\left[\begin{array}{c}-1\\ 4\\ -1\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]$
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Delorenzoz
Step 1
Consider the equation:
${c}_{1}\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]+{c}_{2}\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]+{c}_{3}\left[\begin{array}{c}-1\\ 4\\ -1\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]\dots \left(1\right)$
Step 2
Now, when a matrix is multiplied by a scalar quantity, then all elements of matrix is multiplied by that scalar quantity.
Using this principle, rewrite equation (1) as:
$\left[\begin{array}{c}{c}_{1}\\ 2{c}_{1}\\ -3{c}_{1}\end{array}\right]+\left[\begin{array}{c}-{c}_{2}\\ {c}_{2}\\ {c}_{2}\end{array}\right]+\left[\begin{array}{c}-{c}_{3}\\ 4{c}_{3}\\ -{c}_{3}\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]\dots \left(2\right)$
step 3
Now again,
Addition of matrices can be done by adding same position of elements together, thus creating a new single matrix.
From (2):
$\left[\begin{array}{c}{c}_{1}-{c}_{2}-{c}_{3}\\ 2{c}_{1}+{c}_{2}+4{c}_{3}\\ -3{c}_{1}+{c}_{2}-{c}_{3}\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]$
Step 4
Equating rows from both sides of matrices:
${c}_{1}-{c}_{2}-{c}_{3}=2$
${c}_{1}=2+{c}_{2}+{c}_{3}\dots \left(3\right)$
and
$2{c}_{1}+{c}_{2}+4{c}_{3}=-2$

$3{c}_{2}+6{c}_{3}=-6$
${c}_{2}+2{c}_{3}=-3\dots \left(4\right)$
Step 5
and
$-3{c}_{1}+{c}_{2}-{c}_{3}=3$

$-2{c}_{2}-4{c}_{3}=9$
${c}_{2}+2{c}_{3}=-4.5\dots \left(5\right)$
From (4) and (5): Since L.H.S. of both equations are same but R.H.S. are not same, It is impossible to find the scalar quantity .
Jeffrey Jordon
Jeffrey Jordon