Question

# 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}

Matrices
If possible, find scalars $$c_1, c_2, \text{ and } c_3$$ so that
$$c_1\begin{bmatrix}1 \\2\\-3 \end{bmatrix}+c_2\begin{bmatrix}-1 \\1\\1 \end{bmatrix}+c_3\begin{bmatrix}-1 \\4\\-1 \end{bmatrix}=\begin{bmatrix}2 \\-2\\3 \end{bmatrix}$$

2020-10-28
Step 1
Consider the equation:
$$c_1\begin{bmatrix}1 \\2\\-3 \end{bmatrix}+c_2\begin{bmatrix}-1 \\1\\1 \end{bmatrix}+c_3\begin{bmatrix}-1 \\4\\-1 \end{bmatrix}=\begin{bmatrix}2 \\-2\\3 \end{bmatrix} \dots(1)$$
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:
$$\begin{bmatrix}c_1 \\2c_1\\-3c_1 \end{bmatrix}+\begin{bmatrix}-c_2 \\c_2\\c_2 \end{bmatrix}+\begin{bmatrix}-c_3 \\4c_3\\-c_3 \end{bmatrix}=\begin{bmatrix}2 \\-2\\3 \end{bmatrix} \dots(2)$$
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):
$$\begin{bmatrix}c_1-c_2-c_3 \\2c_1+c_2+4c_3\\-3c_1+c_2-c_3 \end{bmatrix}=\begin{bmatrix}2 \\-2\\3 \end{bmatrix}$$
Step 4
Equating rows from both sides of matrices:
$$c_1-c_2-c_3=2$$
$$c_1=2+c_2+c_3 \dots(3)$$
and
$$2c_1+c_2+4c_3=-2$$
$$2(2+c_2+c_3)+c_2+4c_3=-2 [\because \text{From } (3)]$$
$$3c_2+6c_3=-6$$
$$c_2+2c_3=-3 \dots (4)$$
Step 5
and
$$-3c_1+c_2-c_3=3$$
$$-3(2+c_2+c_3)+c_2-c_3=3 [\because \text{From } (3)]$$
$$-2c_2-4c_3=9$$
$$c_2+2c_3=-4.5 \dots(5)$$
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 $$c_1, c_2 \text{ and } c_3$$.