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}

asked 2020-10-27
If possible, find scalars \(c_1, c_2, \text{ and } c_3\) so that
\(c_1\begin{bmatrix}1 \\2\\-3 \end{bmatrix}+c_2\begin{bmatrix}-1 \\1\\1 \end{bmatrix}+c_3\begin{bmatrix}-1 \\4\\-1 \end{bmatrix}=\begin{bmatrix}2 \\-2\\3 \end{bmatrix}\)

Answers (1)

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:
\(c_1=2+c_2+c_3 \dots(3)\)
\(2(2+c_2+c_3)+c_2+4c_3=-2 [\because \text{From } (3)]\)
\(c_2+2c_3=-3 \dots (4)\)
Step 5
\(-3(2+c_2+c_3)+c_2-c_3=3 [\because \text{From } (3)]\)
\(c_2+2c_3=-4.5 \dots(5)\)
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 \(c_1, c_2 \text{ and } c_3\).
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