Question

Consider A = <0,4,2> , B = <6,-1,0> , and C = <3,0,1>. Find scalars a, b and c such that aA + bB = (c - 1)C.

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Answer 1

$A = \begin{bmatrix} 0\\ 4\\ 2 \end{bmatrix}, B = \begin{bmatrix} 6\\ -1\\ 0 \end{bmatrix}, C = \begin{bmatrix} 3\\ 0\\ 1 \end{bmatrix}$

$aA +bB = (c-1)C$

$a,b,c - ?$

$a\begin{bmatrix} 0\\ 4\\ 2 \end{bmatrix} + b \begin{bmatrix} 6\\ -1\\ 0 \end{bmatrix} = (c-1) \begin{bmatrix} 3\\ 0\\ 1 \end{bmatrix}$

$\begin{cases} 6b = 3c-3 \\ 4a-b = 0\\ 2a = c-1 \end{cases}$ $\begin{cases} 2b = c-1\\ 4a = b\\ 2a = c-1 \end{cases}$

From (1) and (3) $2a = 2b \rightarrow a = b$

From (2) $4a = a \rightarrow 3a = 0 \rightarrow a =0 \rightarrow b = 0$

From (3) $c-1 = 2a \rightarrow c-1 = 0 \rightarrow c =1$

Answer: $a = 0, b = 0, c = 1$