Given that \(p = \begin{pmatrix} 5 \\ 3 \end{pmatrix}, q = \begin{pmatrix} -1 \\ 2
Given that \(p = \begin{pmatrix} 5 \\ 3 \end{pmatrix}, q = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\) and \(r = \begin{pmatrix} 17 \\ 5 \end{pmatrix}\) and \(r = \alpha r + \beta q\), where \(\alpha\) and \(\beta\) are scalars, express q in terms of r and p.
Explanation
\(p = \begin{pmatrix} 5 \\ 3 \end{pmatrix}; q = \begin{pmatrix} -1 \\ 2 \end{pmatrix}; r = \begin{pmatrix} 17 \\ 5 \end{pmatrix}\)
\(r = \alpha p + \beta q\)
\(\begin{pmatrix} 17 \\ 5 \end{pmatrix} = \alpha (\begin{pmatrix} 5 \\ 3 \end{pmatrix}) + \beta (\begin{pmatrix} -1 \\ 2 \end{pmatrix}\)
\(17 = 5\alpha - \beta ... (1); 5 = 3\alpha + 2\beta .... (2)\)
\((1) \times 2 : 34 = 10\alpha - 2\beta ... (3)\)
\((3) + (2) : 39 = 13\alpha \implies \alpha = 3\)
\(5 = 3(3) + 2\beta \implies 2 \beta = 5 - 9 = -4\)
\(\beta = -2\)
\(r = 3p - 2q \implies 2q = 3p - r\)
\(q = \frac{1}{2} (3p - r)\)