\(\begin{pmatrix} \frac{17}{4} \\ 7 \end{pmatrix}\)
\(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\)
\(\begin{pmatrix} \frac{17}{4} \\ 3 \end{pmatrix}\)
\(\begin{pmatrix} \frac{17}{4} \\ 2 \end{pmatrix}\)
Correct answer is B
\(a = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\); \(b = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\)
\(\implies 2 \times a = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\) and \(\frac{1}{4} \times b = \begin{pmatrix} -\frac{1}{4} \\ 1 \end{pmatrix}\)
\(\therefore 2a - \frac{1}{4}b = \begin{pmatrix} 4 - \frac{-1}{4} \\ 6 - 1 \end{pmatrix}\)
= \(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\)
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