x\(\left[\begin{array}{l} 2 \\ 3 \end{array}\right]\) + y\(\left[\begin{array}{l} -1 \\ 1 \end{array}\right]\) = \(\left[\begin{array}{l} 10 \\ 5 \end{array}\right]\).
or \(\left[\begin{array}{l} 2x \\ 3x \end{array}\right]\) + \(\left[\begin{array}{l} -y \\ y \end{array}\right]\) = \(\left[\begin{array}{l} 10 \\ 5 \end{array}\right]\)
or \(\left[\begin{array}{l} 2 x-y \\ 3 x+y \end{array}\right]\) = \(\left[\begin{array}{l} 10 \\ 5 \end{array}\right]\)
Equating the corresponding elements, we get:
2x – y = 10 … (1)
3x + y = 5 … (2)
Adding (1) and (2), we get
5x – 15 ⇒ x = \(\frac { 15 }{ 5 }\) = 3.
Putting value of x in (1), we get
2x – y = 2 x 3 – y = 10
∴ y = 6 – 10 = – 4.
Hence, x = 3, and y = – 4.