R1= 3i-j , R2=i+3j and R3=-2i+j , find scalars a,b such that R3 = aR1 + bR2

  R3 =aR1+bR2
Putting values of R1,R2 and R3
 $$\begin{gathered} -2i+j=a\left(3i-j\right)+b\left(i+3j\right) \\ -2i+j=3ai-aj+bi+3bj=\left(3a+b\right)i+\left(-a+3b\right)j \end{gathered}$$
Comparing both sides
  $$\begin{gathered} 3a+b=-2---\left(1\right) \\ -a+3b=1---\left(2\right) \\ multiplyingequation\left(2\right)by3andthenaddedtoequation\left(1\right) \\ 3a+b=-2 \\ -3a+9b=3 \\ -------- \\ 10b=1 \\ b=\frac{1}{10} \end{gathered}$$
Putting value of b in equation (1)
  $$\begin{gathered} 3a + \frac{1}{10} = -2 \\ we get, a = - \frac{21}{30} \end{gathered}$$