If time $$(t)$$, velocity $$(v)$$, and angular momentum $$(l)$$ are taken as the fundamental units. Then the dimension of mass $$(m)$$ in terms of $$t$$, $$v$$ and $$l$$ is:

We need to express mass $$m$$ in terms of time $$t$$, velocity $$v$$, and angular momentum $$l$$. Let $$m = t^a v^b l^c$$.

Writing dimensions: $$[m] = \text{M}$$, $$[t] = \text{T}$$, $$[v] = \text{LT}^{-1}$$, and $$[l] = \text{ML}^2\text{T}^{-1}$$.

So $$\text{M}^1\text{L}^0\text{T}^0 = \text{T}^a \cdot (\text{LT}^{-1})^b \cdot (\text{ML}^2\text{T}^{-1})^c = \text{M}^c \, \text{L}^{b+2c} \, \text{T}^{a-b-c}$$.

Matching exponents: for M: $$c = 1$$; for L: $$b + 2c = 0 \Rightarrow b = -2$$; for T: $$a - b - c = 0 \Rightarrow a + 2 - 1 = 0 \Rightarrow a = -1$$.

Therefore $$[m] = [t^{-1} v^{-2} l^1]$$.