If force (F), velocity (V) and time (T) are taken as fundamental units, then the dimensions of mass are

Let  $\text{m}\propto {\text{F}}^{\text{x}}{\text{V}}^{\text{y}}{\text{T}}^{\text{z}}$

By substituting the following dimensions

$\left[\text{F}\right]=\left[{\text{MLT}}^{-2}\right],\left[\text{V}\right]=\left[{\text{LT}}^{-1}\right],\left[\text{T}\right]=\left[\text{T}\right]$$\left[{\text{M}}^{\text{1}}{\text{L}}^{\text{0}}{\text{T}}^{\text{0}}\right]={\left[{\text{MLT}}^{-2}\right]}^{\text{x}}{\left[{\text{LT}}^{-1}\right]}^{\text{y}}{\left[\text{T}\right]}^{\text{z}}$$\Rightarrow \left[{\text{M}}^{\text{1}}{\text{L}}^{\text{0}}{\text{T}}^{\text{0}}\right]=\left[{\text{M}}^{\text{x}}{\text{L}}^{\text{x}+\text{y}}{\text{T}}^{-2\text{x}-\text{y}+\text{z}}\right]$$\text{x}=1$

$\text{x}+\text{y}=0\Rightarrow \text{y}=-1$

$-2\text{x}-\text{y}+\text{z}=0$

$\Rightarrow -2\times 1-\left(-1\right)+\text{z}=0$

$\Rightarrow \text{z}=0$

Hence $\left[\text{M}\right]=\left[{\text{F}}^{\text{1}}{\text{V}}^{-1}{\text{T}}^{\text{1}}\right]=\left[{\text{FV}}^{-1}\text{T}\right]$