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When we deal with square matrix $A$ of size $3\times 3$ and $\det(A)=0$, means our output after transformation lies in lesser dimensional space. Instead if deal with a non-square matrix $B$ of size $2\times 3$ means a 3D vector is transformed to 2D vector. What is the intuitive difference in the two formulations?

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    Using your words: What is the intuitive analogy in the two formulations? I don't see any, the first matrix has $9$ entries, while the second has $6$. – 2017-01-19

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In the first case, the union of all outputs is a plane (or a line or a point) in $\mathbb{R}^3$. In the second case, the union of all outputs is $\mathbb{R}^2$ (or a line or a point in $\mathbb{R}^2$). Intuitively, we have a three dimensional space with a plane in it in the first scenario, while we have nothing but a plane in the second scenario.

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    ok.thanks....But can u tell me any specific application where using these different formulations make sense...i mean wen d output of both is same...y two approaches..? – 2017-01-19
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    First of all, the output is not the same. Even if the output can be interpreted as a plane in both cases, there is a fundamental difference: The plane in the first case can have different positions in the 3-dimensional space. – 2017-01-20
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    Secondly, you usually do not choose the matrix, it is determined by the problem (e.g. optimize the usage of 3 resource for the production of 3 different products). And in practice, if there is a $n\times n$-matrix, and you do not know if it could be singular, you won't try do transform it into a matrix with lesser dimensions. You will choose a robust numerical method that can deal with badly conditioned matrices, instead. – 2017-01-20
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    means its application dependent..thanks...:) – 2017-01-20
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In first case, the mapping is necessarily not onto.

In second case, it is not necessary that the mapping is onto.

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    yeah..thanks :) – 2017-01-19