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I have three matrices $A_{p\times q},B_{q\times r},C_{r\times s}$ such that $ABC=M_{p\times s}$ and dimensions of matrices are {$p,q,r,s$}

I am increasing value of $r$ from $2$ to $100$ it means adding one column in B and adding one row in C(means I am adding one row and one column at a time not adding $98$ columns and rows at once). Right now when I increase the value r and thus making $B \rightarrow B_{1}$ and $C \rightarrow C_{1}$, I get $AB_{1}C_{1}$ which is obviously $AB_{1}C_{1} \neq ABC$. How can I modify $B, C$ such that $AB_{1}C_{1} = ABC = M_{p\times s}$.

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    Use `p \times q` rather than `pXq` to get $p \times q$ instead of $pXq$. – 2017-02-03
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    I'm very confused by your statement: *"increasing value of r from 2 to 100 it means adding one column in B and adding one row in C."* I thought increasing $r$ from $r = 2$ to $r = 100$ means adding $98$ columns to $B$, not just one column. – 2017-02-03
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    yes but adding one column at a time , not adding 98 columns at once – 2017-02-03

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It sounds like you want to add columns to $B$ and rows to $C$ in such a way that the augmented matrices $B_1,C_1$ satisfy $B_1C_1 = BC$.

In terms of block matrices, it helps to write $$ B_1 = \pmatrix{B & \hat B} \qquad C_1 = \pmatrix{C\\ \hat C} $$ By block-matrix multiplication, we see that $$ B_1C_1 = BC + \hat B \hat C $$ it suffices, then, to select matrices $\hat B$ and $\hat C$ of an appropriate size such that $\hat B \hat C = 0$. For example: if we modify $B$ by adding only zero-columns (so that $\hat B = 0$), then we'll necessarily have $B_1C_1 = BC$, so that $AB_1C_1 = ABC$.

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    I think that might help,but I cannot add zero column – 2017-02-03