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we have : $$ C = AB$$ where $A,B,C$ are square matrices and A is invertible. given $A$ and $C$, we can solve for $B$ and find $B=A^{-1}C$. now, instead, if the first row and column of $C$ is missing, but we have the first row and column of $B$, how can we solve for the rest of $B$? In fact we know all the elements on the first row and col of B ought to be equal to 1.

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Write $A$ in block form as $$A = \begin{bmatrix} a & a_r \\ a_c & A' \end{bmatrix}$$ where $a$ is a scalar, $a_r$ and $a_c$ are $(n-1)$-dimensional row and column vectors, and $A'$ is an $(n-1)\times(n-1)$ square matrix. Similarly do this for $B$ and $C$. Then performing block matrix multiplication, we find that $$C' = a_c b_r + A'B'.$$ Here everything but $B'$ is known, so you can solve for it.

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I don't think you can in general... certainly not in the case that $A,B,C$ are $1\times 1$ matrices!

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why not? in this case B is known and so is A. You can solve for C. –  Alon Amit Apr 13 '11 at 1:34
    
Hm, yes. My answer appears to be nonsense - apologies! –  mac Apr 13 '11 at 8:56
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You have $c_1 = A b_1$, where $c_1$ is the first column of $C$ and $b_1$ is the first column of $B$. So you can recover $c_1$ from $b_1$ and then you know all of $C$ and from it all of $B$.

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The first row of $C$ is also unknown. –  Rahul Apr 13 '11 at 1:43
    
@Rahul, then transpose everything. Recall that $(A^{-1})^{\top}=(A^{\top})^{-1}$. –  lhf Apr 13 '11 at 1:51
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