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Here is a determinant of a $(k+m) \times (k+m)$ Block matrix.

\begin{align} D=\begin{vmatrix} a_{11} &a_{12} & \cdots & a_{1k} &0 &\cdots &0 \\ a_{21}& a_{22}& \cdots & a_{2k} & 0 &\cdots &0 \\ \vdots& \vdots & & \vdots & \vdots & &\vdots\\ a_{k1} & a_{k2} & \cdots & a_{kk} & 0 &\cdots & 0\\ c_{11}& c_{12} & \cdots& c_{1k} & b_{11} & \cdots & b_{1m}\\ \vdots& \vdots & & \vdots & b_{21}&\cdots & b_{2m}\\ c_{m1}& c_{m2} & \cdots & c_{mk} & b_{m1}& \cdots & b_{mm} \end{vmatrix} \end{align}

If I have got a determinant $$D_1= \begin{vmatrix} 0 &\cdots &0&a_{11} &a_{12} & \cdots & a_{1k} \\ 0 &\cdots &0 &a_{21}& a_{22}& \cdots & a_{2k} \\ \vdots& & \vdots & \vdots & \vdots & &\vdots\\ 0 &\cdots & 0&a_{k1} & a_{k2} & \cdots & a_{kk} \\ b_{11} & \cdots & b_{1m}& c_{11}& c_{12} & \cdots& c_{1k}\\ \vdots& & \vdots & \vdots & \vdots& & \vdots\\ b_{m1}& \cdots & b_{mm}&c_{m1}& c_{m2} & \cdots & c_{mk} \end{vmatrix} $$ Then $D_1$ is equal to $(-1)^{k \times m}$$D$.

I know that the existence of the factor -1 is due to the interchange of 2 row, but i have a question on that $k \times m$.In my book,it said that i have to do $k\times m$ times row operations to transform $D_1$ into $D$.However,i thought only k times is needed for $D_1$ transform into $D$.

If i have done 1 times row operation for $D_1$ $$D_1= \begin{vmatrix} 0 &\cdots &0&a_{11} &a_{12} & \cdots & a_{1k} \\ 0 &\cdots &0 &a_{21}& a_{22}& \cdots & a_{2k} \\ \vdots& & \vdots & \vdots & \vdots & &\vdots\\ 0 &\cdots & 0&a_{k1} & a_{k2} & \cdots & a_{kk} \\ b_{11} & \cdots & b_{1m}& c_{11}& c_{12} & \cdots& c_{1k}\\ \vdots& & \vdots & \vdots & \vdots& & \vdots\\ b_{m1}& \cdots & b_{mm}&c_{m1}& c_{m2} & \cdots & c_{mk} \end{vmatrix} =\begin{vmatrix} a_{11} &\cdots &0&0 &a_{12} & \cdots & a_{1k} \\ a_{21} &\cdots &0 &0& a_{22}& \cdots & a_{2k} \\ \vdots& & \vdots & \vdots & \vdots & &\vdots\\ a_{k1} &\cdots & 0& 0& a_{k2} & \cdots & a_{kk} \\ c_{11} & \cdots & b_{1m}&b_{11} & c_{12} & \cdots& c_{1k}\\ \vdots& & \vdots & \vdots & \vdots& & \vdots\\ c_{m1}& \cdots & b_{mm}&b_{m1}& c_{m2} & \cdots & c_{mk} \end{vmatrix} $$ Correct me if i have made any mistakes

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3 Answers 3

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You can think of it in this way: take column $k+1$ (the first one containing $b$s) and swap it with column $k$. Then swap column $k$ with column $k-1$, etc, until you have "bubbled" the original column $k+1$ all the way to the left, after a total of $k$ swaps.

Repeating this process for all $m$ columns on the right requires a total of $km$ swaps.

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The discrepancy is that you have to interchange two adjacent rows. From this it is quite clear you have to perform $k\cdot m$ such operations.

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  • $\begingroup$ Is that means i can't interchange the 1st row with the kth row directly?I have to interchange 1st with 2nd and then with 3rd....until kth.Is this a special case for row operation in block matrices $\endgroup$
    – Vulcan
    Commented Feb 7, 2013 at 16:06
  • $\begingroup$ If you swap the 1st and $k+1$st row directly, that leaves the $k+1$st row in the right place, but the 1st row in the wrong place (unless $m=k$.) $\endgroup$
    – user7530
    Commented Feb 7, 2013 at 16:09
  • $\begingroup$ I have to think about it.Thank you $\endgroup$
    – Vulcan
    Commented Feb 7, 2013 at 16:26
  • $\begingroup$ When i think of the situation of m<k,i got what you mean finally.Thx again $\endgroup$
    – Vulcan
    Commented Feb 7, 2013 at 16:53
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Apply Laplace expansion http://en.wikipedia.org/wiki/Laplace_expansion. you can directly calculate $D_1$

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