Hey Buddy,<br>This revives old memories of college days. Well I dont remeber the exact formula but I do remeber how I had implemeted it. The trick lies in remebering that one need not worry about adding or deleteing if we consider the matrix to be circular i.e use a lotta mods.<br>I have written a program which I have tested for order 3 and I believe it will work fine for higher order matrices too.<br>------------------------------------------------------<br>/* Increase the value of N for higer order matrix */<br>#define N 3<br>int a[N][N];<br><br>main() {<br> int i,j;<br> /* Fill the values for the matrix */<br> for (i=0;i<N;i++)<br> for(j=0;j<N;j++)<br> a<i>[j]=1;<br> /* Play around with the values for testing purposes */<br> a[1][0]=2;<br> printf("determinant is %d",det(N,N));<br>}<br><br>det(int start,int size)<br>{<br> int i,deter=0;<br> for(i=N-start;i<size;i++) {<br> if (size == 3)<br> deter+= a[(N-size)%N]<i> * det2((N-size+1)%N,(i+1)%N);<br> else<br> deter+= det((start+1)%N,size-1);<br> printf("temp value is %d\n",deter);<br> }<br> return(deter);<br>}<br><br>det2(int starti,int startj) {<br> int endi,endj;<br> endi = (starti+1) %N;<br> endj = (startj+1) %N;<br> return(a[starti][startj]*a[endi][endj] -<br> a[starti][endj]*a[startj][endi]);<br>}<br>-------------------------------------------------------<br>Adios<br>amit <p> amit<br><a href=mailto: crazy_indian@mailcity.com> crazy_indian@mailcity.com</a><br><a href= > </a><br>to bug is human to debug devine