Linear Algebra

The volume will increase in proportion to the increase in absolute temperature.

Graph both and where they cross is the answer to both.

1 × 109

1 yard equals 3 feet so 3/4 is 0.75

The Chinese text The Nine Chapters on the Mathematical Art (Jiu Zhang Suan Shu), from between 300 BC and AD 200, is the first example of the use of matrix methods to solve simultaneous equations. (Wikipedia)

A sparse matrix is one which normally contains a large proportion of elements whose value is 0. There is no exact proportion at which a matrix becomes sparse.

A straight line.

If they are inconsistent and you try to solve them you will get something like:

5=0, which of course isn't true

so... you can't solve them

I do not. I f*cking hate matrices. I multiply sheep.

3. Ayo!

Start at (0, 1 [x, y]). Then, when you go right, for every 1 on the x axis, the line increases 2 on the y axis. The next point would be (1, 3), then (2, 5), then (3, 7), then (4, 9).

Going the other way (to the negatives), you would subtract 2 on the y axis for each 1 on the x axis. The points would be (-1, -1), (-2, -3), (-3, -5), and so on.

// transpose for the sparse matrix void main() { clrscr(); int a[10][10],b[10][10]; int m,n,p,q,t,col; int i,j; printf("enter the no of row and columns :\n"); scanf("%d %d",&m,&n); // assigning the value of matrix for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("a[%d][%d]= ",i,j); scanf("%d",&a[i][j]); } } printf("\n\n"); //displaying the matrix printf("\n\nThe matrix is :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("%d\t",a[i][j]); } printf("\n"); } t=0; printf("\n\nthe non zero value matrix are :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { // accepting only non zero value if(a[i][j]!=0) { t=t+1; b[t][1]=i; b[t][2]=j; b[t][3]=a[i][j]; } } } printf("a[0 %d %d %d\n",m,n,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,b[i][1],b[i][2],b[i][3]); } a[0][1]=n; a[0][2]=m; a[0][3]=t; int s[10],u[10]; if(t>0) { for(i=1;i<=n;i++) { s[i]=0; } for(i=1;i<=t;i++) { s[b[i][2]]=s[b[i][2]]+1; } u[1]=1; for(i=2;i<=n;i++) { u[i]=u[i-1]+s[i-1]; } for(i=1;i<=t;i++) { j=u[b[i][2]]; a[j][1]=b[i][2]; a[j][2]=b[i][1]; a[j][3]=b[i][3]; u[b[i][2]]=j+1; } } printf("\n\n the fast transpose matrix \n\n"); printf("a[0 %d %d %d\n",n,m,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,a[i][1],a[i][2],a[i][3]); } getch(); }

wala...talaga

large numbers

When its determinant is non-zero.

or

When it is a linear transform of the identity matrix.

or

When its rows are independent.

or

When its columns are independent.

These are equivalent statements.

5 of anything are always larger than 1 of the same thing.

Isolate one of the variable using inverse operations. Then solve.

Example: 2x + 4y = 0 --subtract 4y>> 2x = -4y --divide by -4>> -1/2x = y --plug in your variable -->>

2+2

You could write it as X = 2Y, but in that case, you might as well drop one of the variables from the problem and substitute X = 2Y or Y = X/2 throughout.

-6 = n/2 - 10

4 = n/2

2 = n

If one (or more) of the equations can be expressed as a linear combination of the others.

This is equivalent to the statements

the matrix of coefficients does not have an inverse (or is singular),

or

the determinant of the matrix of coefficients is zero.

In a linear inequality the variable is only present raised to the first power (which is usually not explicitly shown). In a quadratic the square of the variable is present (or implied).

The square can be implied in an inequality such as x + 1/x < 6 (x not 0)

This is equivalent to x2 - 6x + 1 < 0

THREE

The mass of a substance can be derived from its density. Density is equal to mass per volume, so if volume is known, divide volume by density to get mass.