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Linear Algebra
Linear algebra is the detailed study of vector spaces. With applications in such disparate fields as sociology, economics, computer programming, chemistry, and physics, including its essential role in mathematically describing quantum mechanics and the theory of relativity, linear algebra has become one of the most essential mathematical disciplines for the modern world. Please direct all questions regarding matrices, determinants, eigenvalues, eigenvectors, and linear transformations into this category.
2,176 Questions
What is the solution to the linear system -8x plus 2y equals 6 and -4x plus y equals 3?
It works out that x = 0 and y = 3
Let M be the subset of 2x2 matrices A such that det(A)=0 and tr(A'A)=4 (I shall use ' to denote transpose).
Recall that one of the three definitions of a k-dimensional manifold is:
M c R^n is a k-dimensional manifold if for any p in M there is a neighborhood W c R^n of p and a smooth function F:W-->R^n-k so that F^-1(0) = M intersection W and rank(DF(x))=n-k for every x in M intersection W.
In short, what we need to do is find a function F so that the inverse image of the zero vector under F gives M and the rank of the derivative of F is equal to the dimension of the codomain of F.
Let an arbitrary 2x2 matrix be written as:
[a c]
[b d]
Then the two constraints that define M are
1) ad-bc=0
2) a^2+b^2+c^2+d^2=norm(a,b,c,d)^2=4
Define F:R^4-->R^2 by F(a, b, c, d)=(ad-bc, norm(a,b,c,d)^2-4). Then clearly F^-1(0,0)=M. Furthermore, F is smooth on M because the multiplication and addition of smooth functions (a, b, c, d) is also smooth. (Note that we have taken the neighborhood W to be some superset of M. This guaranteed to exist because if we interpret the set of 2x2 matrices as R^4, every point in M has norm 2, so any ball centered at the origin with length greater than 2 will contain M).
All that remains to be done is to check that rank(DF(x))=2 for every x in M. Observe that [DF]= [d -c -b a]
[2a 2b 2c 2d]
We shall now argue by contradiction. Suppose rank(DF) did not equal 2 for every x in M. Then we know that the two rows are linearly dependent i.e.
h[d -c -b a] + k[2a 2b 2c 2d] = 0 and h and k are not both 0.
Suppose h is 0. Then we have 2ka = 2kb = 2kc = 2kd = 0, and since k cannot also be 0, this implies that a=b=c=d=0, therefore norm(a,b,c,d)^2=0. But, (a,b,c,d) must be in M, so this is a contradiction. Hence h cannot be 0.
Now suppose h is nonzero. Then we can divide it out and there exists a, b, c, d and a constant k so that
[d -c -b a] + k[2a 2b 2c 2d] = 0 i.e. we have:
d + 2ka = -c + 2kb = -b + 2kc = a + 2kd = 0. From this we can substitute to obtain:
a(1 - 4k^2) = b(4k^2 - 1) = c(4k^2 - 1) = d(1 - 4k^2) = 0 and hence
a^2(1 - 4k^2)^2 = b^2(4k^2 - 1)^2 = c^2(4k^2 - 1)^2 = d^2(1 - 4k^2)^2 = 0.
Note that (1 - 4k^2)^2 = (4k^2 - 1)^2. Now, adding the four above expressions together we get:
(a^2 + b^2 + c^2 + d^2)(1 - 4k^2)^2 = norm(a,b,c,d)^2(1 - 4k^2)^2 = 0. But, since we require (a,b,c,d) to be in M, this reduces to 4(1 - 4k^2)^2 = 0. This implies that
1 - 4k^2 = 0, and hence k = +/-(1/2).
Now, if k=+1/2, then we have a + d = b - c = 0, therefore d = -a and b = c. Since we have det(A) = 0 as one of our constraints on M, this implies that ad - bc =
-(a^2) - (b^2) = -(a^2 + b^2) = 0, which implies a = b = 0, by the property of norms. But, if a = b = 0, then (a,b,c,d) = (0,0,0,0) and hence norm(a,b,c,d)^2 = 0, which is a contradiction.
We can argue analogously for the case where k = -1/2. Hence, assuming that rank(DF) is not 2 for some (a,b,c,d) in M leads to a contradiction, so we conclude that rank(DF)=2 for all x in M.
Finally, from this result, we conclude that since F:R^4-->R^2=R^(4-2), M must be a 2-dimensional manifold.
What would a 7x2 matrice and a 2x6 matrice form?
A 7*2 matrix (not matrice) and a 2*6 matrix, if multiplied together, will from a 7*6 matrix.
What are the coordinates of the roots for 8x2 - 64x 120 0?
If you mean: 8x2-64x+120 = 0 then x = 3 or x = 5
Complex vocal tics involve meaningful words, phrases or sentences
What is the formula for asquare plus bsquare?
a^2+b^2=c^2. This is the Pythagorean theorem. It explains that when you have a right triangle, that is, a triangle in which one angle is exactly 90 degrees, the length of the line opposite the right angle multiplied by itself will equal the sum of the other two lines also multiplied by themselves.
4x+3y=20 / 2x-3y=4 ..... In this case x=4 and y=4/3
How do you write a c program to determine if a matrix is symmetric or not?
A matrix is symmetric when it is a perfect square and the transpose of the matrix is equal to the original matrix. The implementation is reasonably straightforward and can be done without actually creating the transposed matrix. We simply imagine a dividing line through the main diagonal (top-left to bottom right) and compare the corresponding elements on each side of that diagonal.
bool is_symmetric (int* a, int rows, int cols) {
if (rows!=cols) return false;
for (int row=0, row for (int col=row+1; col if (a[row][col] != a[col][row]) return false; return true; } This works because for every element a[row][col] in a square matrix, the corresponding element on the other side of the main diagonal must be a[col][row]. In other words, we simply transpose the subscripts (hence it is called the transpose). Consider a 4x4 matrix of hexadecimal values: 0 1 2 3 4 5 6 7 8 9 A B C D E F The main diagonal line is identified as elements {0, 5, A, F}. Element {7} lies above this line and it corresponds with element {D} below the line. Element {7} is identified with the subscript [1][3] while element {D} is identified with the subscript [3][1]. Note that the inner loop which traverses the columns always starts at col=row+1. This ensures that we always start with the first element after the diagonal element in the current row, and traverse through the remaining elements of that row. In other words, elements in [row][col] are always in the upper-right of the matrix, above the main diagonal. Meanwhile, the corresponding transpose, element [col][row], lies in the lower-left of the matrix, below the main diagonal. There is no need to test the entire array, traversing every column of every row, because there's no point in testing elements on the diagonal itself (we'd be comparing each element to itself and its given that an element is always equal to itself), and there's no point in comparing the lower-left elements to the upper-right elements when we've already performed the reverse comparison. After all, if X==Y then it follows that Y==X because equality is a transitive operation (the result is the same regardless of which side of the operator we place the operands). Note that in order to replace a square matrix with its transpose, we simply swap the corresponding elements unless the matrix is symmetrical (because, by definition, a symmetrical matrix is equal to its own transpose): void transpose (int* a, int rows, int cols) { if (rows!=cols is_symmetric (a, rows, cols)) return; for (int row=0, row for (int col=row+1; col swap (&a[row][col], &a[col][row]); }
draw it out for ease.
data:
ab + 2 = ac
bc-1 = ac
therefore ab +2 = bc - 1
ab - bc = -3
perimeter is 62. /3 as most values are small difference. average distance is ~21.
ab is smallest, guess it is 19. if so, ac would be 21. bc would be 22. does this add up?
it actually does. i guess there is a mathematical way of doing this other than trial and error like the way i just did it, but i cannot see it right now
How many facts need to have when formulating a system of equations?
At least two - otherwise you have just one equation, not a system.
Prove that (axb)n[(bxc)x(cxa)] = [a]n(bxc)]^2 where a,b,and c are all vectors.
First, multiply out the cross products. Since the cross product of two vectors is itself a vector, we'll give the cross products some names to make this a little easier to understand:
(bxc)=(b2c3-b3c2)i-(b1c3-b3c1)j+(b1c2-b2c1)k = vector d
(cxa)=(c2a3-c3a2)i-(c1a3-c3a1)j+(c1a2-c2a1)k = vector v
(axb)=(a2b3-a3b2)i-(a1b3-a3b1)j+(a1b2-a2b1)k = vector u
=> (axb)n[(bxc)x(cxa)] = un[dxv]
(dxv)=(d2v3-d3v2)i-(d1v3-d3v1)j+(d1v2-d2v1)k = vector w
=> un[dxv] = unw = u1w1 + u2w2 + u3w3
Now replace u and w with their vector coordinates (notice that the negative sign is factored into the middle terms, so the variables are switched).
u1w1 + u2w2 + u3w3= (a2b3-a3b2)w1 + (a3b1-a1b3)w2 + (a1b2-a2b1)w3
= (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+ (a1b2-a2b1)(d1v2-d2v1)
Now we need to expand the v terms back out:
(d2v3-d3v2) = d2(c1a2-c2a1) - d3(c3a1-c1a3) = d2c1a2- d2 c2a1- d3c3a1 + d3c1a3
(d3v1-d1v3) = d3(c2a3-c3a2) - d1(c1a2-c2a1) = d3c2a3 - d3c3a2- d1c1a2+ d1c2a1
(d1v2-d2v1) = d1(c3a1-c1a3) - d2(c2a3-c3a2) = d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2
So: (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+ (a1b2-a2b1)(d1v2-d2v1) = (a2b3-a3b2)(d2c1a2- d2 c2a1- d3c3a1 + d3c1a3) + (a3b1 - a1b3)(d3c2a3 - d3c3a2- d1c1a2+ d1c2a1)+ (a1b2-a2b1)(d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2)
= d2c1a2 a2b3- d2 c2a1 a2b3- d3c3a1 a2b3 + d3c1a3 a2b3- d2c1a2 a3b2+ d2 c2a1 a3b2+ d3c3a1 a3b2- d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1- d1c1a2 a3b1+ d1c2a1 a3b1- d3c2a3 a1b3+ d3c3a2 a1b3+ d1c1a2 a1b3- d1c2a1 a1b3+ d1c3a1 a1b2 - d1c1a3 a1b2 - d2c2a3 a1b2 + d2c3a2 a1b2- d1c3a1 a2b1+ d1c1a3 a2b1+ d2c2a3 a2b1- d2c3a2 a2b1
Some of the terms cancel out, leaving us with;
= d2c1a2 a2b3 - d2 c2a1 a2b3 + d3c1a3 a2b3 - d2c1a2 a3b2 + d3c3a1 a3b2 - d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1 + d1c2a1 a3b1 - d3c2a3 a1b3 + d1c1a2 a1b3 - d1c2a1 a1b3 + d1c3a1 a1b2 - d1c1a3 a1b2 + d2c3a2 a1b2 - d1c3a1 a2b1 + d2c2a3 a2b1 - d2c3a2 a2b1
Now factor out d1 , d2 , and d3
= d1(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 - c3a1 a2b1) + d2(c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + d3(c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3)
Now we can factor out a dot product of ( d1 + d2 + d3):
= ( d1 + d2 + d3)n[(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 - c3a1 a2b1) + (c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + (c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3)]
(Remember, to keep from changing the value of the equation we still need to keep the terms grouped together so that they multiply by the correct d components.)
Now factor out all the "a" components within the brackets:
= ( d1 + d2 + d3)n[(a1 a3{c2b1 - c1b2} + a1 a2{c1b3 - c3b1} + a1 a1{c3b2 - c2b3}) + (a1 a2{c3b2 - c2b3} + a2 a2{c1b3 - c3b1} + a2 a3{c2b1 - c1b2}) + (a1 a3{c3b2 - c2b3} + a2 a3{c1b3 - c3b1} + a3 a3{c2b1 - c1b2})]
= dn[( a1 a3+ a1 a2+ a1 a1)n({c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3}) + (a1 a2 + a2 a2 + a2 a3)n({c1b3- c3b1} + {c2b1- c1b2} + {c3b2- c2b3}) + ( a1 a3+ a2 a3 + a3 a3)n({c3b2- c2b3} + {c1b3- c3b1} + {c2b1- c1b2})]
And we know that {c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3} = (bxc), so we factor out (bxc):
= dn[(bxc)n[(a1 a3+ a1 a2+ a1 a1) + (a1 a2 + a2 a2 + a2 a3) + ( a1 a3+ a2 a3 + a3 a3)]
= dn[(bxc)n[a1(a3+ a2 + a1) + a2 (a1 + a2 +a3) + a3(a1+ a2 + a3)]]
= dn[(bxc)n([a1 + a2 +a3]n[a1 + a2 +a3]) = dn[(bxc)n(a n a)]
(from above, remember that d = (bxc) )
= (bxc)n(bxc)n a n a
= [an (bxc)]^2
How to find the Inverse of a square symmetric matrix?
You can factorize the matrix using LU or LDLT factorization algorithm. inverse of a diagonal matrix (D) is really simple. To find the inverse of L, which is a lower triangular matrix, you can find the answer in this link.
www.mcs.csueastbay.edu/~malek/TeX/Triangle.pdf
Since (A T )-1 = (A-1 )T for all matrix, you'll just have to find inverse of L and D.
What is a graphical linear function?
A linear function that is displayed on a graph or a graphical device. Where the function's different values for n variables can be iterated or cross-referenced with other functions.
When is it important for a matrix to be square?
In the context of matrix algebra there are more operations that one can perform on a square matrix. For example you can talk about the inverse of a square matrix (or at least some square matrices) but not for non-square matrices.
No because they are essentially the same line
What do you have to do if it says half?
Divide by 2
For example
Half of 68=34
Half of 34=17
Half of 17=8.5
What is the cross product method?
If a is to b as c is to d, a x d = b x c.
The product of the means (b & c) equals the product of the extremes (a & d).
