First, we need to recall that a linear equation does not involve any products or roots of variables. All variables occur only to the first power and do not appear as arguments for trigonometric, logarithmic, or exponential functions. For example, x + √y = 4, y = sin x, and 2x + y - z + yz = 5 are not linear.
To solve a system of equations such as
3x + y = 5
2x - y = 3
all information required for the solution is emboded in the augmented matrix (imagine that I put those information into a rectangular arrays)
3 1 5
2 -1 3
and that the solution can be obtained by performing appropriate operations on this matrix.
The matrix of this system linear equations is a square matrix A such as
3 1
2 -1
Think this matrix as
a b
c d
To find an inverse of this square matrix A (2 x 2), we need to find a matrix B of the same size such that AB = I and BA = I, then A is said to be invertible and B is called the inverse of A. If no such a matrix can be found, then A is said to be singular.
An invertible matrix has exactly one inverse.
A square matrix A is invertible if ad - bc ≠ 0 (where ad - bc is the determinant)
The formula of finding the inverse of a square matrix A is
A-1 = [1/(ad - bc)][d -b the second row -c a](I'm sorry, I can't draw the arrays)
So let's find the inverse of our example.
A-1 = [1/(-3 -2)][-1 -1 the second row -2 3] = [-1/-5 -1/-5 the sec. row -2/-5 3/-5] =
1/5 1/5
2/5 -3/5
A n x m matrix cannot have an inverse. A n x n matrix may or may not have an inverse.
To find the inverse of a n x n matrix we should to adjoin the identity matrix to the right side of A, thereby producing a matrix of the form [A | I]. Then we should apply row opperations to this matrix until the left side is reduced to I. This opperations will convert the right side to A-1, so the final matrix will have the form [I |A-1 ].
(There are many other methods how to find the inverse of a n x n matrix, but I can't show them by examples. I am so sorry that I can't be so much useful to you).
a linear equation
By elimination and substitution
An independent system of linear equations is a set of vectors in Rm, where any other vector in Rm can be written as a linear combination of all of the vectors in the set. The vector equation and the matrix equation can only have the trivial solution (x=0).
4x + 2y = 6
The main advantage is that many situations cannot be adequately modelled by a system of linear equations. The disadvantage is that the system can often get very difficult to solve.
Arthur Cayley
It is a system of linear equations which does not have a solution.
No....not necessary
If you have a system, which can be expressed as a set of linear equations, then you can utilize matrices to help solve it. One example is an electrical circuit which uses linear devices (example are constant voltage sources and resistive loads). To find the current through each device, a set of linear equations is derived.
Yes
a linear equation
When its matrix is non-singular.
This is a linear algebra question and it is incomplete since there are no equation which have to be solved.
By elimination and substitution
linear system is like a chemistry equation or math equation where on both sides it must balance. Nonlinear is a math equation or physics that does not appear to have a direct answer just like chaos theory. lulu254ever
An independent system of linear equations is a set of vectors in Rm, where any other vector in Rm can be written as a linear combination of all of the vectors in the set. The vector equation and the matrix equation can only have the trivial solution (x=0).
4x + 2y = 6