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## Introduction to Linear Systems in Linear Algebra

Introduction to Linear Systems::How linear systems occurLinear systems of equations naturally take place in many places in engineering such as structural analysis dynamics as well as electric circuits. Computers have made it probable to quickly and accurately solve larger and larger systems of equations. Not merely has this allowed engineers to handle more and more complex problems where linear systems naturally occur but has as well prompted engineers to use linear systems to solve problems where they don’t naturally occur such as internal stress-strain analysis, thermodynamics, fluids and chemical processes. It has become standard practice in several areas to analyze a problem by transforming it into linear systems of equations as well as then solving those equations by computer. In this way computers have made linear systems of equations the most frequently utilized tool in modern engineering.

In Figure we demonstrate a truss with equilateral triangles. In Statics you may perhaps use the ‘method of joints’ to write equations for each node of the truss. This set of equations is an instance of a linear system. Making the estimate √3/2 ≈ .8660 the equations for this truss are:

.5 T

_{1}+ T_{2}= R_{1}= f_{1}.866 T

_{1}= − R_{2}= −.433 f_{1}− .5 f_{2}−.5 T

_{1}+ .5 T_{3}+ T_{4}= − f_{1}.866 T

_{1}+ .866 T_{3}= 0− T

_{2}− .5 T_{3}+ .5 T_{5}+ T_{6}= 0.866 T

_{3}+ .866 T_{5}= f_{2}− T

_{4}− .5 T_{5}+ .5 T_{7}= 0,Where Ti stand for the tension in the i-th member of the truss. You could resolve this system by hand with a little time and patience systematically eliminating variables as well as substituting. Clearly it would be a lot better to place the equations on a computer and let the computer solve it. The initial key to dealing with linear systems is to realize that they are equivalent to matrices which contain numbers not variables.

As we converse various aspects of matrices we wish to keep in mind that the matrices that come up in engineering systems are really large. It isn’t unusual in real engineering to use matrices whose dimensions are in the thousands! It is habitually the case that a method that is fine for a 2×2 or 3×3 matrix is entirely inappropriate for a 2000×2000 matrix. We therefore want to emphasize methods that work for large matrices.

An rhombus truss. Joints or else nodes are labelled alphabetically, A, B, . . . and Members (edges) are labelled numerically: 1, 2, . . . . The forces f1 as well as f2 are applied loads and R1, R2 and R3 are reaction forces applied by the supports.

:Linear systems are equal to matrix equationsA system of linear equations,

x

_{1}− 2x_{2}+ 3x_{3}= 42x

_{1}− 5x_{2}+ 12x_{3}= 152x

_{2}− 10x_{3}= −10,Is equal to the matrix equation,

Which is equal to the augmented matrix,

The benefit of the augmented matrix is that it contains merely numbers not variables. The reason this is improved is because computers are much better in dealing with numbers than variables. To solve this system the major steps are called Gaussian elimination and back substitution.

The augmented matrix for the rhombus truss equations is given by:

Note that plenty of the entries are 0. Matrices similar to this called sparse are common in applications and there are methods specifically designed to efficiently handle sparse matrices.

:Triangular matrices as well as back substitutionRegard as a linear system whose augmented matrix happens to be:

Recall that every row represents an equation and each column a variable. The last row stand for the equation 2x3 = 4. The equation is simply solved that is x

_{3}= 2. The second row stand for the equation −x_{2}+ 6x_{3}= 7 but since we know x_{3}= 2 this simplifies to- −x_{2}+ 12 = 7. This is simply solved giving x_{2}= 5. Ultimately since we know x_{2}and x_{3}the first row simplifies to - x_{1}−10 + 6 = 4. Therefore we have x_{1}= 8 and consequently we know the whole solution vector: x = {8, 5, 2}. The process we merely did is called back substitution which is both efficient as well as easily programmed. The property that made it probable to solve the system so easily is that A in this case is upper triangular. In the next section we demonstrate an efficient way to transform an augmented matrix into an upper triangular matrix.:Gaussian EliminationConsider the matrix:

The first step of Gaussian removal is to get rid of the 2 in the (2,1) position through subtracting 2 times the first row from the second row that is (new 2nd = old 2nd - (2) 1st). We are able to do this because it is essentially the same as adding equations which is a valid algebraic operation. This direct to:

There is previously a zero in the lower left corner thus we don’t need to eliminate anything there. To eradicate the third row second column we need to subtract −2 times the second row from the third row (new 3rd = old 3rd - (-2) 2nd):

This is now just exactly the matrix in equation (9.3), which we can now solve by back substitution.

:Mat lab’s matrix solve commandIn Mat lab the standard means to solve a system Ax = b is by the command:

> x = A\b

This command carries out Gaussian removal and back substitution. We are able to do the above computations as follows:

> A = [1 -2 3 ; 2 -5 12 ; 0 2 -10]

> b = [4 15 -10]’

> x = A\b

Next utilize the Mat lab commands above to solve Ax = b when the augmented matrix for the system is:

By entering:

>x1 = A\b Check the result by entering:

> A*x1 - b

You will observe that the resulting answer satisfies the equation exactly. Next try solving utilize the inverse of A:

>x2 = inv(A)*b

This answer is able to be seen to be inaccurate by checking

> A*x2 - b

Therefore we see one of the reasons why the inverse is never utilized for actual computations only for theory.

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