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Solving Systems of Equations using Matrices
A common application of statics is the analysis of structures, which gen- erally involves computing a large number of forces or moments. For instance, say we would like to determine the tensile or compressive force in each mem- ber of a truss (e.g. a railroad bridge). Applying the basic static equilibrium equations (
F = 0,
M = 0) to each member or connection point yields a set (or ”system”) of linear equations, which can usually be solved for each of the unknown variables. A simple example will be used to illustrate the matrix formulation and solution.
Figure 1: Simply Supported Beam
Figure 1 shows a simply supported beam with an applied force of 100 N and two reaction forces R 1 and R 2. Unknown are the two reaction forces. Our goal will be to formulate the system equations, find the equivalent matrix representation, and then solve for the unknowns using two different methods.
1 Formulate the System of Equations
The matrix solution is not magical – we still have to get from the problem statement to a set of linear equations. This first requires identifying the number of unknown quantities we want to solve for. Given n unknowns, then, we must come up with n independent equations in order to solve for everything.
Need n independent equations to solve for n unknowns
A set of n linear equations, each involving n unknowns is shown below in Eq. 1. The coefficients (aij ) are scalar (as opposed to vector) values and must be determined ahead of time (e.g. by applying equilibrium equations to a structure). The unknowns (xi) are what you’re trying to solve for. The solutions (bi) are either given in the problem statement or determined by basic static analysis.
a 11 x 1 + a 12 x 2 + · · · + a 1 nxn = b 1 a 21 x 1 + a 22 x 2 + · · · + a 2 nxn = b 2 .. . an 1 x 1 + an 2 x 2 + · · · + annxn = bn (1)
In our example there are two unknowns, namely the two reaction forces R 1 and R 2. Therefore, we need to find two (independent) equations to solve for both values. Summing forces in the positive y-direction (they’re all aligned with the y-axis in this case), we arrive at the first equation:
R 1 + R 2 = 100 (2)
Summing moments about the left end of the beam, we arrive at the second equation: 5 · R 2 = 2 · 100 (3) Eqs. 2 and 3 now represent a system of equations:
R 1 + R 2 = 100 5 · R 2 = 200
Or, to be more general:
1 · R 1 + 1 · R 2 = 100 0 · R 1 + 5 · R 2 = 200 (4)
First, the coefficients multiplying the unknowns are placed in to the ma- trix A as follows:
A =
[
]
Next, the unknowns are collected in the vector x:
x =
[
R 1
R 2
]
Finally, the solutions form the vector b:
b =
[
]
Then the equivalent matrix representation of Eq. 4 is given by: [ 1 1 0 5
] [
R 1
R 2
]
[
]
3 Solve the Matrix Equation
3.1 Matrix Inversion
The most straightforward method of solving the matrix equation given by Eq. 6 is to rearrange the equation so that the vector of unknowns (x) is left on one side of the equals sign. Starting from Eq. 6, Ax = b, we multiply both sides of the equals sign by the inverse of A. It is important to realize that the inverse of a matrix cannot (in general) be found by taking the reciprocal of each element – rather, it is typically performed by a computer, especially for systems of 3 or more equations. A−^1 Ax = A−^1 b (8) Now this looks a little more complicated at first, except that the left hand side of the equation can be simplified by noting that A−^1 A = I, where I is the identity matrix (ones down the diagonal and zeros everywhere else). The identity matrix, which is basically the matrix equivalent of the number 1, has the nice property that it doesn’t change a vector under multiplication. That is, Ix = xI = x for all x. Therefore, Eq. 8 reduces to:
x = A−^1 b (9)
This can now be entered into a calculator and the solution will be a vector of the n unknowns.
In the beam example, we begin with the matrix equation: [ 1 1 0 5
] [
R 1
R 2
]
[
]
The inverse of A is found either by hand or by using a calculator:
A−^1 =
[
]
A−^1 is then substituted into Eq. 9 and post-multiplied by b to obtain:
x = A−^1 b
[
] [
]
[
]
[
]
Therefore, the reaction forces in the case of the simply supported beam of Figure 1 are given by:
R 1 = 60N R 2 = 40N
3.2 Row Reduction
The other method of finding the solution, x, is to systematically reduce the equations until you’re left with a value for each variable. Doing this by hand can be tedious for large matrices (anything bigger than 2x2, in my opinion). If you reduce a large matrix by hand, it’s possible you’ll make a small algebraic mistake due to the large number of calculations involved. Fortunately, we have computers and calculators, which can reduce a matrix automatically, sparing you the trouble. The matrix reduction technique works as follows: We construct an aug- mented matrix consisting of the original A-matrix side by side with the col-
4 Using a Calculator to Solve Systems of Equa-
tions
Most graphing calculators can solve systems of equations. This includes: HP’s, TI-81, 82, 83, 85, 86, 89, 92. The syntax for each calculator is differ- ent, so I’ve provided links to websites which document how to find an inverse of a matrix and compute row-reduced echelon form on several different kinds of calculators.
Matrix Inversion:
- TI-81 and TI-
- TI-
- TI-
- TI-
- TI-89 and TI-
- HP
Row Reduction:
- TI-81 and TI-82 (source code for program to do RREF)
- TI-
- TI-
- TI-
- TI-89 and TI-
- HP
