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Orthogonal Matrix
A matrix \underline { T } is an orthogonal matrix if
\underline {{T}} ^ {T} \underline {{T}} = \underline {{T}} \underline {{T}} ^ {T} = \underline {{I}} \tag {A.2.17}
Hence, for an orthogonal matrix, we have
\underline {{{T}}} ^ {- 1} = \underline {{{T}}} ^ {T} \tag {A.2.18}
An orthogonal matrix frequently used is the transformation or rotation matrix T. In two-dimensional space, the transformation matrix relates components of a vector in one coordinate system to components in another system. For instance, the displacement (and force as well) vector components of d expressed in the x-y system are related to those in the \hat { x } { - } \hat { y } system (Figure A–1 and Section 3.3) by
\underline {{\hat {d}}} = \underline {{T}} \underline {{d}} \tag {A.2.19}
or \left\{ \begin{array} { c } { { \hat { d } _ { x } } } \\ { { \hat { d } _ { y } } } \end{array} \right\} = \left[ \begin{array} { c c } { { \cos \theta } } & { { \sin \theta } } \\ { { - \sin \theta } } & { { \cos \theta } } \end{array} \right] \left\{ \begin{array} { c } { { d _ { x } } } \\ { { d _ { y } } } \end{array} \right\} ðA:2:20Þ
where \underline { T } is the square matrix on the right side of Eq. (A.2.20).
Another use of an orthogonal matrix is to change from the local stiffness matrix to a global stiffness matrix for an element. That is, given a local stiffness matrix \underline { { \hat { k } } } for an element, if the element is arbitrarily oriented in the x-y plane, then
\underline {{k}} = \underline {{T}} ^ {T} \hat {\underline {{k}}} \underline {{T}} = \underline {{T}} ^ {- 1} \hat {\underline {{k}}} \underline {{T}} \tag {A.2.21}
Equation (A.2.21) is used throughout this text to express the stiffness matrix \underline { { k } } in the x-y plane.
By further examination of T , we see that the trigonometric terms in \underline { T } can be interpreted as the direction cosines of lines Ox^ and Oy^ with respect to the x-y axes. Thus for Ox^ or \hat { d } _ { x } , we have from Eq. (A.2.20)
\left\langle t _ {1 1} \quad t _ {1 2} \right\rangle = \left\langle \cos \theta \quad \sin \theta \right\rangle \tag {A.2.22}
text_image
y \hat{y} \bar{d} \theta -\hat{d}_y \hat{x} \hat{d}_y \cos \theta d_x \sin \theta \hat{d}_x \hat{d}_y \sin \theta \theta \hat{d}_x \cos \theta \theta O d_x x
Figure A–1 Components of a vector in x-y and x^-y^ coordinates
and for O \hat { y } or \hat { d } _ { y } , we have
\langle t _ {2 1} \quad t _ {2 2} \rangle = \langle - \sin \theta \quad \cos \theta \rangle \tag {A.2.23}
or unit vectors \bar { \bf i } and \bar { \bf j } can be represented in terms of unit vectors \hat { \bar { \bf i } } and \hat { \bar { \bf j } } [also see Section 3.3 for proof of Eq. (A.2.24)] as
\hat {\mathbf {i}} = \mathbf {i} \cos \theta + \mathbf {j} \sin \theta \tag {A.2.24}
\hat {\mathbf {j}} = - \mathbf {i} \sin \theta + \mathbf {j} \cos \theta
and hence
t _ {1 1} ^ {2} + t _ {1 2} ^ {2} = 1 \quad t _ {2 1} ^ {2} + t _ {2 2} ^ {2} = 1 \tag {A.2.25}
and since these vectors ( \hat { \bar { \bf i } } \mathrm { a n d } \hat { \bar { \bf j } } ) are orthogonal, by the dot product, we have
\left\langle t _ {1 1} \overline {{{\mathbf {i}}}} + t _ {1 2} \overline {{{\mathbf {j}}}} \right\rangle \cdot \left\langle t _ {2 1} \overline {{{\mathbf {i}}}} + t _ {2 2} \overline {{{\mathbf {j}}}} \right\rangle
or t _ { 1 1 } t _ { 2 1 } + t _ { 1 2 } t _ { 2 2 } = 0 ðA:2:26Þ
or we say \underline { T } is orthogonal and therefore \underline { { T } } ^ { T } \underline { { T } } = \underline { { T } } \underline { { T } } ^ { T } = \underline { { I } } and that the transpose is its inverse. That is,
\underline {{T}} ^ {T} = \underline {{T}} ^ {- 1} \tag {A.2.27}
Differentiating a Matrix
A matrix is differentiated by differentiating every element in the matrix in the conventional manner. For example, if
\underline {{a}} = \left[ \begin{array}{c c c} x ^ {3} & 2 x ^ {2} & 3 x \\ 2 x ^ {2} & x ^ {4} & x \\ 3 x & x & x ^ {5} \end{array} \right] \tag {A.2.28}
the derivative d \underline { { a } } / d x is given by
\frac {d \underline {{a}}}{d x} = \left[ \begin{array}{c c c} 3 x ^ {2} & 4 x & 3 \\ 4 x & 4 x ^ {3} & 1 \\ 3 & 1 & 5 x ^ {4} \end{array} \right] \tag {A.2.29}
Similarly, the partial derivative of a matrix is illustrated as follows:
\frac {\partial \underline {{a}}}{\partial x} = \frac {\partial}{\partial x} \left[ \begin{array}{l l l} x ^ {2} & x y & x z \\ x y & y ^ {2} & y z \\ x z & y z & z ^ {2} \end{array} \right] = \left[ \begin{array}{l l l} 2 x & y & z \\ y & 0 & 0 \\ z & 0 & 0 \end{array} \right] \tag {A.2.30}
In structural analysis theory, we sometimes differentiate an expression of the form
U = \frac {1}{2} [ x \quad y ] \left[ \begin{array}{l l} a _ {1 1} & a _ {1 2} \\ a _ {1 2} & a _ {2 2} \end{array} \right] \left\{ \begin{array}{l} x \\ y \end{array} \right\} \tag {A.2.31}
where U might represent the strain energy in a bar. Expression (A.2.31) is known as a quadratic form. By matrix multiplication of Eq. (A.2.31), we obtain
U = \frac {1}{2} \left(a _ {1 1} x ^ {2} + 2 a _ {1 2} x y + a _ {2 2} y ^ {2}\right) \tag {A.2.32}
Differentiating U now yields
\frac {\partial U}{\partial x} = a _ {1 1} x + a _ {1 2} y \tag {A.2.33}
\frac {\partial U}{\partial y} = a _ {1 2} x + a _ {2 2} y
Equation (A.2.33) in matrix form becomes
\left\{ \begin{array}{l} \frac {\partial U}{\partial x} \\ \frac {\partial U}{\partial y} \end{array} \right\} = \left[ \begin{array}{l l} a _ {1 1} & a _ {1 2} \\ a _ {1 2} & a _ {2 2} \end{array} \right] \left\{ \begin{array}{l} x \\ y \end{array} \right\} \tag {A.2.34}
A general form of Eq. (A.2.31) is
U = \frac {1}{2} \{X \} ^ {T} [ a ] \{X \} \tag {A.2.35}
Then, by comparing Eq. (A.2.31) and (A.2.34), we obtain
\frac {\partial U}{\partial x _ {i}} = [ a ] \{X \} \tag {A.2.36}
where xi denotes x and y. Here Eq. (A.2.36) depends on matrix a in Eq. (A.2.35) being symmetric.
Integrating a Matrix
Just as in matrix differentiation, to integrate a matrix, we must integrate every element in the matrix in the conventional manner. For example, if
\underline {{a}} = \left[ \begin{array}{c c c} 3 x ^ {2} & 4 x & 3 \\ 4 x & 4 x ^ {3} & 1 \\ 3 & 1 & 5 x ^ {4} \end{array} \right]
we obtain the integration of a as
\int \underline {{a}} d x = \left[ \begin{array}{c c c} x ^ {3} & 2 x ^ {2} & 3 x \\ 2 x ^ {2} & x ^ {4} & x \\ 3 x & x & x ^ {5} \end{array} \right]
In our finite element formulation of equations, we often integrate an expression of the form
\iint [ X ] ^ {T} [ A ] [ X ] d x d y \tag {A.2.37}
The triple product in Eq. (A.2.37) will be symmetric if \varunderline { A } is symmetric. The form \left[ X \right] ^ { T } [ A ] [ X ] is also called a quadratic form. For example, letting
[ A ] = \left[ \begin{array}{c c c} 9 & 2 & 3 \\ 2 & 8 & 0 \\ 3 & 0 & 5 \end{array} \right] \quad [ X ] = \left\{ \begin{array}{l} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right\}
we obtain
\begin{array}{l} \{X \} ^ {T} [ A ] \{X \} = \left[ \begin{array}{c c c} x _ {1} & x _ {2} & x _ {3} \end{array} \right] \left[ \begin{array}{c c c} 9 & 2 & 3 \\ 2 & 8 & 0 \\ 3 & 0 & 5 \end{array} \right] \left\{ \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right\} \\ = 9 x _ {1} ^ {2} + 4 x _ {1} x _ {2} + 6 x _ {1} x _ {3} + 8 x _ {2} ^ {2} + 5 x _ {3} ^ {2} \\ \end{array}
which is in quadratic form.
A.3 Cofactor or Adjoint Method to Determine the Inverse of a Matrix
We will now introduce a method for finding the inverse of a matrix. This method is useful for longhand determination of the inverse of smaller-order square matrices (preferably of order 4 \times 4 or less). A matrix a must be square for us to determine its inverse.
We must first define the determinant of a matrix. This concept is necessary in determining the inverse of a matrix by the cofactor method. A determinant is a square array of elements expressed by
| \underline {{a}} | = | \underline {{a}} _ {i j} | \tag {A.3.1}
where the straight vertical bars, j j, on each side of the array denote the determinant. The resulting determinant of an array will be a single numerical value when the array is evaluated.
To evaluate the determinant of \underline { { \boldsymbol { a } } } , we must first determine the cofactors of [ a _ { i j } ] . The cofactors of [ a _ { i j } ] are given by
C _ {i j} = (- 1) ^ {i + j} | \underline {{d}} | \tag {A.3.2}
where the matrix \underline { { d } } , called the first minor of [ a _ { i j } ] , is matrix \underline { { \boldsymbol { a } } } with row i and column j deleted. The inverse of matrix \underline { { \boldsymbol { a } } } is then given by
\underline {{a}} ^ {- 1} = \frac {\underline {{C}} ^ {T}}{| \underline {{a}} |} \tag {A.3.3}
where \underline { { \boldsymbol { C } } } is the cofactor matrix and | \underline { { a } } | is the determinant of { \underline { { a } } } . To illustrate the method of cofactors, we will determine the inverse of a matrix \underline { { \boldsymbol { a } } } given by
\underline {{a}} = \left[ \begin{array}{r r r} - 1 & 3 & - 2 \\ 2 & - 4 & 2 \\ 0 & 4 & 1 \end{array} \right] \tag {A.3.4}
Using Eq. (A.3.2), we find that the cofactors of matrix \underline{a} are
\begin{array}{l} C _ {1 1} = (- 1) ^ {1 + 1} \left| \begin{array}{c c} - 4 & 2 \\ 4 & 1 \end{array} \right| = - 1 2 \\ C _ {1 2} = (- 1) ^ {1 + 2} \left| \begin{array}{c c} 2 & 2 \\ 0 & 1 \end{array} \right| = - 2 \\ C _ {1 3} = (- 1) ^ {1 + 3} \left| \begin{array}{c c} 2 & - 4 \\ 0 & 4 \end{array} \right| = 8 \tag {A.3.5} \\ C _ {2 1} = (- 1) ^ {2 + 1} \left| \begin{array}{c c} 3 & - 2 \\ 4 & 1 \end{array} \right| = - 1 1 \\ C _ {2 2} = (- 1) ^ {2 + 2} \left| \begin{array}{c c} - 1 & - 2 \\ 0 & 1 \end{array} \right| = - 1 \\ C _ {2 3} = (- 1) ^ {2 + 3} \left| \begin{array}{c c} - 1 & 3 \\ 0 & 4 \end{array} \right| = 4 \\ \end{array}
Similarly, C_{31} = -2 C_{32} = -2 C_{33} = -2 (A.3.6)
Therefore, from Eqs. (A.3.5) and (A.3.6), we have
\underline {{{C}}} = \left[ \begin{array}{c c c} - 1 2 & - 2 & 8 \\ - 1 1 & - 1 & 4 \\ - 2 & - 2 & - 2 \end{array} \right] \tag {A.3.7}
The determinant of \underline{a} is then
| \underline {{a}} | = \sum_ {j = 1} ^ {n} a _ {i j} C _ {i j} \quad \text { with } i \text { any row number } (1 \leqslant i \leqslant n) \tag {A.3.8}
or |\underline{a}| = \sum_{j=1}^{n} a_{ji} C_{ji} with i any column number ( 1 \leqslant i \leqslant n ) (A.3.9)
For instance, if we choose the first rows of \underline{a} and \underline{C} , then i = 1 in Eq. (A.3.8), and j is summed from 1 to 3 such that
\begin{array}{l} | \underline {{{a}}} | = a _ {1 1} C _ {1 1} + a _ {1 2} C _ {1 2} + a _ {1 3} C _ {1 3} \\ = (- 1) (- 1 2) + (3) (- 2) + (- 2) (8) = - 1 0 \tag {A.3.10} \\ \end{array}
Using the definition of the inverse given by Eq. (A.3.3), we have
\underline {{a}} ^ {- 1} = \frac {\underline {{C}} ^ {T}}{| \underline {{a}} |} = \frac {1}{- 1 0} \left[ \begin{array}{c c c} - 1 2 & - 1 1 & - 2 \\ - 2 & - 1 & - 2 \\ 8 & 4 & - 2 \end{array} \right] \tag {A.3.11}
We can then check that
\underline {{a}} \underline {{a}} ^ {- 1} = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]
The transpose of the cofactor matrix is often defined as the adjoint matrix; that is,
\operatorname{adj} \underline {{a}} = \underline {{C}} ^ {T}
Therefore, an alternative equation for the inverse of \underline { { \boldsymbol { a } } } is
\underline {{a}} ^ {- 1} = \frac {\operatorname{adj} \underline {{a}}}{| \underline {{a}} |} \tag {A.3.12}
An important property associated with the determinant of a matrix is that if the determinant of a matrix is zero—that is, jaj ¼ 0—then the matrix is said to be singular. A singular matrix does not have an inverse. The stiffness matrices used in the finite element method are singular until sufficient boundary conditions (support conditions) are applied. This characteristic of the stiffness matrix is further discussed in the text.
d A.4 Inverse of a Matrix by Row Reduction
The inverse of a nonsingular square matrix a can be found by the method of row reduction (sometimes called the Gauss–Jordan method ) by performing identical simultaneous operations on the matrix \underline { { \boldsymbol { a } } } and the identity matrix I (of the same order as \underline { { a } } ) such that the matrix \underline { { \boldsymbol { a } } } becomes an identity matrix and the original identity matrix becomes the inverse of \underline { { a } } . .
A numerical example will best illustrate the procedure. We begin by converting matrix a to an upper triangular form by setting all elements below the main diagonal equal to zero, starting with the first column and continuing with succeeding columns. We then proceed from the last column to the first, setting all elements above the main diagonal equal to zero.
We will invert the following matrix by row reduction.
\underline {{a}} = \left[ \begin{array}{l l l} 2 & 2 & 1 \\ 2 & 1 & 0 \\ 1 & 1 & 1 \end{array} \right] \tag {A.4.1}
To find \underline { { \boldsymbol { a } } } ^ { - 1 } , we need to find \underline { x } such that \underline { { { a } } } \underline { { { x } } } = \underline { { { I } } } . , where
\underline {{x}} = \left[ \begin{array}{c c c} x _ {1 1} & x _ {1 2} & x _ {1 3} \\ x _ {2 1} & x _ {2 2} & x _ {2 3} \\ x _ {3 1} & x _ {3 2} & x _ {3 3} \end{array} \right]
That is, solve { \left[ \begin{array} { l l l } { 2 } & { 2 } & { 1 } \\ { 2 } & { 1 } & { 0 } \\ { 1 } & { 1 } & { 1 } \end{array} \right] } { \underline { { x } } } = { \left[ \begin{array} { l l l } { 1 } & { 0 } & { 0 } \\ { 0 } & { 1 } & { 0 } \\ { 0 } & { 0 } & { 1 } \end{array} \right] }
We begin by writing a and I side by side as
\left[ \begin{array}{c c c c c c} 2 & 2 & 1 & 1 & 0 & 0 \\ 2 & 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right] \tag {A.4.2}
where the vertical dashed line separates \underline { { \boldsymbol { a } } } and \underline { { I } } .
- Divide the first row of Eq. (A.4.2) by 2.
\left[ \begin{array}{c c c c c c} 1 & 1 & \frac {1}{2} & \frac {1}{2} & 0 & 0 \\ 2 & 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right] \tag {A.4.3}
- Multiply the first row of Eq. (A.4.3) by 2 and add the result to the second row.
\left[ \begin{array}{c c c c c c} 1 & 1 & \frac {1}{2} & \frac {1}{2} & 0 & 0 \\ 0 & - 1 & - 1 & - 1 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right] \tag {A.4.4}
- Subtract the first row of Eq. (A.4.4) from the third row.
\left[ \begin{array}{c c c c c c} 1 & 1 & \frac {1}{2} & \frac {1}{2} & 0 & 0 \\ 0 & - 1 & - 1 & - 1 & 1 & 0 \\ 0 & 0 & \frac {1}{2} & - \frac {1}{2} & 0 & 1 \end{array} \right] \tag {A.4.5}
- Multiply the second row of Eq. (A.4.5) by 1 and the third row by 2.
\left[ \begin{array}{c c c c c c} 1 & 1 & \frac {1}{2} & \frac {1}{2} & 0 & 0 \\ 0 & 1 & 1 & 1 & - 1 & 0 \\ 0 & 0 & 1 & - 1 & 0 & 2 \end{array} \right] \tag {A.4.6}
- Subtract the third row of Eq. (A.4.6) from the second row.
\left[ \begin{array}{c c c c c c} 1 & 1 & \frac {1}{2} & \frac {1}{2} & 0 & 0 \\ 0 & 1 & 0 & 2 & - 1 & - 2 \\ 0 & 0 & 1 & - 1 & 0 & 2 \end{array} \right] \tag {A.4.7}
- Multiply the third row of Eq. (A.4.7) by
- \frac 1 2and add the result to the first row.
\left[ \begin{array}{c c c c c c} 1 & 1 & 0 & 1 & 0 & - 1 \\ 0 & 1 & 0 & 2 & - 1 & - 2 \\ 0 & 0 & 1 & - 1 & 0 & 2 \end{array} \right] \tag {A.4.8}
- Subtract the second row of Eq. (A.4.8) from the first row.
\left[ \begin{array}{c c c c c c} 1 & 0 & 0 & - 1 & 1 & 1 \\ 0 & 1 & 0 & 2 & - 1 & - 2 \\ 0 & 0 & 1 & - 1 & 0 & 2 \end{array} \right] \tag {A.4.9}
The replacement of a by the inverse matrix is now complete. The inverse of a is then the right side of Eq. (A.4.9); that is,
\underline {{a}} ^ {- 1} = \left[ \begin{array}{c c c} - 1 & 1 & 1 \\ 2 & - 1 & - 2 \\ - 1 & 0 & 2 \end{array} \right] \tag {A.4.10}
For additional information regarding matrix algebra, consult References [1] and [2].
d References
[1] Gere, J. M., and Weaver, W., Jr., Matrix Algebra for Engineers, Van Nostrand, Princeton, NJ, 1966.
[2] Jennings, A., Matrix Computation for Engineers and Scientists, Wiley, New York, 1977.
Problems
Solve Problems A.1–A.6 using matrices A, B, C, D, and E given by
\underline {{A}} = \left[ \begin{array}{c c} 1 & 0 \\ - 1 & 4 \end{array} \right] \qquad \underline {{B}} = \left[ \begin{array}{c c} 2 & 0 \\ - 2 & 8 \end{array} \right] \qquad \underline {{C}} = \left[ \begin{array}{c c c} 3 & 1 & 0 \\ - 1 & 0 & 3 \end{array} \right]
\underline {{{D}}} = \left[ \begin{array}{l l l} 3 & 1 & 2 \\ 1 & 4 & 0 \\ 2 & 0 & 3 \end{array} \right] \quad \underline {{{E}}} = \left\{ \begin{array}{l} 1 \\ 2 \\ 3 \end{array} \right\}
(Write ‘‘nonsense’’ if the operation cannot be performed.)
A.1 (a) (b) \underline { { A } } + \underline { { C } }
(c) (d) \underline { { D } } \underline { { E } }
(e) (f ) \underline { { C D } }
A.2 Determine \underline { { \boldsymbol { A } } } ^ { - 1 } by the cofactor method.
A.3 Determine \underline { { \boldsymbol { D } } } ^ { - 1 } by the cofactor method.
A.4 Determine \underline { { C } } ^ { - 1 } .
A.5 Determine \underline { { B } } ^ { - 1 } by row reduction.
A.6 Determine \underline { { \boldsymbol { D } } } ^ { - 1 } by row reduction.
A.7 Show that \left( \underline { { { A } } } \underline { { { B } } } \right) ^ { T } = \underline { { { B } } } ^ { T } \underline { { { A } } } ^ { T } by using
\underline {{A}} = \left[ \begin{array}{c c} a _ {1 1} & a _ {1 2} \\ a _ {2 1} & a _ {2 2} \end{array} \right] \quad \underline {{B}} = \left[ \begin{array}{c c c} b _ {1 1} & b _ {1 2} & b _ {1 3} \\ b _ {2 1} & b _ {2 2} & b _ {2 3} \end{array} \right]
A.8 Find \underline { { T } } ^ { - 1 } given that
\underline {{T}} = \left[ \begin{array}{c c} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{array} \right]
and show that \underline { { T } } ^ { - 1 } = \underline { { T } } ^ { T } and hence that \underline { T } is an orthogonal matrix.
A.9 Given the matrices
\underline {{X}} = \left[ \begin{array}{c c} x & y \\ 1 & x \end{array} \right] \qquad \underline {{A}} = \left[ \begin{array}{c c} a & b \\ b & c \end{array} \right]
show that the triple matrix product \underline { { X } } ^ { T } \underline { { A } } \underline { { X } } is symmetric.
A.10 Evaluate the following integral in explicit form:
\underline {{k}} = \int_ {0} ^ {L} \underline {{B}} ^ {T} E \underline {{B}} d x
where
\underline {{B}} = \left[ - \frac {1}{L} \quad \frac {1}{L} \right]
[Note: This is the step needed to obtain Eq. (10.1.16) from Eq. (10.1.15).]
A.11 The following integral represents the strain energy in a bar:
U = \frac {A}{2} \int_ {0} ^ {L} \underline {{d}} ^ {T} \underline {{B}} ^ {T} \underline {{D}} \underline {{B}} \underline {{d}} d x
where
\underline {{d}} = \left\{ \begin{array}{c} d _ {1} \\ d _ {2} \end{array} \right\} \quad \underline {{B}} = \left[ \begin{array}{c c} - \frac {1}{L} & \frac {1}{L} \end{array} \right] \quad \underline {{D}} = E.
Show that d U / d \{ \underline { { d } } \} yields kd, where \underline { { k } } is the bar stiffness matrix given by
\underline {{k}} = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right]
Methods for Solution of Simultaneous Linear Equations
Introduction
Many problems in engineering and mathematical physics require the solution of a system of simultaneous linear algebraic equations. Stress analysis, heat transfer, and vibration analysis are engineering problems for which the finite element formulation for solution typically involves the solving of simultaneous linear equations. This appendix introduces methods applicable to both longhand and computer solutions of simultaneous linear equations. Many methods are available for the solution of equations; for brevity’s sake, we will discuss only some of the more common methods.
B.1 General Form of the Equations
In general, the set of equations will have the form
a _ {1 1} x _ {1} + a _ {1 2} x _ {2} + \dots + a _ {1 n} x _ {n} = c _ {1}
a _ {2 1} x _ {1} + a _ {2 2} x _ {2} + \dots + a _ {2 n} x _ {n} = c _ {2} \tag {B.1.1}
\begin{array}{c c c c} \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \end{array}
a _ {n 1} x _ {1} + a _ {n 2} x _ {2} + \dots + a _ {n n} x _ {n} = c _ {n}
where the \boldsymbol { a _ { i j } } ^ { \prime } \mathbf { s } are the coefficients of the unknown x _ { j } ^ { \prime } \mathbf { s } , and the c _ { i } { ' } \mathbf { s } are the known right-side terms. In the structural analysis problem, the \boldsymbol { a _ { i j } } ^ { \prime } \mathbf { s } are the stiffness coefficients k _ { i j } { ' } \mathbf { s } _ { \mathrm { ~ } } , the \boldsymbol { x _ { j } } ^ { \prime } \mathbf { s } are the unknown nodal displacements d _ { i } { \bf \dot { s } } . , and the c _ { i } { ' } \{ s are the known nodal forces F _ { i } ^ { \mathrm { ~ > ~ } } \mathrm { { s } } .
If the c { \mathrm { { s } } } are not all zero, the set of equations is nonhomogeneous, and all equations must be independent to yield a unique solution. Stress analysis problems typically involve solving sets of nonhomogeneous equations.