LECTURE 11-25-13 MATRICES Matrices: a rectangular ...
LECTURE 11-25-13 MATRICES Matrices: a rectangular array of numbers that ovey certain rules of algebra. An m x n matrix consists of m rows and n columns. Example 1: 2 x 3 matrix: a = 88- 2, 1, 3
The eigenvalue problem to be solved becomes evident when we write the equations of motion in teh form: xA '' HtL = -w2 xA HtL xB '' HtL = -w2 xB HtL xC '' HtL = -w2 xC HtL Now we construct a matrix of the coefficients of xA , xB , xC in the equations of motion. Printed by Wolfram Mathematica Student Edition
The frequencies are the square root of the negative of the eigenvalues, and the vibrational modes are the eigenvectors of the resulting matrix
form: xA '' HtL = -w2 xA HtL xB '' HtL = -w2 xB HtL
10 - matrices and eigens.nb
xC '' HtL = -w2 xC HtL Now we construct a matrix of the coefficients of xA , xB , xC in the equations of motion. The frequencies are the square root of the negative of the eigenvalues, and the vibrational modes are the eigenvectors of the resulting matrix Clear@k, mA, mB, mCD; k k k k k k k matrix = ::, , 0>, : , -2 * , >, :0, ,>>; mA mA mB mB mB mC mC matrix êê MatrixForm k
- mA k mB
0
k mA 2k - mB k mC
0 k mB k - mC
Now we enter the data for our model CO2 and solve the eigenvalue problem: mA = 16.0; mB = 12.0; mC = 16.0; a = 2.5; b = 1.162; d = 7.65; k = 2 * a2 * d; frequencies = Sqrt@- Chop@Eigenvalues@matrixDDD 84.68125, 2.4447, 0< modes = Chop@Eigenvectors@matrixDD 88- 0.331295, 0.883452, - 0.331295
The eigenvalue problem to be solved becomes evident when we write the equations of motion in teh form: xA '' HtL = -w2 xA HtL xB '' HtL = -w2 xB HtL xC '' HtL = -w2 xC HtL Now we construct a matrix of the coefficients of xA , xB , xC in the equations of motion. Printed by Wolfram Mathematica Student Edition
The frequencies are the square root of the negative of the eigenvalues, and the vibrational modes are the eigenvectors of the resulting matrix
form: xA '' HtL = -w2 xA HtL xB '' HtL = -w2 xB HtL
10 - matrices and eigens.nb
xC '' HtL = -w2 xC HtL Now we construct a matrix of the coefficients of xA , xB , xC in the equations of motion. The frequencies are the square root of the negative of the eigenvalues, and the vibrational modes are the eigenvectors of the resulting matrix Clear@k, mA, mB, mCD; k k k k k k k matrix = ::, , 0>, : , -2 * , >, :0, ,>>; mA mA mB mB mB mC mC matrix êê MatrixForm k
- mA k mB
0
k mA 2k - mB k mC
0 k mB k - mC
Now we enter the data for our model CO2 and solve the eigenvalue problem: mA = 16.0; mB = 12.0; mC = 16.0; a = 2.5; b = 1.162; d = 7.65; k = 2 * a2 * d; frequencies = Sqrt@- Chop@Eigenvalues@matrixDDD 84.68125, 2.4447, 0< modes = Chop@Eigenvectors@matrixDD 88- 0.331295, 0.883452, - 0.331295
