Difference between revisions of "TDSM 8.11"
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(Created page with "Proof for <math> 2\times 2</math> square matrix (higher degree matrixes proof is similar): Let <math> L = \begin{bmatrix} 1 & 0\\ l & 1 \end{bmatrix} </math>, <math> U = \...") |
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</math> | </math> | ||
| − | + | <math> M = | |
\begin{bmatrix} | \begin{bmatrix} | ||
m_1 & m_3\\ | m_1 & m_3\\ | ||
m_2 & m_4 | m_2 & m_4 | ||
\end{bmatrix} | \end{bmatrix} | ||
| − | </math> is | + | </math> is a non-invertable matrix <math>\Rightarrow m_4m_1 - m_2m_3 = 0</math> |
<math>\iff (lu_3+u_2)u_1 - lu_1u_3 = 0</math> | <math>\iff (lu_3+u_2)u_1 - lu_1u_3 = 0</math> | ||
Latest revision as of 13:15, 7 December 2017
Proof for [math] 2\times 2[/math] square matrix (higher degree matrixes proof is similar):
Let [math] L = \begin{bmatrix} 1 & 0\\ l & 1 \end{bmatrix} [/math], [math] U = \begin{bmatrix} u_1 & u_3\\ 0 & u_2 \end{bmatrix} [/math]
[math] \Rightarrow LU = \begin{bmatrix} u_1 & u_3\\ lu_1 & lu_3+u_2 \end{bmatrix} [/math]
[math] M = \begin{bmatrix} m_1 & m_3\\ m_2 & m_4 \end{bmatrix} [/math] is a non-invertable matrix [math]\Rightarrow m_4m_1 - m_2m_3 = 0[/math]
[math]\iff (lu_3+u_2)u_1 - lu_1u_3 = 0[/math]
[math]\iff u_2 u_1 = 0[/math]
[math]\iff u_1 = 0[/math] or [math]u_2 = 0[/math] (proved)
