Question

How do you find the LU factorization \[A = \left( {\begin{array}{*{20}{c}}
  1&1 \\
  { - 3}&1
\end{array}} \right)\]such that L is a unit diagonal?

Answer 1

Hint: LU decomposition of a matrix is the factorization of a given square matrix into two triangular matrices, one upper and one lower triangular matrix. Here in the given question we need to break the matrix into two forms and then simplify accordingly to get the result.

Formulae Used:
LU decomposition formulae for a matrix A
\[
 A = L \times U \\
   \Rightarrow \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&0 \\
  {{l_{21}}}&1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  {{u_{11}}}&{{u_{12}}} \\
  0&{{u_{22}}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {{u_{11}}}&{{u_{12}}} \\
  {{l_{21}}{u_{11}}}&{{l_{21}}{u_{12}} + {u_{22}}}
\end{array}} \right) \\
 \]

Complete step by step solution:
To solve the given matrix we are going to use the LU factorization formulae which states that for any matrix say A, LU matrix can be determined as:
\[
   A = L \times U \\
   \Rightarrow \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&0 \\
  {{l_{21}}}&1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  {{u_{11}}}&{{u_{12}}} \\
  0&{{u_{22}}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {{u_{11}}}&{{u_{12}}} \\
  {{l_{21}}{u_{11}}}&{{l_{21}}{u_{12}} + {u_{22}}}
\end{array}} \right) \\
 \]
Now, here we know the value for the matrix A, hence after putting values in the upper equation, which is the formulae for the LU decomposition, we can get the answer after comparing on the both side of the elements of the given matrix, on solving we get:
\[ \Rightarrow \left( {\begin{array}{*{20}{c}}
  1&1 \\
  { - 3}&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {{u_{11}}}&{{u_{12}}} \\
  {{l_{21}}{u_{11}}}&{{l_{21}}{u_{12}} + {u_{22}}}
\end{array}} \right)\]
Here on comparison we get:
\[
   \Rightarrow {u_{11}} = 1 \\
   \Rightarrow {u_{12}} = 1 \\
   \Rightarrow {l_{21}}{u_{11}} = - 3 \\
   \Rightarrow {l_{21}} \times 1 = - 3 \\
   \Rightarrow {l_{21}} = - 3 \\
   \Rightarrow {l_{21}}{u_{12}} + {u_{22}} = 1 \\
   \Rightarrow ( - 3) \times 1 + {u_{22}} = 1 \\
   \Rightarrow {u_{22}} = 1 + 3 = 4 \\
 \]
Now writing LU decomposition we have to put these values in the formulae, on solving we get:
\[ \Rightarrow \left( {\begin{array}{*{20}{c}}
  1&1 \\
  { - 3}&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&0 \\
  {{l_{21}}}&1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  {{u_{11}}}&{{u_{12}}} \\
  0&{{u_{22}}}
\end{array}} \right)\]
\[ \Rightarrow \left( {\begin{array}{*{20}{c}}
  1&1 \\
  { - 3}&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&0 \\
  { - 3}&1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  1&1 \\
  0&4
\end{array}} \right)\]
This is our required matrix, after LU decomposition of the given matrix in the question.

Note: The given question is of matrix, in which we have to decompose the given matrix into two matrix, known as LU decomposition, in this we know that for decomposition of matrix we assume that the two decomposed matrix are upper and lower triangular matrix, after splitting the matrix and writing in the equation form, we can easily compare and get the decomposed matrix.