Question: "Find the LU-factorization of following matrix. A=[12438142613]"

High School
Algebra
Algebra I
Algebra foundations

Oberlaudacu

Answered question

2021-12-28

Find the LU-factorization of following matrix.

A = [\begin{matrix} 1 & 2 & 4 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{matrix}]

Answer & Explanation

trisanualb6

Beginner2021-12-29Added 32 answers

Step 1
$A = [\begin{matrix} 1 & 2 & 4 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{matrix}]$
LU factorization of A
$A = L U$
Where L is lower triangular matrix.
U is upper triangular matrix.
We obtain upper mangular matrix U by Gaussian Elimination method and L is made upd multiplies we used in Gaussian Elimination with 1 at diagonal enties.
Step 2
$A = [\begin{matrix} 1 & 2 & 4 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{matrix}] R_{2} - 3 R_{1} \Rightarrow L_{2, 1} = 3 [\begin{matrix} 1 & 2 & 4 \\ 0 & 2 & 2 \\ 2 & 6 & 13 \end{matrix}]$
$[\begin{matrix} 1 & 2 & 4 \\ 0 & 2 & 2 \\ 2 & 6 & 13 \end{matrix}] R_{3} - 2 R_{1} \Rightarrow L_{3, 1} = 2 [\begin{matrix} 1 & 2 & 4 \\ 0 & 2 & 2 \\ 0 & 2 & 5 \end{matrix}] R_{3} - R_{2} \Rightarrow L_{3, 2} = 1 [\begin{matrix} 1 & 2 & 4 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix}]$
$∴ U = [\begin{matrix} 1 & 2 & 4 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix}]$
and $L = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 2 & 1 & 1 \end{matrix}] [\begin{matrix} L_{2, 1} = 3 \\ L_{3, 1} = 2 \\ L_{3, 2} = 1 \end{matrix}]$
$A = L U = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 2 & 1 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 4 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix}]$

Barbara Meeker

Beginner2021-12-30Added 38 answers

$A = [\begin{matrix} 1 & 2 & 4 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{matrix}]$
and $L = [\begin{matrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{matrix}]$
and $K U = [\begin{matrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{matrix}]$
Now $A = L U$
$[\begin{matrix} 1 & 2 & 4 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{matrix}] [\begin{matrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{matrix}]$
$[\begin{matrix} 1 & 2 & 4 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{matrix}] = [\begin{matrix} u_{11} & u_{12} & u_{13} \\ l_{21} u_{11} & l_{21} u_{22} + u_{22} & l_{22} u_{23} + u_{23} \\ l_{31} u_{11} & l_{31} u_{12} + l_{32} u_{22} & l_{31} u_{13} + l_{32} u_{23} + u_{33} \end{matrix}]$
Now, compaing both the sides, then
$u_{11} = 1, u_{12} = 2, u_{13} = 4$
$l_{21} u_{11} = 3 \Rightarrow l_{21} \times 1 = 3 \Rightarrow l_{21} = 3$
$l_{21} u_{12} + u_{22} = 8 \Rightarrow 3 \times 2 + u_{22} = 8$
$u_{22} = 2$
$l_{21} u_{13} + u_{23} = 14 \Rightarrow 3 \times 4 + u_{23} = 14$
$u_{23} = 14 - 12 = 2$
$u_{23} = 2$
$l_{31} u_{11} = 2 \Rightarrow l_{31} \times 1 = 2 \Rightarrow l_{31} = 2$
$l_{31} u_{12} + l_{32} u_{22} = 6$
$2 \times 2 + l_{}$

karton Expert2022-01-05Added 613 answers

Let

\begin{matrix} A = [\begin{matrix} 1 & 2 & 4 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{matrix}] \\ A = L U \\ L = [\begin{matrix} 1 & 0 & 0 \\ L_{21} & 1 & 0 \\ L_{31} & L_{32} & 1 \end{matrix}], U = [\begin{matrix} U_{11} & U_{12} & U_{13} \\ 0 & U_{22} & U_{23} \\ 0 & 0 & U_{33} \end{matrix}] \\ [\begin{matrix} u_{11} & u_{12} & u_{13} \\ l_{21} u_{11} & l_{21} u_{12} + u_{22} & l_{21} u_{13} + u_{23} \\ l_{31} u_{11} & l_{31} u_{12} + l_{32} u_{22} & l_{31} u_{13} + l_{32} u_{23} + u_{33} \end{matrix}] \\ = [\begin{matrix} 1 & 2 & 4 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{matrix}] \\ u_{11} = 1 \\ u_{12} = 2 \\ u_{13} = 4 \\ l_{21} u_{11} = 3, l_{21} x_{1} = 3 \Rightarrow l_{21} = 3 \\ l_{21} u_{12} + u_{22} = 8 ∴ 3 \times 2 + u_{22} = 8 \\ u_{22} = 2 \\ l_{21} u_{13} + u_{23} = 14 \Rightarrow 3 \times 4 + u_{23} = 14 \\ u_{23} = 2 \\ l_{31} u_{11} = 2 \Rightarrow l_{31} = 2 \\ l_{31} u_{12} + l_{32} u_{32} = 6 \Rightarrow 2 \times 2 + l_{32} \times 2 = 6 \\ l_{32} = 1 \\ l_{31} u_{13} + l_{32} u_{23} + u_{33} = 13 \Rightarrow 8 + 2 + u_{33} = 13 \\ u_{33} = 3 \\ A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 2 & 1 & 1 \end{matrix}] = L, [\begin{matrix} 1 & 2 & 4 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix}] = U \end{matrix}

This is the LU decomposition.

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