Find the LU-Factorization of the matrix A below A=[21−1−20321−441−465−2]

Step 1 Given [21−1−20321−441−465−2] Step 2 Solution L=Lower triangular matrix U=Upper triangular matrix A=[21−1−20321−441−465−2]∼[21−101200−30−1−2021](R2→R2+R1R3→R3−R1R4→R4−2R1R5→R5−3R1) ∼[21−101200−300000−3]R4→R4+R2,R5→R5−2R2 ∼[21−101200−3000000]R5→R5−R3 U=[21−101200−3000000] Step 3 And lower triangular matrix that is L is calculated by the entries are the number in the 0 respectively. i.e., L=[100−1101012−10321]

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High School
Algebra
Algebra II
Polynomial factorization

Kaycee Roche

Answered question

2021-02-24

Find the LU-Factorization of the matrix A below

A = [\begin{array}{ccc} 2 & 1 & - 1 \\ - 2 & 0 & 3 \\ 2 & 1 & - 4 \\ 4 & 1 & - 4 \\ 6 & 5 & - 2 \end{array}]

Answer & Explanation

escumantsu

Skilled2021-02-25Added 98 answers

Step 1
Given

[\begin{array}{ccc} 2 & 1 & - 1 \\ - 2 & 0 & 3 \\ 2 & 1 & - 4 \\ 4 & 1 & - 4 \\ 6 & 5 & - 2 \end{array}]

Step 2
Solution
L=Lower triangular matrix
U=Upper triangular matrix

A = [\begin{array}{ccc} 2 & 1 & - 1 \\ - 2 & 0 & 3 \\ 2 & 1 & - 4 \\ 4 & 1 & - 4 \\ 6 & 5 & - 2 \end{array}] \sim [\begin{array}{ccc} 2 & 1 & - 1 \\ 0 & 1 & 2 \\ 0 & 0 & - 3 \\ 0 & - 1 & - 2 \\ 0 & 2 & 1 \end{array}] (\begin{matrix} R_{2} \to R_{2} + R_{1} \\ R_{3} \to R_{3} - R_{1} \\ R_{4} \to R_{4} - 2 R_{1} \\ R_{5} \to R_{5} - 3 R_{1} \end{matrix})

\sim [\begin{array}{ccc} 2 & 1 & - 1 \\ 0 & 1 & 2 \\ 0 & 0 & - 3 \\ 0 & 0 & 0 \\ 0 & 0 & - 3 \end{array}] R_{4} \to R_{4} + R_{2}, R_{5} \to R_{5} - 2 R_{2}

\sim [\begin{array}{ccc} 2 & 1 & - 1 \\ 0 & 1 & 2 \\ 0 & 0 & - 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] R_{5} \to R_{5} - R_{3}

U = [\begin{array}{ccc} 2 & 1 & - 1 \\ 0 & 1 & 2 \\ 0 & 0 & - 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]

Step 3
And lower triangular matrix that is L is calculated by the entries are the number in the 0 respectively.
i.e.,

L = [\begin{array}{ccc} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 1 & 0 & 1 \\ 2 & - 1 & 0 \\ 3 & 2 & 1 \end{array}]

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