find an LU factorization of the given matrix. [22−1404344]

Step 1 To solve, we use the multiplier method. In an elementary row operation is of the form Ri−kRj , where k is called the scalar multiplier. A=[22−1404344] =[22−10−4601112]R2→R2−2R1 R3→R3−32R1 =[22−10−46007]R3→R3−(−14)R2 Step 2 The multipliers were I21=2,I31=32andI32=−14 , so we get L=[11021032−141]

"find an LU factorization of the given matrix...." solved and explained

High School
Algebra
Algebra II
Polynomial factorization

tricotasu

Answered question

2021-03-08

find an LU factorization of the given matrix.

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

Answer & Explanation

Jozlyn

Skilled2021-03-09Added 85 answers

Step 1
To solve, we use the multiplier method.
In an elementary row operation is of the form

R_{i} - k R_{j}

, where k is called the scalar multiplier.

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

= [\begin{array}{ccc} 2 & 2 & - 1 \\ 0 & - 4 & 6 \\ 0 & 1 & \frac{11}{2} \end{array}] R_{2} \to R_{2} - 2 R_{1}

R_{3} \to R_{3} - \frac{3}{2} R_{1}

= [\begin{array}{ccc} 2 & 2 & - 1 \\ 0 & - 4 & 6 \\ 0 & 0 & 7 \end{array}] R_{3} \to R_{3} - (- \frac{1}{4}) R_{2}

Step 2
The multipliers were

I_{21} = 2, I_{31} = \frac{3}{2} and I_{32} = - \frac{1}{4}

, so we get

L = [\begin{array}{ccc} 1 & 1 & 0 \\ 2 & 1 & 0 \\ \frac{3}{2} & - \frac{1}{4} & 1 \end{array}]

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