Question

4. Using a computer with four significant digits with chopping, Gaussian elimination with partial pivoting solution to \[ \begin{array}{l} 0.0030 x_{1}+55.23 x_{2}=58.12 \\ 6.239 x_{1}-7.123 x_{2}=47.23 \end{array} \] is (A) \( x_{1}=26.66 ; x_{2}=1.051 \) (B) \( x_{1}=8.769 ; x_{2}=1.051 \) (C) \( x_{1}=8.800 ; x_{2}=1.000 \) (D) \( x_{1}=8.771 ; x_{2}=1.052 \)

          4. Using a computer with four significant digits with chopping, Gaussian elimination with partial pivoting solution to
\[
\begin{array}{l}
0.0030 x_{1}+55.23 x_{2}=58.12 \\
6.239 x_{1}-7.123 x_{2}=47.23
\end{array}
\]
is
(A) \( x_{1}=26.66 ; x_{2}=1.051 \)
(B) \( x_{1}=8.769 ; x_{2}=1.051 \)
(C) \( x_{1}=8.800 ; x_{2}=1.000 \)
(D) \( x_{1}=8.771 ; x_{2}=1.052 \)

4. Using a computer with four significant digits with chopping, Gaussian elimination with partial pivoting solution to

0.0030 x1+55.23 x2=58.12

6.239 x1-7.123 x2=47.23

is
(A) x1=26.66 ; x2=1.051
(B) x1=8.769 ; x2=1.051
(C) x1=8.800 ; x2=1.000
(D) x1=8.771 ; x2=1.052

Added by Darnell C.

Linear Algebra and Its Applications

David C. Lay, Steven R. Lay, Judi J.… 5th Edition

Chapter 2

Instant Answer

Solved by Expert Maitreya E

07/05/2022

Step 1

The given system is: 0.0030x₁ + 55.23x₂ = 58.12 6.239x₁ - 7.123x₂ = 47.23 This can be written in matrix form as: \[ \begin{array}{cc|c} 0.0030 & 55.23 & 58.12 \\ 6.239 & -7.123 & 47.23 \\ \end{array} \] Show more…

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4. Using a computer with four significant digits with chopping, Gaussian elimination with partial pivoting solution to \[ \begin{array}{l} 0.0030 x_{1}+55.23 x_{2}=58.12 \\ 6.239 x_{1}-7.123 x_{2}=47.23 \end{array} \] is (A) \( x_{1}=26.66 ; x_{2}=1.051 \) (B) \( x_{1}=8.769 ; x_{2}=1.051 \) (C) \( x_{1}=8.800 ; x_{2}=1.000 \) (D) \( x_{1}=8.771 ; x_{2}=1.052 \)