Quiz Chapter 04.06: Gaussian Elimination

MULTIPLE CHOICE TEST

(All Tests)

GAUSSIAN ELIMINATION

(More on Gaussian Elimination)

SIMULTANEOUS LINEAR EQUATIONS

(More on Simultaneous Linear Equations)


Pick the most appropriate answer


1. The goal of forward elimination steps in the Naïve Gauss elimination method is to reduce the coefficient matrix to a(n) _____ matrix.

 
 
 
 

2. Division by zero during forward elimination steps in Naïve Gaussian elimination of the set of equation \left[ A \right] \left[ X \right] = \left[ C \right] implies the coefficient matrix \left[ A \right] is

 
 
 
 

3. Using a computer with four significant digits with chopping, Naïve Gauss elimination solution to

      • 0.0030x_{1} + 55.23x_{2} = 58.12
      • 6.239x_{1} - 7.123x_{2} = 47.23

is

 
 
 
 

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

      • 0.0030x_{1} + 55.23x_{2} = 58.12
      • 6.239x_{1} - 7.123x_{2} = 47.23

is

 
 
 
 

5. At the end of forward elimination steps of Naïve Gauss Elimination method of the following equations

\begin{bmatrix} 4.2857 \times 10^{7}&-9.2307 \times 10^{5}&0&0 \\ 4.2857 \times 10^{7}&-5.4619 \times 10^{5}&-4.2857 \times 10^{7}&5.4619 \times 10^{5} \\ -6.5&-0.15384&6.5&0.15384 \\ 0&0&4.2857 \times 10^{7}&-3.6057 \times 10^{5} \\ \end{bmatrix} \begin{bmatrix} c_{1} \\ c_{2} \\ c_{3} \\ c_{4} \\ \end{bmatrix} = \begin{bmatrix}-7.887 \times 10^{3} \\ 0 \\ 0.007 \\ 0 \\ \end{bmatrix}

the resulting equations in the matrix form are given by

\begin{bmatrix} 4.2857 \times 10^{7}&-9.2307 \times 10^{5}&0&0 \\ 0&3.7688 \times 10^{5}&-4.2857 \times 10^{7}&5.4619 \times 10^{5} \\ 0&0&-26.9140&0.579684 \\ 0&0&0&5.62500 \times 10^{5} \\ \end{bmatrix} \begin{bmatrix} c_{1} \\ c_{2} \\ c_{3} \\ c_{4} \\ \end{bmatrix} = \begin{bmatrix}-7.887 \times 10^{3} \\ 7.887 \times 10^{3} \\1.19530 \times 10^{-2} \\ 1.90336 \times 10^{4} \\ \end{bmatrix}

The determinant of the original coefficient matrix is

 
 
 
 

6. The following data is given for the velocity of the rocket as a function of time. To find the velocity at t=21 s, you are asked to use a quadratic polynomial, v(t) = at^{2} + bt + c to approximate the velocity profile.

t (s) 0 14 15 20 30 35
v \left ( t \right) m/s 0 227.04 362.78 517.35 602.97 901.67

The correct set of equations will find a, b, and c are