Question-37

Consider the following system of linear equations:

\[ \begin{equation*} \begin{aligned} x_{1} +x_{2} -2x_{3} & =4\\ x_{1} +3x_{2} -x_{3} & =7\\ 2x_{1} +x_{2} -5x_{3} & =7 \end{aligned} \end{equation*} \]

The LU decomposition of the coefficient matrix is given as:

\[ \begin{equation*} L=\begin{bmatrix} L_{11} & 0 & 0\\ L_{21} & L_{22} & 0\\ L_{31} & L_{32} & L_{33} \end{bmatrix} ,U=\begin{bmatrix} U_{11} & U_{12} & U_{13}\\ 0 & U_{22} & U_{23}\\ 0 & 0 & U_{33} \end{bmatrix} \end{equation*} \]

Which one of the following is the correct combination of values for \(\displaystyle L_{32} ,U_{33}\) and \(\displaystyle x_{1}\)?

• \(L_{32} = 2, U_{33} = -0.5, x_{1} = -1\)
• \(L_{32} = 2, U_{33} = 2, x_1 = -1\)
• \(L_{32} = -0.5, U_{33} = 2, x_1 = 0\)
• \(L_{32} = -0.5, U_{33} = -0.5, x_1 = 0\)

Answer

• \(L_{32} = 2, U_{33} = -0.5, x_{1} = -1\)
• \(L_{32} = 2, U_{33} = 2, x_1 = -1\)
• \(L_{32} = -0.5, U_{33} = 2, x_1 = 0\)
• \(L_{32} = -0.5, U_{33} = -0.5, x_1 = 0\)

Solution

We have:

\[ \begin{equation*} A=\begin{bmatrix} 1 & 1 & -2\\ 1 & 3 & -1\\ 2 & 1 & -5 \end{bmatrix} ,\ x=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} ,\ b=\begin{bmatrix} 4\\ 7\\ 7 \end{bmatrix} \end{equation*} \]

We proceed with Gaussian elimination:

\[ \begin{equation*} \begin{bmatrix} 1 & 1 & -2 & & 4\\ 1 & 3 & -1 & & 7\\ 2 & 1 & -5 & & 7 \end{bmatrix}\xrightarrow{R_{2}\rightarrow R_{2} -R_{1}}\begin{bmatrix} 1 & 1 & -2 & & 4\\ 0 & 2 & 1 & & 3\\ 2 & 1 & -5 & & 7 \end{bmatrix} \end{equation*} \]

\[ \begin{equation*} \begin{bmatrix} 1 & 1 & -2 & & 4\\ 0 & 2 & 1 & & 3\\ 2 & 1 & -5 & & 7 \end{bmatrix}\xrightarrow{R_{3}\rightarrow R_{3} -2R_{1}}\begin{bmatrix} 1 & 1 & -2 & & 4\\ 0 & 2 & 1 & & 3\\ 0 & -1 & -1 & & -1 \end{bmatrix} \end{equation*} \]

\[ \begin{equation*} \begin{bmatrix} 1 & 1 & -2 & & 4\\ 0 & 2 & 1 & & 3\\ 0 & -1 & -1 & & -1 \end{bmatrix}\xrightarrow{R_{3}\rightarrow R_{3} +0.5R_{2}}\begin{bmatrix} 1 & 1 & -2 & & 4\\ 0 & 2 & 1 & & 3\\ 0 & 0 & -0.5 & & 0.5 \end{bmatrix} \end{equation*} \]

Gathering the multipliers in the row reduction and transferring them to \(\displaystyle L\), we have:

\[ \begin{equation*} L=\begin{bmatrix} 1 & 0 & 0\\ 1 & 1 & 0\\ 2 & -0.5 & 1 \end{bmatrix} ,\ U=\begin{bmatrix} 1 & 1 & -2\\ 0 & 2 & 1\\ 0 & 0 & -0.5 \end{bmatrix} \end{equation*} \]

The RREF of the augmented matrix is \(\displaystyle \begin{bmatrix} 1 & 0 & 0 & & 0\\ 0 & 1 & 0 & & 2\\ 0 & 0 & 1 & & -1 \end{bmatrix}\). We see that:

\[ \begin{equation*} x_{1} =0,x_{2} =2,x_{3} =-1 \end{equation*} \]