Definition 3.2.56. Elementary matrices.
A elementary matrix, denoted by \(\boldsymbol{E}_i\text{,}\) derives from applying one elementary row operation to the identity matrix of the same size.
Subsection 3.2 Elementary matrices
Example 3.2.57. Some elementary matrices.
- \begin{equation*} R_1 \rightarrow \displaystyle \frac{1}{2} R_1 \hspace{2cm} \boldsymbol{E} = \left[ \begin{array}{rrr} \displaystyle \frac{1}{2} \amp 0 \amp 0\\ 0 \amp 1 \amp 0\\ 0 \amp 0 \amp 1 \end{array}\right]. \end{equation*}
- \begin{equation*} R_2 \rightarrow R_2 - R_1 \hspace{2cm} \boldsymbol{E} = \left[ \begin{array}{rrr} 1 \amp 0 \amp 0\\ -1 \amp 1 \amp 0\\ 0 \amp 0 \amp 1 \end{array}\right]. \end{equation*}
- \begin{equation*} R_3 \rightarrow R_3 + 2 R_2 \hspace{2cm} \boldsymbol{E} = \left[ \begin{array}{rrrr} 1 \amp 0 \amp 0 \amp 0\\ 0 \amp 1 \amp 0 \amp 0\\ 0 \amp 2 \amp 1 \amp 0\\ 0 \amp 0 \amp 0 \amp 1 \end{array}\right]. \end{equation*}
- \begin{equation*} R_2 \leftrightarrow R_4 \hspace{2cm} \boldsymbol{E} = \boldsymbol{P}_{24} = \left[ \begin{array}{rrrr} 1 \amp 0 \amp 0 \amp 0\\ 0 \amp 0 \amp 0 \amp 1\\ 0 \amp 0 \amp 1 \amp 0\\ 0 \amp 1 \amp 0 \amp 0 \end{array}\right]. \end{equation*}
Subsubsection 3.2.1 Using elementary matrices
Let \(\boldsymbol{A}\) be a \(m\times n\) matrix and let \(\boldsymbol{E} \) be an \(m\times m\) elementary matrix. The matrix multiplication \(\boldsymbol{E}\,\boldsymbol{A}\) is the matrix that results when the same row operation is performed on \(\boldsymbol{A}\) as that performed to produce the elementary matrix \(\boldsymbol{E} \text{.}\)
Example 3.2.58. Applying elementary matrices.
Let
\begin{equation*}
\boldsymbol{A} = \left[ \begin{array}{rrr}
2 \amp 5 \amp 10\\
2 \amp -5 \amp 12\\
1 \amp 10 \amp 5
\end{array}\right],
\hspace{1cm}
\boldsymbol{B} = \left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
2 \amp -5 \amp 12\\
2 \amp 5 \amp 10
\end{array}\right],
\hspace{1cm}
\boldsymbol{C} = \left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0 \amp -10 \amp 2\\
2 \amp 5 \amp 10
\end{array}\right].
\end{equation*}
\(\,\)
-
Find the elementary matrix \(\boldsymbol{E}_1 \) such that \(\boldsymbol{E}_1 \boldsymbol{A} = \boldsymbol{B}.\)\(\boldsymbol{B} \) results from interchanging rows 1 and 3 on \(\boldsymbol{A} \text{,}\) then the elementary matrix is\begin{equation*} \boldsymbol{E}_1 = \left[ \begin{array}{rrr} 0 \amp 0 \amp 1\\ 0 \amp 1 \amp 0\\ 1 \amp 0 \amp 0 \end{array}\right]. \end{equation*}so that\begin{equation*} \boldsymbol{E}_1 \boldsymbol{A} = \left[ \begin{array}{rrr} 0 \amp 0 \amp 1\\ 0 \amp 1 \amp 0\\ 1 \amp 0 \amp 0 \end{array}\right] \left[ \begin{array}{rrr} 2 \amp 5 \amp 10\\ 2 \amp -5 \amp 12\\ 1 \amp 10 \amp 5 \end{array}\right] = \left[ \begin{array}{rrr} 1 \amp 10 \amp 5\\ 2 \amp -5 \amp 12\\ 2 \amp 5 \amp 10 \end{array}\right]. \end{equation*}\(\, \)
-
Find the elementary matrix \(\boldsymbol{E}_2 \) such that \(\boldsymbol{E}_2 \boldsymbol{B} = \boldsymbol{C}.\)\(\boldsymbol{C} \) results from the row operation \(R_2 \rightarrow R_2 - R_3\) applied to \(\boldsymbol{B} \text{,}\) then the elementary matrix is\begin{equation*} \boldsymbol{E}_2 = \left[ \begin{array}{rrr} 1 \amp 0 \amp 0\\ 0 \amp 1 \amp -1\\ 0 \amp 0 \amp 1 \end{array}\right]. \end{equation*}so that\begin{equation*} \boldsymbol{E}_2 \boldsymbol{B} = \left[ \begin{array}{rrr} 1 \amp 0 \amp 0\\ 0 \amp 1 \amp -1\\ 0 \amp 0 \amp 1 \end{array}\right] \left[ \begin{array}{rrr} 1 \amp 10 \amp 5\\ 2 \amp -5 \amp 12\\ 2 \amp 5 \amp 10 \end{array}\right] = \left[ \begin{array}{rrr} 1 \amp 10 \amp 5\\ 0 \amp -10 \amp 2\\ 2 \amp 5 \amp 10 \end{array}\right]. \end{equation*}
Subsubsection 3.2.2 Row reduction with elementary matrices
The row reduction of a matrix \(\boldsymbol{A} \) to row-echelon form can be obtained by successive multiplication on the left by elementary matrices,
\begin{equation*}
\boldsymbol{A}_{\text{REF}} = \boldsymbol{E}_r \boldsymbol{E}_{r-1} \cdots \boldsymbol{E}_3 \boldsymbol{E}_2 \boldsymbol{E}_1 \boldsymbol{A}.
\end{equation*}
We denote the product of all elementary matrices as the elimination matrix, \(\boldsymbol{E}\text{,}\)
\begin{equation*}
\boldsymbol{E} = \boldsymbol{E}_r \boldsymbol{E}_{r-1} \cdots \boldsymbol{E}_3 \boldsymbol{E}_2 \boldsymbol{E}_1 \hspace{1cm} \Rightarrow \hspace{1cm} \boldsymbol{A}_{\text{REF}}=\boldsymbol{E} \boldsymbol{A}.
\end{equation*}
Example 3.2.59. Applying more than one elementary matrix.
Let
\begin{equation*}
\boldsymbol{A} = \left[ \begin{array}{rrr}
2 \amp 5 \amp 10\\
2 \amp -5 \amp 12\\
1 \amp 10 \amp 5
\end{array}\right],
\hspace{1cm}
\boldsymbol{B} = \left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
2 \amp -5 \amp 12\\
2 \amp 5 \amp 10
\end{array}\right],
\hspace{1cm}
\boldsymbol{C} = \left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0 \amp -10 \amp 2\\
2 \amp 5 \amp 10
\end{array}\right].
\end{equation*}
\(\,\)
Find the elementary matrix \(\boldsymbol{E}_3 \) such that \(\boldsymbol{E}_3 \boldsymbol{A} = \boldsymbol{C}.\)
\(\, \)
From the previous example we have
\begin{equation*}
\boldsymbol{B} = \boldsymbol{E}_1 \boldsymbol{A}, \hspace{1cm} \text{with} \hspace{1cm}
\boldsymbol{E}_1 = \left[ \begin{array}{rrr}
0 \amp 0 \amp 1\\
0 \amp 1 \amp 0\\
1 \amp 0 \amp 0
\end{array}\right].
\end{equation*}
And \(\boldsymbol{C} \) from \(\boldsymbol{B}\) as,
\begin{equation*}
\boldsymbol{C} = \boldsymbol{E}_2 \boldsymbol{B}, \hspace{1cm} \text{with} \hspace{1cm}
\boldsymbol{E}_2 = \left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp -1\\
0 \amp 0 \amp 1
\end{array}\right].
\end{equation*}
This implies that one can obtain \(\boldsymbol{C} \) from \(\boldsymbol{A}\) as
\begin{equation*}
\begin{array}{ccc}
\boldsymbol{C} \amp = \amp \boldsymbol{E}_2 \boldsymbol{B}\\
\amp = \amp \boldsymbol{E}_2 \left(\boldsymbol{E}_1 \boldsymbol{A} \right)\\
\amp =\amp \boldsymbol{E}_2 \boldsymbol{E}_1 \boldsymbol{A}\\
\amp \amp\\
\amp = \amp \left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp -1\\
0 \amp 0 \amp 1
\end{array}\right]
\left[ \begin{array}{rrr}
0 \amp 0 \amp 1\\
0 \amp 1 \amp 0\\
1 \amp 0 \amp 0
\end{array}\right]
\left[ \begin{array}{rrr}
2 \amp 5 \amp 10\\
2 \amp -5 \amp 12\\
1 \amp 10 \amp 5
\end{array}\right]\\
\amp \amp\\
\amp = \amp \left[ \begin{array}{rrr}
0 \amp 0 \amp 1\\
-1 \amp 1 \amp 0\\
1 \amp 0 \amp 0
\end{array}\right]
\left[ \begin{array}{rrr}
2 \amp 5 \amp 10\\
2 \amp -5 \amp 12\\
1 \amp 10 \amp 5
\end{array}\right]\\
\amp \amp\\
\amp = \amp \left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0 \amp -10 \amp 2\\
2 \amp 5 \amp 10
\end{array}\right]
\end{array}
\end{equation*}
\(\, \)
Example 3.2.60. Row reduction.
Find \(\boldsymbol{A}_{\text{RREF}} \) where
\begin{equation*}
\boldsymbol{A} = \left[ \begin{array}{rrr}
2 \amp 5 \amp 10\\
2 \amp -5 \amp 12\\
1 \amp 10 \amp 5
\end{array}\right].
\end{equation*}
\(\, \)
We will take matrices \(\boldsymbol{E}_1 \) and \(\boldsymbol{E}_2 \) from the previous example, so that, so far, our elimination matrix is
\begin{equation*}
\boldsymbol{E} = \boldsymbol{E}_2 \boldsymbol{E}_1 =
\left[ \begin{array}{rrr}
0 \amp 0 \amp 1\\
-1 \amp 1 \amp 0\\
1 \amp 0 \amp 0
\end{array}\right],
\end{equation*}
and the current matrix to be reduced is
\begin{equation*}
\left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0 \amp -10 \amp 2\\
2 \amp 5 \amp 10
\end{array}\right],
\end{equation*}
we continue with the reduction:
\begin{equation*}
\begin{array}{ccc}
\color{Maroon}{\left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0 \amp -10 \amp 2\\
2 \amp 5 \amp 10
\end{array}\right]}
\amp \hspace{0.5cm} \color{Maroon}{R_3 \rightarrow R_3 - 2 R_1} \hspace{0.5cm} \amp
\color{Maroon}{
\left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0 \amp -10 \amp 2\\
0 \amp -15 \amp 0
\end{array}\right]
}
\color{black}{, \hspace{1cm}
\boldsymbol{E}_3 =
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 0\\
-2 \amp 0 \amp 1
\end{array}\right]}\\
\amp \amp \\
\amp\boldsymbol{E} = \boldsymbol{E}_3 \boldsymbol{E}_2 \boldsymbol{E}_1 = \amp
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 0\\
-2 \amp 0 \amp 1
\end{array}\right]
\left[ \begin{array}{rrr}
0 \amp 0 \amp 1\\
-1 \amp 1 \amp 0\\
1 \amp 0 \amp 0
\end{array}\right] =
\left[ \begin{array}{rrr}
0 \amp 0 \amp 1\\
-1 \amp 1 \amp 0\\
1 \amp 0 \amp -2
\end{array}\right] \\
\amp \amp \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccc}
\color{white}{
\left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0\amp -10 \amp 2\\
2\amp 4 \amp -1
\end{array}\right]
}
\amp \hspace{0.5cm} \color{Maroon}{R_1 \rightarrow R_1 + R_2} \hspace{0.5cm} \amp
\color{Maroon}{
\left[ \begin{array}{rrr}
1 \amp 0 \amp 7\\
0 \amp -10 \amp 2\\
0 \amp -15 \amp 0
\end{array}\right]
} \color{black}{
, \hspace{1cm}
\boldsymbol{E}_4 =
\left[ \begin{array}{rrr}
1 \amp 1 \amp 0\\
0 \amp 1 \amp 0\\
0 \amp 0 \amp 1
\end{array}\right]}\\
\amp \amp \\
\amp\boldsymbol{E} = \boldsymbol{E}_4 \boldsymbol{E}_3 \boldsymbol{E}_2 \boldsymbol{E}_1 = \amp
\left[ \begin{array}{rrr}
1 \amp 1 \amp 0\\
0 \amp 1 \amp 0\\
0 \amp 0 \amp 1
\end{array}\right]
\left[ \begin{array}{rrr}
0 \amp 0 \amp 1\\
-1 \amp 1 \amp 0\\
1 \amp 0 \amp -2
\end{array}\right] =
\left[ \begin{array}{rrr}
-1 \amp 1 \amp 1\\
-1 \amp 1 \amp 0\\
1 \amp 0 \amp -2
\end{array}\right] \\
\amp \amp \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccc}
\color{white}{
\left[ \begin{array}{rrr}
1 \amp 8 \amp 5\\
0\amp -10 \amp 2\\
2\amp 4 \amp -1
\end{array}\right]
}
\amp \hspace{0.5cm} \color{Maroon}{R_2 \rightarrow -\frac{1}{10}R_2} \hspace{0.5cm} \amp
\color{Maroon}{
\left[ \begin{array}{rrr}
1 \amp 0 \amp 7\\
0 \amp 1 \amp -\frac{1}{5}\\
0 \amp -15 \amp 0
\end{array}\right]
} \color{black}{
, \hspace{1cm}
\boldsymbol{E}_5 =
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp -\frac{1}{10} \amp 0\\
0 \amp 0 \amp 1
\end{array}\right]}\\
\amp \amp \\
\amp\boldsymbol{E} = \boldsymbol{E}_5 \cdots \boldsymbol{E}_1 = \amp
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp -\frac{1}{10} \amp 0\\
0 \amp 0 \amp 1
\end{array}\right]
\left[ \begin{array}{rrr}
-1 \amp 1 \amp 1\\
-1 \amp 1 \amp 0\\
1 \amp 0 \amp -2
\end{array}\right] =
\left[ \begin{array}{rrr}
-1 \amp 1 \amp 1\\
\frac{1}{10} \amp -\frac{1}{10} \amp 0\\
1 \amp 0 \amp -2
\end{array}\right] \\
\amp \amp \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccc}
\color{white}{
\left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0\amp -10 \amp 2\\
2\amp 4 \amp -1
\end{array}\right]
}
\amp \hspace{0.5cm} \color{Maroon}{R_3 \rightarrow R_3 + 15 R_2} \hspace{0.5cm} \amp
\color{Maroon}{
\left[ \begin{array}{rrr}
1 \amp 0 \amp 7\\
0 \amp 1 \amp -\frac{1}{5}\\
0 \amp 0 \amp -3
\end{array}\right]
} \color{black}{
, \hspace{1cm}
\boldsymbol{E}_6 =
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 0\\
0 \amp 15 \amp 1
\end{array}\right]}\\
\amp \amp \\
\amp\boldsymbol{E} = \boldsymbol{E}_6 \cdots \boldsymbol{E}_1 = \amp
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 0\\
0 \amp 15 \amp 1
\end{array}\right]
\left[ \begin{array}{rrr}
-1 \amp 1 \amp 1\\
\frac{1}{10} \amp -\frac{1}{10} \amp 0\\
1 \amp 0 \amp -2
\end{array}\right] =
\left[ \begin{array}{rrr}
-1 \amp 1 \amp 1\\
\frac{1}{10} \amp -\frac{1}{10} \amp 0\\
\frac{5}{2} \amp -\frac{3}{2} \amp -2
\end{array}\right] \\
\amp \amp \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccc}
\color{white}{
\left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0\amp -10 \amp 2\\
2\amp 4 \amp -1
\end{array}\right]
}
\amp \hspace{0.5cm} \color{Maroon}{R_3 \rightarrow -\frac{1}{3} R_3} \hspace{0.5cm} \amp
\color{Maroon}{
\left[ \begin{array}{rrr}
1 \amp 0 \amp 7\\
0 \amp 1 \amp -\frac{1}{5}\\
0 \amp 0 \amp 1
\end{array}\right]
} \color{black}{
, \hspace{1cm}
\boldsymbol{E}_7 =
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 0\\
0 \amp 0 \amp -\frac{1}{3}
\end{array}\right]}\\
\amp \amp \\
\amp\boldsymbol{E} = \boldsymbol{E}_7 \cdots \boldsymbol{E}_1 = \amp
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 0\\
0 \amp 0 \amp -\frac{1}{3}
\end{array}\right]
\left[ \begin{array}{rrr}
-1 \amp 1 \amp 1\\
\frac{1}{10} \amp -\frac{1}{10} \amp 0\\
\frac{5}{2} \amp -\frac{3}{2} \amp -2
\end{array}\right] =
\left[ \begin{array}{rrr}
-1 \amp 1 \amp 1\\
\frac{1}{10} \amp -\frac{1}{10} \amp 0\\
-\frac{5}{6} \amp \frac{1}{2} \amp \frac{2}{3}
\end{array}\right] \\
\amp \amp \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccc}
\color{white}{
\left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0\amp -10 \amp 2\\
2\amp 4 \amp -1
\end{array}\right]
}
\amp \hspace{0.5cm} \color{Maroon}{R_1 \rightarrow R_1 -7 R_3} \hspace{0.5cm} \amp
\color{Maroon}{
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp -\frac{1}{5}\\
0 \amp 0 \amp 1
\end{array}\right]
} \color{black}{
, \hspace{1cm}
\boldsymbol{E}_8 =
\left[ \begin{array}{rrr}
1 \amp 0 \amp -7\\
0 \amp 1 \amp 0\\
0 \amp 0 \amp 1
\end{array}\right]}\\
\amp \amp \\
\amp\boldsymbol{E} = \boldsymbol{E}_8 \cdots \boldsymbol{E}_1 = \amp
\left[ \begin{array}{rrr}
1 \amp 0 \amp -7\\
0 \amp 1 \amp 0\\
0 \amp 0 \amp 1
\end{array}\right]
\left[ \begin{array}{rrr}
-1 \amp 1 \amp 1\\
\frac{1}{10} \amp -\frac{1}{10} \amp 0\\
-\frac{5}{6} \amp \frac{1}{2} \amp \frac{2}{3}
\end{array}\right] =
\left[ \begin{array}{rrr}
\frac{29}{6} \amp -\frac{5}{2} \amp -\frac{11}{3}\\
\frac{1}{10} \amp -\frac{1}{10} \amp 0\\
-\frac{5}{6} \amp \frac{1}{2} \amp \frac{2}{3}
\end{array}\right] \\
\amp \amp \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccc}
\color{white}{
\left[ \begin{array}{rrr}
1 \amp 10 \amp 5\\
0\amp -10 \amp 2\\
2\amp 4 \amp -1
\end{array}\right]
}
\amp \hspace{0.5cm} \color{Maroon}{R_2 \rightarrow R_2 + \frac{1}{5} R_3} \hspace{0.5cm} \amp
\color{Maroon}{
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 0\\
0 \amp 0 \amp 1
\end{array}\right]
} \color{black}{
, \hspace{1cm}
\boldsymbol{E}_9 =
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp \frac{1}{5}\\
0 \amp 0 \amp 1
\end{array}\right]}\\
\amp \amp \\
\amp\boldsymbol{E} = \boldsymbol{E}_9 \cdots \boldsymbol{E}_1 = \amp
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp \frac{1}{5}\\
0 \amp 0 \amp 1
\end{array}\right]
\left[ \begin{array}{rrr}
\frac{29}{6} \amp -\frac{5}{2} \amp -\frac{11}{3}\\
\frac{1}{10} \amp -\frac{1}{10} \amp 0\\
-\frac{5}{6} \amp \frac{1}{2} \amp \frac{2}{3}
\end{array}\right] =
\left[ \begin{array}{rrr}
\frac{29}{6} \amp -\frac{5}{2} \amp -\frac{11}{3}\\
-\frac{1}{15} \amp 0 \amp \frac{2}{15}\\
-\frac{5}{6} \amp \frac{1}{2} \amp \frac{2}{3}
\end{array}\right] \\
\amp \amp \\
\amp \amp \\
\end{array}
\end{equation*}
We can verify the elimination matrix by performing the following operation
\begin{equation*}
\begin{array}{ccc}
\boldsymbol{E}\cdot\boldsymbol{A} \amp = \amp
\left[ \begin{array}{rrr}
\frac{29}{6} \amp -\frac{5}{2} \amp -\frac{11}{3}\\
-\frac{1}{15} \amp 0 \amp \frac{2}{15}\\
-\frac{5}{6} \amp \frac{1}{2} \amp \frac{2}{3}
\end{array}\right]
\left[ \begin{array}{rrr}
2 \amp 5 \amp 10\\
2 \amp -5 \amp 12\\
1 \amp 10 \amp 5
\end{array}\right] \\
\amp \amp\\
\amp = \amp
\left[ \begin{array}{rrr}
\frac{29}{6}\cdot 2 + \frac{-5}{2} \cdot 2 + \frac{-11}{3} \cdot 1 \amp
\frac{29}{6} \cdot 5 + \frac{-5}{2} \cdot (-5) + \frac{-11}{3}\cdot 10 \amp
\frac{29}{6} \cdot 10 + \frac{-5}{2} \cdot 12 + \frac{-11}{3}\cdot 5\\
\frac{-1}{15}\cdot 2 + 0 \cdot 2 + \frac{2}{15}\cdot 1 \amp
\frac{-1}{15}\cdot 5 + 0 \cdot (-5) + \frac{2}{15}\cdot 10 \amp
\frac{-1}{15}\cdot 10 + 0 \cdot 12 + \frac{2}{15}\cdot 5\\
\frac{-5}{2}\cdot 2 + \frac{1}{2}\cdot 2 + \frac{2}{3}\cdot 1 \amp
\frac{-5}{2}\cdot 5 + \frac{1}{2}\cdot (-5) + \frac{2}{3}\cdot 10 \amp
\frac{-5}{2}\cdot 10 + \frac{1}{2}\cdot 12 + \frac{2}{3}\cdot 5
\end{array}\right] \\
\amp \amp \\
\amp = \amp
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 0\\
0 \amp 0 \amp 1
\end{array}\right]
\end{array}
\end{equation*}
Subsubsection 3.2.3 Solving systems of equations with elimination matrices
Assume that we have a system of equations represented as a matrix equation
\begin{equation*}
\boldsymbol{A} \boldsymbol{x} = \boldsymbol{b}.
\end{equation*}
Next, assume that we find an elimination matrix, \(\boldsymbol{E} \) such that,
\begin{equation*}
\boldsymbol{E} \cdot \boldsymbol{A} = \boldsymbol{I}.
\end{equation*}
Now, if we multiply both sides of the matrix equation by \(\boldsymbol{E} \) we obtain,
\begin{equation*}
\begin{array}{ccc}
\boldsymbol{A} \boldsymbol{x} \amp = \amp \boldsymbol{b}\\
\boldsymbol{E} \boldsymbol{A} \boldsymbol{x} \amp = \amp \boldsymbol{E}\boldsymbol{b}\\
\boldsymbol{I}\boldsymbol{x} \amp = \amp \boldsymbol{E}\boldsymbol{b}\\
\boldsymbol{x} \amp = \amp \boldsymbol{E}\boldsymbol{b}.\\
\end{array}
\end{equation*}
This implies that for this system of equations the solution can be found by multiplying the elimination matrix to the right-hand-side vector.
Consider the following system of equations,
\begin{equation*}
\begin{array}{lll}
\color{white}{2} \color{black} x_1 + 2x_2 - \color{white}{2} \color{black}x_3 \amp = \amp 4\\
\color{white}{1} \color{black} x_1 + 3x_2 \color{white}{- 2x_3} \color{black} \amp = \amp 5\\
2x_1 + 7x_2 + 2x_3 \amp = \amp 9\\
\end{array}
\end{equation*}
The matrix representation of the system is
\begin{equation*}
\left[ \begin{array}{rrrr}
1 \amp 2 \amp -1 \\
1 \amp 3 \amp 0 \\
2 \amp 7 \amp 2
\end{array}\right] \,
\left[ \begin{array}{c}
x_1 \\ x_2 \\x_3
\end{array}\right]
=\left[ \begin{array}{c}
4 \\ 5 \\ 9
\end{array}\right],
\end{equation*}
and augmented matrix,
\begin{equation*}
\left[ \begin{array}{rrrr}
1 \amp 2 \amp -1 \amp 4\\
1 \amp 3 \amp 0 \amp 5\\
2 \amp 7 \amp 2 \amp 9
\end{array}\right]
\end{equation*}
We can find the elimination matrix for the augmented matrix as follows,
\begin{equation*}
\begin{array}{ccccc}
\left[ \begin{array}{rrrr}
1 \amp 2 \amp -1 \amp 4\\
1 \amp 3 \amp 0 \amp 5\\
2 \amp 7 \amp 2 \amp 9
\end{array}\right]
\amp \hspace{0.5cm} R_2 \rightarrow R_2 - R_1 \hspace{0.5cm} \amp
\left[ \begin{array}{rrrr}
1 \amp 2 \amp -1 \amp 4\\
0 \amp 1 \amp 1 \amp 1\\
2 \amp 7 \amp 2 \amp 9
\end{array}\right] \amp
\hspace{1cm} \boldsymbol{E}_1 \amp =
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
-1 \amp 1 \amp 0 \\
0 \amp 0 \amp 1
\end{array}\right] \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccccc}
\color{white}{\left[ \begin{array}{rrrr}
1 \amp 2 \amp -1 \amp 4\\
1 \amp 3 \amp 0 \amp 5\\
2 \amp 7 \amp 2 \amp 9
\end{array}\right] }
\amp \hspace{0.5cm} R_3 \rightarrow R_3 - 2R_1 \hspace{0.5cm} \amp
\left[ \begin{array}{rrrr}
1 \amp 2 \amp -1 \amp 4\\
0 \amp 1 \amp 1 \amp 1\\
0 \amp 3 \amp 4 \amp 1
\end{array}\right] \amp
\hspace{1cm} \boldsymbol{E}_2 \amp =
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 0 \\
-2 \amp 0 \amp 1
\end{array}\right] \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccccc}
\color{white}{
\left[ \begin{array}{rrrr}
1 \amp 2 \amp -1 \amp 4\\
1 \amp 3 \amp 0 \amp 5\\
2 \amp 7 \amp 2 \amp 9
\end{array}\right] }
\amp \hspace{0.5cm} R_3 \rightarrow R_1 - 2R_2 \hspace{0.5cm} \amp
\left[ \begin{array}{rrrr}
1 \amp 0 \amp -3 \amp 2\\
0 \amp 1 \amp 1 \amp 1\\
0 \amp 3 \amp 4 \amp 1
\end{array}\right] \amp
\hspace{1cm} \boldsymbol{E}_3 \amp =
\left[ \begin{array}{rrr}
1 \amp -2 \amp 0\\
0 \amp 1 \amp 0 \\
0 \amp 0\amp 1
\end{array}\right] \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccccc}
\color{white}{
\left[ \begin{array}{rrrr}
1 \amp 2 \amp -1 \amp 4\\
1 \amp 3 \amp 0 \amp 5\\
2 \amp 7 \amp 2 \amp 9
\end{array}\right] }
\amp \hspace{0.5cm} R_3 \rightarrow R_3 - 3R_2 \hspace{0.5cm} \amp
\left[ \begin{array}{rrrr}
1 \amp 0 \amp -3 \amp 2\\
0 \amp 1 \amp 1 \amp 1\\
0 \amp 0 \amp 1 \amp -2
\end{array}\right] \amp
\hspace{1cm} \boldsymbol{E}_4 \amp =
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 0 \\
0 \amp -3 \amp 1
\end{array}\right] \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccccc}
\color{white}{
\left[ \begin{array}{rrrr}
1 \amp 2 \amp -1 \amp 4\\
1 \amp 3 \amp 0 \amp 5\\
2 \amp 7 \amp 2 \amp 9
\end{array}\right] }
\amp \hspace{0.5cm} R_1 \rightarrow R_1 + 3R_3 \hspace{0.5cm} \amp
\left[ \begin{array}{rrrr}
1 \amp 0 \amp 0 \amp -4\\
0 \amp 1 \amp 1 \amp 1\\
0 \amp 0 \amp 1 \amp -2
\end{array}\right] \amp
\hspace{1cm} \boldsymbol{E}_5 \amp =
\left[ \begin{array}{rrr}
1 \amp 0 \amp 3\\
0 \amp 1 \amp 0 \\
0 \amp 0 \amp 1
\end{array}\right] \\
\amp \amp \\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccccc}
\color{white}{
\left[ \begin{array}{rrrr}
1 \amp 2 \amp -1 \amp 4\\
1 \amp 3 \amp 0 \amp 5\\
2 \amp 7 \amp 2 \amp 9
\end{array}\right] }
\amp \hspace{0.5cm} R_2 \rightarrow R_2 - R_3 \hspace{0.5cm} \amp
\left[ \begin{array}{rrrr}
1 \amp 0 \amp 0 \amp -4\\
0 \amp 1 \amp 0 \amp 3\\
0 \amp 0 \amp 1 \amp -2
\end{array}\right] \amp
\hspace{1cm} \boldsymbol{E}_6 \amp =
\left[ \begin{array}{rrr}
1 \amp 0 \amp 0\\
0 \amp 1 \amp -1 \\
0 \amp 0 \amp 1
\end{array}\right] \\
\amp \amp \\
\end{array}
\end{equation*}
The resulting elimination matrix is
\begin{equation*}
\begin{array}{ccc}
\boldsymbol{E} \amp = \amp \boldsymbol{E}_6 \cdot \boldsymbol{E}_5 \cdot \boldsymbol{E}_4 \cdot \boldsymbol{E}_3 \cdot \boldsymbol{E}_2 \cdot \boldsymbol{E}_1\\
\amp \amp\\
\amp = \amp\left[ \begin{array}{rrr}
6 \amp -11 \amp 3\\
-2 \amp 4 \amp -1 \\
1 \amp -3 \amp 1
\end{array}\right]
\end{array}
\end{equation*}
