There are three elementary row operations we perform on a matrix:
- Interchanging two rows (\(R_i\leftrightarrow R_j\))
- Multiplying a row by a nonzero scalar (\(cR_i\), \(c\neq 0\))
- Adding a scalar multiple of row \(i\) to row \(j\) (\(cR_i+R_j\))
Steps of solving a linear system \(A\overrightarrow{x}=\overrightarrow{b}\) are equivalent to elementary
row operations on the augmented matrix \([A\:\overrightarrow{b}]\) as illustrated by the following example.
Example.
\[\begin{eqnarray*}
\begin{array}{rcrcrcr}
&&2x_2 &-&8x_3&=&8\;\; \; (2.1)\\
x_1&-&2x_2 &+&x_3&=&0\;\; \; (2.2)\\
-4x_1&+&5x_2&+&9x_3&=&-9\;\; \; (2.3)
\end{array}
\end{eqnarray*}\]
We do the following steps to solve the above system:
- Interchange (2.1) and (2.2):
\[\begin{eqnarray*}
\begin{array}{rcrcrcr}
x_1&-&2x_2 &+&x_3&=&0\;\; \; (3.1)\\
&&2x_2 &-&8x_3&=&8\;\; \; (3.2)\\
-4x_1&+&5x_2&+&9x_3&=&-9\;\; \; (3.3)
\end{array}
\end{eqnarray*}\]
Corresponding row operation is
\[[A\:\overrightarrow{b}]=\left[\begin{array}{rrr|r}0&2&-8&8\\1&-2&1&0\\-4&5&9&-9\end{array} \right]\xrightarrow{R_1\leftrightarrow R_2}
\left[\begin{array}{rrrr}1&-2&1&0\\0&2&-8&8\\-4&5&9&-9\end{array} \right].\]
- Replace (3.3) by 4(3.1)+(3.3):
\[\begin{eqnarray*}
\begin{array}{rcrcrcr}
x_1&-&2x_2 &+&x_3&=&0\;\; \; (4.1)\\
&&2x_2 &-&8x_3&=&8\;\; \; (4.2)\\
&-&3x_2&+&13x_3&=&-9\;\; \; (4.3)
\end{array}
\end{eqnarray*}\]
Corresponding row operation is
\[\left[\begin{array}{rrrr}1&-2&1&0\\0&2&-8&8\\-4&5&9&-9\end{array} \right]
\xrightarrow{4R_1+ R_3}
\left[\begin{array}{rrrr}1&-2&1&0\\0&2&-8&8\\0&-3&13&-9\end{array} \right].\]
- Scale \(\frac{1}{2}\)(4.2):
\[\begin{eqnarray*}
\begin{array}{rcrcrcr}
x_1&-&2x_2 &+&x_3&=&0\;\; \; (5.1)\\
&&x_2 &-&4x_3&=&4\;\; \; (5.2)\\
&-&3x_2&+&13x_3&=&-9\;\; \; (5.3)
\end{array}
\end{eqnarray*}\]
Corresponding row operation is
\[\left[\begin{array}{rrrr}1&-2&1&0\\0&2&-8&8\\0&-3&13&-9\end{array} \right]
\xrightarrow{\frac{1}{2}R_2}
\left[\begin{array}{rrrr}1&-2&1&0\\0&1&-4&4\\0&-3&13&-9\end{array} \right].\]
- Replace (5.3) by 3(5.2)+(5.3):
\[\begin{eqnarray*}
\begin{array}{rcrcrcr}
x_1&-&2x_2 &+&x_3&=&0\;\; \; (6.1)\\
&&x_2 &-&4x_3&=&4\;\; \; (6.2)\\
&&&&x_3&=&3\;\; \; (6.3)
\end{array}
\end{eqnarray*}\]
Corresponding row operation is
\[\left[\begin{array}{rrrr}1&-2&1&0\\0&1&-4&4\\0&-3&13&-9\end{array} \right]
\xrightarrow{3R_2+R_3}
\left[\begin{array}{rrr|r}1&-2&1&0\\0&1&-4&4\\0&0&1&3\end{array} \right].\]
- Back substitutions:
\[\begin{array}{rcl}
(6.3) &\implies& x_3=3\\
(6.2) &\implies& x_2=4+4x_3=4+4\cdot 3=16\\
(6.1) &\implies& x_1=0+2x_2-x_3=2\cdot 16 -3=29
\end{array}\]
So the solution set is \(\{ (29,16,3) \}\).
Remark.
- Two matrices are row equivalent if we can transform one matrix to another by elementary
row operations. If two linear systems have row equivalent augmented matrices, then they have the same solution
set.
- To solve \(A\overrightarrow{x}=\overrightarrow{b}\), using row operations we transform the augmented
matrix \([A\:\overrightarrow{b}]\) into an "upper-triangular" form called echelon form and then use back
substitutions.
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