A matrix is a rectangular array of numbers. A matrix can be used to represent a system of linear equations.
Consider the system of equations
\( \mathbb{S}_1\)
\(\begin{eqnarray} x + y + z &=& ~~~0\\ 2 y - z &=& -1\\ x\quad\,\,\,\, + z &=& ~~~1 \end{eqnarray}\)
This system can be represented by the coefficient matrix
\(\left|\begin{array}~ 1 & 1 & ~~~1 & ~~~0\\ 0 & 2 & -1 & -1\\ 1 & 0 & ~~~1 & ~~~1 \end{array}\right|\)
This matrix has three rows and four columns. The equivalent system corresponding to the solution will look like
\( \mathbb{S}\)
\(\begin{eqnarray} x\quad\quad &=& a\\ y \quad &=& b\\ z &=& c \end{eqnarray}\)
which will have a corresponding coefficient matrix
\(\left|\begin{array}~ 1 & 0 & 0 & a\\ 0 & 1 & 0 & b\\ 0 & 0 & 1 & c \end{array}\right|\)
The zeros below the main diagonal will be achieved using the operations discussed in Section 7.2 which result in equivalent linear systems.
The zeros above the main diagonal will be achieved by back substitution.
\(\left|\begin{array}~ 1 & 1 & ~~~1 & ~~~0\\ 0 & 2 & -1 & -1\\ 1 & 0 & ~~~1 & ~~~1 \end{array}\right|\)
\(E_3 \rightarrow - E_1 + E_3\)
\(\left|\begin{array}~ 1 & ~~~1 & ~~~1 & ~~~0\\ 0 & ~~~2 & -1 & -1\\ 0 & -1 & ~~~0 & ~~~1 \end{array}\right|\)
\(E_2 \rightarrow E_2 + E_3\)
\(\left|\begin{array}~ 1 & ~~~1 & ~~~1 & 0\\ 0 & ~~~1 & -1 & 0\\ 0 & -1 & ~~~0 & 1 \end{array}\right|\)
\(E_3 \rightarrow E_2 + E_3\)
\(\left|\begin{array}~ 1 & 1 & ~~~1 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & -1 & 1 \end{array}\right|\)
\(E_3 \rightarrow - E_3\)
\(\left|\begin{array}~ 1 & 1 & ~~~1 & ~~~0\\ 0 & 1 & -1 & ~~~0\\ 0 & 0 & ~~~1 & -1 \end{array}\right|\)
Now that the system is triangularized, we begin the back substitution phase:
\(E_2 \rightarrow E_3 + E_2\)
\(\left|\begin{array}~ 1 & 1 & 1 & ~~~0\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & -1 \end{array}\right|\)
\(E_1 \rightarrow - E_3 + E_1\)
\(\left|\begin{array}~ 1 & 1 & 0 & ~~~1\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & -1 \end{array}\right|\)
\(E_1 \rightarrow - E_2 + E_1\)
\(\left|\begin{array}~ 1 & 0 & 0 & ~~~2\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & -1 \end{array}\right|\)
Which corresponds to the system
\( \mathbb{S}\)
\(\begin{eqnarray} x\quad\quad &=& ~~~2\\ y\quad &=& -1\\ z &=& -1 \end{eqnarray}\)
Thus the fourth column of the matrix represents the solution \((2,-1,-1)\)
The matrix method is just triangularization followed by back-substitution, but using the coefficient matrix rather than the equations.
Use the matrix method to solve the system
\(\mathbb{S}\)
\(\begin{eqnarray} x + y + z &=& ~~~0\\ -x + 2 y \quad &=& -1\\ x\quad\,\,\,\, + z &=& ~~~1 \end{eqnarray}\)