Elimination with Matrices

Method of Elimination

Linear systems are solved by the Matrix-powered method of elimination.

\left{\begin{matrix} x + 2y + z = 2 \\ 3x+8y+z=12 \\ 4y+z=2 \end{matrix}\right} =================== \underbrace{ \begin{bmatrix} 1&2&1 \\ 3&8&1 \\ 0&4&1 \end{bmatrix} ============= \begin{bmatrix} 2 \\ 12 \\ 2 \end{bmatrix} }\_{ \text{Initial Matrix }Ax = b } \rightarrow \underbrace{ \begin{bmatrix} \boxed{1}& 2& 1 \\ 0&\boxed{2}& -2 \\ 0&0&\boxed{5} \end{bmatrix} ============= \begin{bmatrix} 2 \\ 6 \\ -10 \end{bmatrix} }\_{ \text{Eliminated Matrix }Ux = c } \implies \underbrace{ \left{\begin{matrix} x + 2y+ z = 2\\ 2y-2z=6 \\ 5z=-10 \end{matrix}\right} }\_{ \text{Can easily be solved now} }

With $\boxed{n}$ being pivots that cannot equal 0. If there is a zero-pivot, exchanging rows can fix the problem. Complete failure arises when a pivot cannot be altered to not be 0. Then the matrix is uninvertible.

Elimination Matrices

Store the elimination steps that have been taken to solve the system.

$$\underset{\text{Matrix }{E}_{2,1}}{\underbrace{\left[\begin{array}{ccc}1& 0& 0\\ -3& 1& 0\\ 0& 0& 1\end{array}\right]}}\underbrace{\left[\begin{array}{ccc}1& 2& 1\\ 3& 8& 1\\ 0& 4& 1\end{array}\right]}\_\text{Initial Matrix A}============================================\underset{\text{Eliminated Matrix}}{\underbrace{\left[\begin{array}{ccc}\boxed{1}& 2& 1\\ 0& \boxed{2}& -2\\ 0& 4& 1\end{array}\right]}}$$

Matrix $E_{2,1}$ takes \left\[ 1, 0, 0 \right] rows from $A$, hence one time the first row and zero times the others to produce the first row of the eliminated Matrix . It has no change on the row. The second row of $E\_2,1$ subtracts $-3$ times the first row from the second row of the matrix, eliminating the item at position $a\_2,1$ with that.

To change $A\to U$ we need to have two eliminations $E\_2,1$, $E\_3,2$ that take care of row 2 and 3.

$$(E\_3,2E\_2,1)A=U$$

Important Properties

• Elimination does not change the Row Space of a matrix because it is just taking linear combinations of rows, staying in the same subspace.