Gaussian Elimination/Examples/Arbitrary Matrix 6

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Example of Use of Gaussian Elimination

Let $\mathbf A$ denote the matrix:

$\mathbf A = \begin {bmatrix}
   i &  1 - i & i & 0 \\

1 & -2 & 0 & i \\ 1 - i & -1 + i & 1 & 1 \\ \end {bmatrix}$

The reduced echelon form of $\mathbf A$ is:

$\mathbf E = \begin {bmatrix}

1 & 0 & 1 + i & 1 \\ 0 & 1 & \dfrac {1 + i} 2 & \dfrac {1 - i} 2 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$


Proof

In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\mathbf A$.

The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$.


$e_1 := r_2 \leftrightarrow r_1$

Hence:

$\mathbf A_1 = \begin {bmatrix}

1 & -2 & 0 & i \\

   i &  1 - i & i & 0 \\

1 - i & -1 + i & 1 & 1 \\ \end {bmatrix}$


$e_2 := r_3 \to r_3 - r_1$

$e_3 := r_3 \to r_3 + r_2$

$\mathbf A_3 = \begin {bmatrix}

1 & -2 & 0 & i \\

   i &  1 - i &     i & 0     \\

0 & 2 & 1 + i & 1 - i \\ \end {bmatrix}$


$e_4 := r_2 \to r_2 - i r_1$

$\mathbf A_4 = \begin {bmatrix}

1 & -2 & 0 & i \\ 0 & 1 + i & i & 1 \\ 0 & 2 & 1 + i & 1 - i \\ \end {bmatrix}$


$e_5 := r_1 \to r_1 + r_3$

$\mathbf A_5 = \begin {bmatrix}

1 & 0 & 1 + i & 1 \\ 0 & 1 + i & i & 1 \\ 0 & 2 & 1 + i & 1 - i \\ \end {bmatrix}$


$e_6 := r_3 \to \dfrac {r_3} 2$

$\mathbf A_6 = \begin {bmatrix}

1 & 0 & 1 + i & 1 \\ 0 & 1 + i & i & 1 \\ 0 & 1 & \dfrac {1 + i} 2 & \dfrac {1 - i} 2 \\ \end {bmatrix}$


$e_7 := r_2 \to r_2 - \paren {1 + i} r_3$

$\mathbf A_7 = \begin {bmatrix}

1 & 0 & 1 + i & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & \dfrac {1 + i} 2 & \dfrac {1 - i} 2 \\ \end {bmatrix}$


$e_8 := r_2 \leftrightarrow r_3$

$\mathbf A_8 = \begin {bmatrix}

1 & 0 & 1 + i & 1 \\ 0 & 1 & \dfrac {1 + i} 2 & \dfrac {1 - i} 2 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$


and it is seen that $\mathbf A_8$ is the required reduced echelon form.

$\blacksquare$


Sources