Matrix Inverse Algorithm/Examples/Arbitrary Matrix 3

Examples of use of Matrix Inverse Algorithm
Let $\mathbf A$ be the (square) matrix defined as:


 * $\mathbf A = \begin {pmatrix}

1 & 2 & 0 \\ 0 & -1 & 2 \\ -1 &  2 & 0 \\ \end {pmatrix}$

Then its inverse $\mathbf A^{-1}$ is:


 * $\mathbf A^{-1} = \begin {pmatrix}

\dfrac 1 2 & 0 & -\dfrac 1 2 \\ \dfrac 1 4 & 0 & \dfrac 1 4 \\ \dfrac 1 8 & \dfrac 1 2 & \dfrac 1 8 \\ \end {pmatrix}$

Proof
We construct $\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix}$:


 * $\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix} = \paren {\begin {array} {ccc|ccc}

1 & 2 & 0 & 1 & 0 & 0 \\ 0 & -1 & 2 & 0 & 1 & 0 \\ -1 &  2 & 0 & 0 & 0 & 1 \\ \end {array} }$

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

The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf B_k \end {pmatrix}$.

$e_1 := r_3 \to r_3 + r_1$

Hence:
 * $\begin {pmatrix} \mathbf A_1 & \mathbf B_1 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}

1 & 2 & 0 & 1 & 0 & 0 \\ 0 & -1 & 2 & 0 & 1 & 0 \\ 0 &  4 & 0 & 1 & 0 & 1 \\ \end {array} }$

$e_2 := r_3 \to r_3 + 4 r_2$


 * $\begin {pmatrix} \mathbf A_2 & \mathbf B_2 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}

1 & 2 & 0 & 1 & 0 & 0 \\ 0 & -1 & 2 & 0 & 1 & 0 \\ 0 &  0 & 8 & 1 & 4 & 1 \\ \end {array} }$

$e_3 := r_2 \to -r_2$

$e_4 := r_3 \to \dfrac {r_3} 8$


 * $\begin {pmatrix} \mathbf A_4 & \mathbf B_4 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}

1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & -2 & 0 & -1 & 0 \\ 0 & 0 & 1 & \dfrac 1 8 & \dfrac 1 2 & \dfrac 1 8 \\ \end {array} }$

$e_4 := r_2 \to r_2 + 2 r_3$


 * $\begin {pmatrix} \mathbf A_4 & \mathbf B_4 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}

1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & \dfrac 1 4 & 0 & \dfrac 1 4 \\ 0 & 0 & 1 & \dfrac 1 8 & \dfrac 1 2 & \dfrac 1 8 \\ \end {array} }$

$e_5 := r_1 \to r_1 - 2 r_2$


 * $\begin {pmatrix} \mathbf A_5 & \mathbf B_5 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}

1 & 0 & 0 & \dfrac 1 2 & 0 & -\dfrac 1 2 \\ 0 & 1 & 0 & \dfrac 1 4 & 0 & \dfrac 1 4 \\ 0 & 0 & 1 & \dfrac 1 8 & \dfrac 1 2 & \dfrac 1 8 \\ \end {array} }$

and it is seen that $\begin {pmatrix} \mathbf A_5 & \mathbf B_5 \end {pmatrix}$ is the required reduced echelon form:
 * $\mathbf A_5 = \mathbf I$

and so by the Matrix Inverse Algorithm:
 * $\mathbf B_5 = \mathbf A^{-1}$