# Row Operation to Clear First Column of Matrix/Examples/Arbitrary Matrix 2

## Example of Use of Row Operation to Clear First Column of Matrix

Let $\mathbf A$ be the matrix:

$\mathbf A = \begin {pmatrix} 1 & 0 & 1 & 1 \\ -1 & 0 & 2 & 1 \\ -1 & 1 & 5 & 9 \end {pmatrix}$

The matrix $\mathbf R$ corresponding to the row operation to clear the first column is:

$\mathbf R = \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$

and the matrix $\mathbf B$ which results from the row operation $\mathbf R$ is:

$\mathbf B = \begin {pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & 3 & 2 \\ 0 & 1 & 6 & 10 \end {pmatrix}$

## Proof

We use Row Operation to Clear First Column of Matrix as follows:

$(1): \quad$ Apply the elementary row operation $r_2 \to r_2 + r_1$ to to add $1$ times row $1$ to row $2$.

From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, this is accomplished by pre-multiplying by the matrix:

$\mathbf E_1 := \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix}$

which converts $\mathbf A$ to $\mathbf A_1$ as follows:

 $\ds \mathbf A_1$ $=$ $\ds \mathbf E_1 \mathbf A$ $\ds$ $=$ $\ds \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix} \begin {pmatrix} 1 & 0 & 1 & 1 \\ -1 & 0 & 2 & 1 \\ -1 & 1 & 5 & 9 \end {pmatrix}$ $\ds$ $=$ $\ds \begin {pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & 3 & 2 \\ -1 & 1 & 5 & 9 \end {pmatrix}$

$(2): \quad$ Apply the elementary row operation $r_3 \to r_3 + r_1$ to add $1$ times row $1$ to row $3$.

From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, this is accomplished by pre-multiplying by the matrix:

$\mathbf E_2 := \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$

which converts $\mathbf A_1$ to $\mathbf A_2$ as follows:

 $\ds \mathbf A_2$ $=$ $\ds \mathbf E_2 \mathbf A_1$ $\ds$ $=$ $\ds \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix} \begin {pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & 3 & 2 \\ -1 & 1 & 5 & 9 \end {pmatrix}$ $\ds$ $=$ $\ds \begin {pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & 3 & 2 \\ 0 & 1 & 6 & 10 \end {pmatrix}$

The matrix is in the correct form, and so:

$\mathbf B = \mathbf A_2$

$\Box$

Hence we can calculate $\mathbf R$:

 $\ds \mathbf R$ $=$ $\ds \mathbf E_2 \mathbf E_1$ $\ds$ $=$ $\ds \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix} \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix}$ $\ds$ $=$ $\ds \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$

$\blacksquare$