91Ó°ÊÓ

Gaussian elimination method is used to solve a system of linear equations.

Gaussian elimination provides a relatively efficient way of constructing the inverse to a matrix.

Interchanging two rows.Multiplying a row by a constant (any constant which is not zero).

Given Matrix,

A=$\mathbf{\left[}\begin{array}{cccc}\mathbf{1}& \mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& \mathbf{1}& \mathbf{4}& \mathbf{4}\\ \mathbf{1}& \mathbf{-}\mathbf{1}& \mathbf{2}& \mathbf{-}\mathbf{2}\\ \mathbf{1}& \mathbf{-}\mathbf{1}& \mathbf{8}& \mathbf{-}\mathbf{8}\end{array}\mathbf{\right]}$

First, we multiply the first row by -1 and add it to the second, third and fourth row.

We get

$A_1=\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& 0& 3& 3\\ 0& -2& 1& -3\\ 0& -2& 7& -9\end{array}\right]$

Then, we switch the second and third row, and right afterwards, the third and fourth row, too.

We get

$A_2=\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& -2& 1& -3\\ 0& -2& 7& -9\\ 0& 0& 3& 3\end{array}\right]$

Now, we multiply the second row by -1 and add it to the third row.

We get

$A_3=\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& -2& 1& -3\\ 0& 0& 6& -6\\ 0& 0& 3& 3\end{array}\right]$

Lastly, we multiply the third row by $-\frac{1}{2}$and add it to the fourth row.

We get

$A_4=\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& -2& 1& -3\\ 0& 0& 6& -6\\ 0& 0& 0& 6\end{array}\right]$

We had two row interchange, so

$$\begin{gathered} \mathrm{detA}={\left(-1\right)}^2·1·-2·6·6 \\ \mathrm{detA}=-72 \end{gathered}$$