Question

Use Gaussian elimination to find the determinant of the matrices A in Exercises 1 through 10.

4.$\left[\begin{array}{cccc}1& -1& 2& -2\\ -1& 2& 1& 6\\ 2& 1& 14& 10\\ -2& 6& 10& 33\end{array}\right]$

Short answer

Therefore, the determinant of given matrix is given by,

$detA=9$.

Step-by-step solution

Definition

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

Given Matrix,

$A=\left[\begin{array}{cccc}1& -1& 2& -2\\ -1& 2& 1& 6\\ 2& 1& 14& 10\\ -2& 6& 10& 33\end{array}\right]$

To find determinant by using Gaussian Eliminations

First, we multiply the first row by and add it to the second row. Then, we multiply the first row by 2 and add it to the third row. By - and add it to the 4^th row.

We get,

$A_1=\left[\begin{array}{cccc}1& -1& 2& -2\\ 0& 1& 3& 4\\ 0& 3& 10& 14\\ 0& 4& 14& 29\end{array}\right]$

Now, we multiply the first row by 3 and add it to the third one. Then, we multiply it by 4 and add it to the fourth one.

We get

$A_2=\left[\begin{array}{cccc}1& -1& 2& -2\\ 0& 1& 3& 4\\ 0& 0& 1& 2\\ 0& 0& 2& 13\end{array}\right]$

In the end, we multiply the third row by -2 and add it to the fourth one. We get

$A_3=\left[\begin{array}{cccc}1& -1& 2& -2\\ 0& 1& 3& 4\\ 0& 0& 1& 2\\ 0& 0& 0& 9\end{array}\right]$

We had zero row swaps, so

localid="1659502945985" $\begin{array}{l}detA={(-1)}^0.1.1.1.9\\ detA=9.\end{array}$