Chapter 10.6, Problem 26E

To determine

With the help of row and column operation the determinant of the given matrix is to be calculated:

  $\left[\begin{array}{cccc}1& 2& 0& 2\\ 3& -4& 0& 4\\ 0& 1& 6& 0\\ 1& 0& 2& 0\end{array}\right]$

Answer to Problem 26E

The given matrix $\left[\begin{array}{cccc}1& 2& 0& 2\\ 3& -4& 0& 4\\ 0& 1& 6& 0\\ 1& 0& 2& 0\end{array}\right]$ contains the inverse matrix.

Explanation of Solution

Given:

  $\left[\begin{array}{cccc}1& 2& 0& 2\\ 3& -4& 0& 4\\ 0& 1& 6& 0\\ 1& 0& 2& 0\end{array}\right]$

Concept Used:

The determinant of every square matrix with the order $n$ is defined as $A=\left[a_{ij}\right]$ where,

  $a={\left(i,j\right)}^{th}$ of element $A$ .

Calculations:

The determinant of the given matrix is calculated as:

  $\begin{array}{l}\left[\begin{array}{cccc}1& 2& 0& 2\\ 3& -4& 0& 4\\ 0& 1& 6& 0\\ 1& 0& 2& 0\end{array}\right]\\ R_3\leftrightarrow R_2\left[\begin{array}{cccc}1& 2& 0& 2\\ 3& -4& 0& 4\\ -3& 1& 0& 0\\ 1& 0& 2& 0\end{array}\right]\end{array}$

Now expand by $C3$

  $\begin{array}{l}C3=-2\left[\begin{array}{ccc}1& 2& 2\\ 3& -4& 4\\ -3& 1& 0\end{array}\right]\\ C3=-2\left[2\left(3-12\right)-9\left(1+6\right)\right]\\ C3=-2\left(-18-28\right)\\ C3=92\ne 0\end{array}$

Now, its determinant value is $D=92$ or non-zero value which clearly shows that this matrix have the inverse matrix.

Conclusion:

Hence, the given matrix has an inverse.