Chapter 8.3, Problem 33E

To determine

The inverse of the matrix.

Answer to Problem 33E

  $A^{-1}=\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 2& 0& 1& 0\\ 0& 1& 0& 2\end{array}\right]$ .

Explanation of Solution

Given information:

The given matrix is:

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

Calculation:

The given matrix:

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

To find the inverse matrix, the identity matrix and perform row operations trying to make the identity matrix to the left.

So, the given matrix with identity matrix:

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

Make zero in column one except the entry at row one, column $1$ :

Add row one multiplied by $2$ to $3$

  $(R_3=R_3+(2)R_1):$

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

Multiply row $1$ by $-1$

  $(R_1=-1\cdot R_1)$ and divide row $2$ by $2\left(R_2=\frac{R_2}{2}\right)$ :

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

Add row $2$ to row $4$

  $(R_4=R_4+R_2):$

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

Make zeros in column $4$ except the entry at row $4$ , column $4$ .

Add row $4$ to row $2$

  $(R_2=R_2+R_4)$ and multiply row $4$ by $2(R_4=(2)R_4)$ :

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

Therefore, the solution is $A^{-1}=\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 2& 0& 1& 0\\ 0& 1& 0& 2\end{array}\right]$ .