Chapter 8.3, Problem 26E

To determine

The inverse of the matrix.

Answer to Problem 26E

  $A^{-1}=\left[\begin{array}{ccc}-10& -4& 27\\ 2& 1& -5\\ -13& -5& 35\end{array}\right]$ .

Explanation of Solution

Given information:

The given matrix is:

  $\left[\begin{array}{ccc}10& 5& -7\\ -5& 1& 4\\ 3& 2& -2\end{array}\right]$

Calculation:

The given matrix:

  $\left[\begin{array}{ccc}10& 5& -7\\ -5& 1& 4\\ 3& 2& -2\end{array}\right]$

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:

  $\left[\begin{array}{ccc}10& 5& -7\\ -5& 1& 4\\ 3& 2& -2\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Make zero in column one except the entry at row one, column $1$ . Divide row $1$ by $10\left(R_1=\frac{R_1}{10}\right)$ :

  $\left[\begin{array}{ccc}1& 1/2& -7/10\\ -5& 1& 4\\ 3& 2& -2\end{array}\right]\left[\begin{array}{ccc}1/10& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Add row $1$ mltiplied by $5$ to row two $(R_2=R_2+(5)R_1)$ and subtract row one multiplied by three from row three $(R_3=R_3-(3)R_1):$

  $\left[\begin{array}{ccc}1& 1/2& -7/10\\ 0& 7/2& 1/2\\ 0& 1/2& 1/10\end{array}\right]\left[\begin{array}{ccc}1/10& 0& 0\\ 1/2& 1& 0\\ -3/10& 0& 1\end{array}\right]$

Make zeroes in column two except the entry at row $2$ ,column $2$ :

Multiply row $2$ multiplied by $\frac{2}{7}$

  $(R_2=(\frac{2}{7})R_2):$

  $\left[\begin{array}{ccc}1& 1/2& -7/10\\ 0& 1& 1/7\\ 0& 1/2& 1/10\end{array}\right]\left[\begin{array}{ccc}1/10& 0& 0\\ 1/7& 2/7& 0\\ -3/10& 0& 1\end{array}\right]$

Subtract row $2$ multiplied by $\frac{1}{2}$ from row $1(R_1=R_1-(\frac{1}{2})R_2)$ and subtract row $2$ multiplied by $\frac{1}{2}$ from row $3(R_3=R_3-(\frac{1}{2})R_2)$ :

  $\left[\begin{array}{ccc}1& 0& -27/35\\ 0& 1& 1/7\\ 0& 0& 1/35\end{array}\right]\left[\begin{array}{ccc}1/35& -1/7& 0\\ 1/7& 2/7& 0\\ -13/35& -1/7& 1\end{array}\right]$

Add row $3$ by $27$ to row one $\left(R_1=R_1+(27)R_3\right)$ and subtract row $3$ multiplied by $5$ from row $2$

  $(R_2=R_2-(5)R_3)$ :

  $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-10& -4& 27\\ 2& 1& -5\\ -13& -5& 35\end{array}\right]$

Therefore, the solution is $A^{-1}=\left[\begin{array}{ccc}-10& -4& 27\\ 2& 1& -5\\ -13& -5& 35\end{array}\right]$ .