Question 9.17: (a) Find the inverse of the matrix D = ( 1 0 -2 2 2 3 1 3 2)......
(a) Find the inverse of the matrix
D = \left(\begin{matrix} 1 & 0 & -2 \\ 2 & 2 & 3 \\ 1 & 3 & 2 \end{matrix} \right)
by Gauss–Jordan elimination.
(b) Show that DD^{–1} = I.
(a) Write out the augmented matrix consisting of the matrix D and the unit matrix of the same dimension:
\left(\begin{array}{ccc|ccc} 1 & 0 & -2&1 & 0 & 0 \\ 2 & 2 & 3& 0 & 1 & 0 \\ 1 & 3 & 2& 0 & 0 & 1 \end{array} \right)
Carry out Gauss–Jordan elimination on the matrix D by the method given in Worked Example 9.9.
The original matrix, D, is now reduced to the unit matrix. The inverse of D is given by the transformed unit matrix: the 4th, 5th and 6th columns of the augmented matrix.
D^{-1} = \left(\begin{matrix} \frac {5}{13} & \frac {6}{13} & -\frac {4}{13} \\ \frac {1}{13} & -\frac {4}{13} & \frac {7}{13} \\ – \frac {4}{13} & \frac {3}{13} & -\frac {2}{13}\end{matrix} \right) = \frac {1}{13} \left(\begin{matrix} 5 & 6 & -4 \\ 1 & -4 & 7 \\ -4 & 3 & -2 \end{matrix} \right)Note: If division by zero arises at any stage, then D does not have an inverse.
(b) Every element in D^{−1} is multiplied by 1/13. Before multiplying D^{−1} by D, factor out this scalar to simplify the arithmetic involved
DD^{−1} = \left(\begin{matrix} 1 & 0 & -2 \\ 2 & 2 & 3 \\ 1 & 3 & 2 \end{matrix} \right) \frac {1}{13} \left(\begin{matrix} 5 & 6 & -4 \\ 1 & -4 & 7 \\ -4 & 3 & -2 \end{matrix} \right) = \frac {1}{13} \left(\begin{matrix} 13 & 0 & 0 \\ 0 & 13 & 0 \\ 0 & 0 & 13 \end{matrix} \right) = \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)| Action | Augmented matrix | Calculations |
| \left(\begin{array}{ccc|ccc} 1 & 0 & -2& 1 & 0 & 0 \\ 2 & 2 & 3 &0 & 1 & 0\\ 1 & 3 & 2&0 & 0 & 1 \end{array} \right)\begin{matrix} (1) \\ (2) \\ (3) \end{matrix} | ||
| row 2 + (–2 × row 1) row 3 + (–1 × row 1) |
\left(\begin{array}{ccc|ccc} 1 & 0 & -2& 1 & 0 & 0 \\ 0 & 2 & 7 &-2 & 1 & 0 \\ 0 & 3 & 4&-1 & 0 & 1 \end{array} \right)\begin{matrix} (1) \\ (2)¹ \\ (3)¹ \end{matrix} | Calculate (–2 × row 1) →(–2 0 4 –2 0 0) Calculate (–1 × row 1) →(–1 0 2 –1 0 0) |
| row 2¹ × \frac {1}{2} | \left(\begin{array}{ccc|ccc} 1 & 0 & -2&1 & 0 & 0 \\ 0 & 1 & \frac {7}{2}&-1 & \frac {1}{2} & 0 \\ 0 & 3 & 4& 0 & 0 & 1 \end{array}\right)\begin{matrix} (1) \\ (2)² \\ (3)¹ \end{matrix} | |
| row 3¹ + (–1 × row 2²) | \left(\begin{array}{ccc|ccc} 1 & 0 & -2&1 & 0 & 0 \\ 0 & 1 & \frac {7}{2}&-1 & \frac {1}{2} & 0 \\ 0 & 0 & -\frac {13}{2}&2 & -\frac {3}{2} & 1 \end{array} \right)\begin{matrix} (1) \\ (2)² \\ (3)² \end{matrix} | Calculate (–3 × row 2²) →(0 -3 –\frac {21}{2} 3 –\frac {3}{2} 0) |
| row 3² × \frac {2}{13} | \left(\begin{array}{ccc|ccc} 1 & 0 & -2& 1 & 0 & 0 \\ 0 & 1 & \frac {7}{2}&\ -1 & \frac {1}{2} & 0 \\ 0 & 0 & 1 &-\frac {4}{13} & \frac {3}{13} & -\frac {2}{13}\end{array}\right)\begin{matrix} (1) \\ (2)² \\ (3)³ \end{matrix} | |
| row 1 + (2 × row 3³) row 2² + (−\frac {7}{2} × row 3³) |
\left(\begin{array}{ccc|ccc} 1 & 0 & 0&\frac {5}{13} & \frac {6}{13} & -\frac {4}{13} \\ 0 & 1 & 0 &\frac {1}{13} & -\frac {4}{13} & \frac {7}{13} \\ 0 & 0 & 1&- \frac {4}{13} & \frac {3}{13} & -\frac {2}{13} \end{array} \right)\begin{matrix} (1)¹ \\ (2)² \\ (3)³ \end{matrix} | Calculate (–2 × row 3³) →(0 0 2 -\frac {8}{13} \frac {6}{13} -\frac {4}{13}) Calculate (-\frac {7}{2} × row 3³) →(0 0 -\frac {7}{2} \frac {14}{13} \frac {10.5}{13} \frac {7}{13}) |