- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.20 Exercises
Determine the inverse of each of the matrices in Exercises 12 through 16.
12. $\quad$ \begin{align*} \begin{bmatrix} 2 & 3 & 4 \\ 2 & 1 & 1 \\ -1 & 1 & 2 \end{bmatrix} \end{align*}
Solution. $\quad$ First, we initialize the following augmented matrix. \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 2 & 3 & 4 & 1 & 0 & 0\\ 2 & 1 & 1 & 0 & 1 & 0\\ -1 & 1 & 2 & 0 & 0 & 1 \end{array} \end{bmatrix} \end{align*} Then, we apply the Gauss-Jordan elimination process to turn the left side of the array into the $3 \times 3$ identity matrix, and the right side into the inverse of the given matrix. \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 2 & 3 & 4 & 1 & 0 & 0\\ 2 & 1 & 1 & 0 & 1 & 0\\ -1 & 1 & 2 & 0 & 0 & 1 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 2 & 3 & 4 & 1 & 0 & 0\\ 0 & 3 & 5 & 0 & 1 & 2\\ -1 & 1 & 2 & 0 & 0 & 1 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 2 & 0 & -1 & 1 & -1 & -2\\ 0 & 3 & 5 & 0 & 1 & 2\\ -1 & 1 & 2 & 0 & 0 & 1 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 0 & 2 & 3 & 1 & -1 & 0\\ 0 & 3 & 5 & 0 & 1 & 2\\ -1 & 1 & 2 & 0 & 0 & 1 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 0 & 2 & 3 & 1 & -1 & 0\\ 0 & 3 & 5 & 0 & 1 & 2\\ 1 & -3 & -5 & -1 & 1 & -1 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 0 & 2 & 3 & 1 & -1 & 0\\ 0 & 3 & 5 & 0 & 1 & 2\\ 1 & 0 & 0 & -1 & 2 & 1 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 0 & 2 & 3 & 1 & -1 & 0\\ 0 & 1 & 2 & -1 & 2 & 2\\ 1 & 0 & 0 & -1 & 2 & 1 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 0 & 0 & -1 & 3 & -5 & -4\\ 0 & 1 & 2 & -1 & 2 & 2\\ 1 & 0 & 0 & -1 & 2 & 1 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 0 & 0 & -1 & 3 & -5 & -4\\ 0 & 1 & 2 & -1 & 2 & 2\\ 1 & 0 & 0 & -1 & 2 & 1 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 0 & 0 & -1 & 3 & -5 & -4\\ 0 & 1 & 0 & 5 & -8 & -6\\ 1 & 0 & 0 & -1 & 2 & 1 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|ccc} 1 & 0 & 0 & -1 & 2 & 1 \\ 0 & 1 & 0 & 5 & -8 & -6\\ 0 & 0 & 1 & -3 & 5 & 4 \end{array} \end{bmatrix} \end{align*} This gives us the inverse $\begin{bmatrix} \begin{array}{ccc} -1 & 2 & 1 \\ 5 & -8 & -6\\ -3 & 5 & 4 \end{array} \end{bmatrix}. \quad \blacksquare$