In the previous lesson we saw that reducing a matrix \(A\) to the identity \(I\) is the same as multiplying \(A\) by a sequence of matrices \(S_n \dots S_2 S_1\) which means \(S_n \dots S_2 S_1 = A^{-1}.\)
Using the fact that for any matrix \(B,\), \(BI = B,\) we have \[S_n \dots S_2 S_1 I = S_n \dots S_2 S_1 = A^{-1}\]
Multiplying \(I\) by \(S_n \dots S_2 S_1\) is the same as starting with \(I\) and doing all the same steps that were used to reduce \(A.\)
Therefore, a method to find \(A^{-1}\) is to start with the augmented matrix \([A|I]\) and reduce to \([I|S_n \dots S_2 S_1 I] = [I|A^{-1}].\)
We will find the inverse of \[ A = \begin{bmatrix} 2 & 6 \\ 1 & 2 \end{bmatrix} \]
Start with the augmented matrix, \([A|I]:\) \[ \left[ \begin{array}{cc|cc} 2 & 6 & 1 & 0 \\ 1 & 2 & 0 & 1 \end{array} \right] \]First we will get a \(1\) in the top left by swapping the rows. \[ \left[ \begin{array}{cc|cc} 1 & 2 & 0 & 1 \\ 2 & 6 & 1 & 0 \end{array} \right] \]
Next we cancel the \(2\) in row 2, column 1 by adding \(-2\) times row 1 to row 2. \[ \left[ \begin{array}{cc|cc} 1 & 2 & 0 & 1 \\ 0 & 2 & 1 & -2 \end{array} \right] \]
Get a \(1\) in the bottom right by dividing row 2 by \(2.\) \[ \left[ \begin{array}{cc|cc} 1 & 2 & 0 & 1 \\ 0 & 1 & 0.5 & -1 \end{array} \right] \]
Finally, cancel the \(2\) in row 1, column 2 by adding \(-2\) times row 2 to row 1. \[ \left[ \begin{array}{cc|cc} 1 & 0 & -1 & 3 \\ 0 & 1 & 0.5 & -1 \end{array} \right] \]