how to find $A^{-1}$ in a matrix

Question

I have been given this matrix

$$A=\begin{pmatrix} 3 & 9 & 4 \\ 5 & 6 & 8 \\ 1 & 0 & 2 \end{pmatrix}$$

and I have to find $A^{-1}$, how would I go about doing this?

Answer 1

One of the first methods commonly taught is to row reduce $\left[\begin{array}{ccc|ccc}3&9&4&1&0&0\\5&6&8&0&1&0\\1&0&2&0&0&1\end{array}\right]$ and put this into reduced row echelon form

Assuming the inverse actually exists where initially we begin with $[A\mid I]$ the result of the row reduction will be $[I\mid A^{-1}]$. If it is impossible to row reduce in such a way that the left-half of the augmented matrix reduces to the identity then the matrix is not invertible.

Other methods exist as well and can be found on the wiki page.

Answer 2

As pointed out by @JMoravitz, one strategy is to use elementary operations to reduce the augmented matrix $$ \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I} \end{array} \right] \to \left[ \begin{array}{c|c} \mathbf{I} & \mathbf{A}^{-1} \end{array} \right]. $$

1. Swap the first and third rows, this makes the top row look more like the target identity matrix. 2 Row 2: subtract 5 $\times$ row 1. Row 3: subtract 3 $\times$ row 1.
2. Normalize rows 2 and 3.
3. Subtract row 2 from row 3.
4. Normalize row 3.
5. Row 1: subtract 2 $\times$ row 3. Row 2: add one third of row 3.

$$ \left[ \begin{array}{rrr|rrr} 3 & 9 & 4 & 1 & 0 & 0 \\ 5 & 6 & 8 & 0 & 1 & 0 \\ 1 & 0 & 2 & 0 & 0 & 1 \\ \end{array} \right] % \overset{1}{\longrightarrow} % \left[ \begin{array}{rrr|rrr} 1 & 0 & 2 & 0 & 0 & 1 \\ 5 & 6 & 8 & 0 & 1 & 0 \\ 3 & 9 & 4 & 1 & 0 & 0 \\ \end{array} \right] % \overset{2}{\longrightarrow} % \left[ \begin{array}{rrr|rrr} 1 & 0 & 2 & 0 & 0 & 1 \\ 0 & 6 & -2 & 0 & 1 & -5 \\ 0 & 9 & -2 & 1 & 0 & -3 \\ \end{array} \right] % \overset{3}{\longrightarrow} % \left[ \begin{array}{rrr|rrr} 1 & 0 & 2 & 0 & 0 & 1 \\ 0 & 1 & -\frac{1}{3} & 0 & \frac{1}{6} & -\frac{5}{6} \\ 0 & 1 & -\frac{2}{9} & \frac{1}{9} & 0 & -\frac{1}{3} \\ \end{array} \right] % \overset{4}{\longrightarrow} % \left[ \begin{array}{rrr|rrr} 1 & 0 & 2 & 0 & 0 & 1 \\ 0 & 1 & -\frac{1}{3} & 0 & \frac{1}{6} & -\frac{5}{6} \\ 0 & 0 & \frac{1}{9} & \frac{1}{9} & -\frac{1}{6} & \frac{1}{2} \\ \end{array} \right] % \overset{5}{\longrightarrow} % \left[ \begin{array}{rrr|rrr} 1 & 0 & 2 & 0 & 0 & 1 \\ 0 & 1 & -\frac{1}{3} & 0 & \frac{1}{6} & -\frac{5}{6} \\ 0 & 0 & 1 & 1 & -\frac{3}{2} & \frac{9}{2} \\ \end{array} \right] % \overset{6}{\longrightarrow} % \left[ \begin{array}{rrr|rrr} 1 & 0 & 0 & -2 & 3 & -8 \\ 0 & 1 & 0 & \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \\ 0 & 0 & 1 & 1 & -\frac{3}{2} & \frac{9}{2} \\ \end{array} \right] $$

Harvest the matrix inverse $$ \mathbf{A}^{-1} = \frac{1}{6} \left[ \begin{array}{rrr} -12 & 18 & -48 \\ 2 & -2 & 4 \\ 6 & -9 & 27 \\ \end{array} \right] $$

You should verify that $\mathbf{A} \, \mathbf{A}^{-1}= \mathbf{A}^{-1}\, \mathbf{A}=\mathbf{I}_{3}.$