# Inverse of 3-by-3 matrix

Hi, so this question is taken straight from khan academy help exercises, i know how to do it dynamically meaning using the determinant and the adjugate how i was trying to do it using guass bla bla way with help of RREF but i somehow never managed to find the inverse. my second question would be is there anyway that i can find the whether or not the matrix is invertable without trying to find the determinant i mean also using Gauss bla bla way i use the word bla bla because i dont know what it is actually called :p

• You probably want Gaussian elimination. – Pedro M. Mar 2 '15 at 14:22
• Well you could try writing down the nine equations in nine unknowns and solve the old fashioned hard way. – Gregory Grant Mar 2 '15 at 14:22
• @PedroM. yea its a little bit of guassian eli. and RREF too but if someone knows how to do it and shows me it'd be much appreciated – Reddevil Mar 2 '15 at 14:33
• You want to find the bla bla RREF of $(D,I)$, giving you bla bla $(I,D^{-1})$ if $D^{-1}$ exists. If you can't find bla bla $(I,D^{-1})$, the matrix is not invertible or how is it actually called. – Algebraic Pavel Mar 2 '15 at 14:33
• @AlgebraicPavel hahah what? you over used bla bla in your comment made it hard for me to follow along – Reddevil Mar 2 '15 at 14:35

## 2 Answers

Using Gauss-Jordan elimination: $$\left[\begin{array}{ccc|ccc} 0&1&2 &1&0&0\\ 1&0&1 &0&1&0\\ 0&1&0 &0&0&1 \end{array}\right] \to \\ \left[\begin{array}{ccc|ccc} 1&0&1 &0&1&0\\ 0&1&0 &0&0&1\\ 0&1&2 &1&0&0\\ \end{array}\right] \to\\ \left[\begin{array}{ccc|ccc} 1&0&1 &0&1&0\\ 0&1&0 &0&0&1\\ 0&0&2 &1&0&-1\\ \end{array}\right] \to \\ \left[\begin{array}{ccc|ccc} 1&0&1 &0&1&0\\ 0&1&0 &0&0&1\\ 0&0&1 &1/2&0&-1/2\\ \end{array}\right] \to \\ \left[\begin{array}{ccc|ccc} 1&0&0 &-1/2&1&1/2\\ 0&1&0 &0&0&1\\ 0&0&1 &1/2&0&-1/2\\ \end{array}\right]$$ So, we conclude $$D^{-1} = \pmatrix{ -1/2&1&1/2\\ 0&0&1\\ 1/2&0&-1/2}$$

bla bla bla bla : $$|{\rm D}|=2$$ blaaa blabla blabla: $${\rm adj\; A}=\left[\begin{matrix}-1&2&1\\0&0&2\\1&0&-1\end{matrix}\right]$$ blah blehblaqa bla: $${\rm A}^{-1}=\left[\begin{matrix}-1/2&1&1/2\\0&0&1\\1/2&0&-1/2\end{matrix}\right]$$

• Bla bla bla bla. Bla bla, bla bla bla using Gauss bla bla, not bla bla adjugate blabla. – Ben Grossmann Mar 2 '15 at 15:06