Inverse of 3-by-3 matrix

Question

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

Answer 1

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 $$ D^{-1} = \pmatrix{ -1/2&1&1/2\\ 0&0&1\\ 1/2&0&-1/2} $$

Answer 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: $${\rm A}^{-1}=\left[\begin{matrix}-1/2&1&1/2\\0&0&1\\1/2&0&-1/2\end{matrix}\right]$$