Taken from khan academy  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

  • $\begingroup$ You probably want Gaussian elimination. $\endgroup$
    – Pedro M.
    Mar 2 '15 at 14:22
  • $\begingroup$ Well you could try writing down the nine equations in nine unknowns and solve the old fashioned hard way. $\endgroup$ Mar 2 '15 at 14:22
  • $\begingroup$ @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 $\endgroup$
    – Reddevil
    Mar 2 '15 at 14:33
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 2 '15 at 14:33
  • $\begingroup$ @AlgebraicPavel hahah what? you over used bla bla in your comment made it hard for me to follow along $\endgroup$
    – Reddevil
    Mar 2 '15 at 14:35

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

  • 3
    $\begingroup$ Bla bla bla bla. Bla bla, bla bla bla using Gauss bla bla, not bla bla adjugate blabla. $\endgroup$ Mar 2 '15 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.