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$$ \begin{pmatrix} 5 & 6 & 6 & 8 \\ 2 & 2 & 2 & 8 \\ 6 & 6 & 2 & 8 \\ 2 & 3 & 6 & 7 \\ \end{pmatrix} $$

Is this matrix invertible? I would like to show that it is invertible but first I should find the det(Matrix) which should not be equal to zero. To find the determinant, maybe the best idea is to use row operations and find an upper triangular of zeroes and then multiply the numbers on the diagonal to get the determinant. I have been doing some row operations and get this: $$ \begin{pmatrix} 5 & 6 & 6 & 8 \\ 0 & -1 & -4 & 1 \\ 0 & 0 & 2 & 6 \\ -1 & 0 & 0 & -12 \\ \end{pmatrix} $$
I just need to get rid of the -1 on the last row. But I am stuck. Thank you for your assistance.

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    $\begingroup$ Yes ${}{}{}{}{}$ Write the augmented matrix $$[~A~ |~ I~]$$ and then perform operations on it to find the inverse. $\endgroup$
    – Moo
    Commented Dec 20, 2017 at 20:26
  • $\begingroup$ If you applied the Gauss Algorithm you did something wrong. Maybe restart with Gauss paying attention to the right steps. $\endgroup$
    – user507623
    Commented Dec 20, 2017 at 20:27
  • $\begingroup$ $$\left( \begin{array}{cccc} 2 & 2 & 2 & 8 \\ 0 & 1 & 1 & -12 \\ 0 & 0 & 3 & 11 \\ 0 & 0 & 0 & -\frac{4}{3} \\ \end{array} \right)$$ $\endgroup$
    – Raffaele
    Commented Dec 20, 2017 at 20:37
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    $\begingroup$ Please do not change the matrix we are dealing with, since it might invalidate some of the answers already provided. Have a look at this trick, too. $\endgroup$ Commented Dec 20, 2017 at 20:59

5 Answers 5

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A combination of Gaussian elimination and a well-known trick readily gives the answer. $$\det \begin{pmatrix} 5 & 6 & 6 & 8 \\ 2 & 2 & 2 & 8 \\ 6 & 6 & 2 & 8 \\ 2 & 3 & 6 & 7 \\ \end{pmatrix} =\det\begin{pmatrix} 5 & 6 & 6 & 2 \\ 2 & 2 & 2 & 6 \\ 6 & 6 & 2 & 2 \\ 2 & 3 & 6 & 4 \\ \end{pmatrix} =4\det\begin{pmatrix} 5 & 6 & 6 & 2 \\ 1 & 1 & 1 & 3 \\ 3 & 3 & 1 & 1 \\ 2 & 3 & 6 & 4 \\ \end{pmatrix} \\[0.5cm]=4\det\begin{pmatrix} 5 & 6 & 6 & 2 \\ 1 & 1 & 1 & 3 \\ 2 & 2 & 0 & -2 \\ 2 & 3 & 6 & 4 \\ \end{pmatrix}=8\det\begin{pmatrix} 5 & 6 & 6 & 2 \\ 1 & 1 & 1 & 3 \\ 1 & 1 & 0 & -1 \\ 2 & 3 & 6 & 4 \\ \end{pmatrix} $$ and the last determinant is an odd integer, hence the original matrix is invertible. $$ \det\begin{pmatrix} 5 & 6 & 6 & 2 \\ 1 & 1 & 1 & 3 \\ 1 & 1 & 0 & -1 \\ 2 & 3 & 6 & 4 \\ \end{pmatrix}\equiv \det\begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix}\equiv 1\pmod{2}.$$

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Proof only by determinant

We have $$\begin{vmatrix} 5 & 6 & 6 & 8 \\ 2 & 2 & 2 & 8 \\ 6 & 6 & 2 & 8 \\ 2 & 3 & 6 & 7 \\ \end{vmatrix}=A-B+C-D$$ where $$A=5\begin{vmatrix} 2 & 2 & 8 \\ 6 & 2 & 8 \\ 3 & 6 & 7 \\ \end{vmatrix}=5\left(2\begin{vmatrix} 2 & 8 \\ 6 & 7 \\ \end{vmatrix}-2\begin{vmatrix} 6 & 8 \\ 3 & 7 \\ \end{vmatrix}+8\begin{vmatrix} 6 & 2 \\ 3 & 6 \\ \end{vmatrix}\right)=5(-68-36+240)=680,$$ $$B=6\begin{vmatrix} 2 & 2 & 8 \\ 6 & 2 & 8 \\ 2 & 6 & 7 \\ \end{vmatrix}=6\left(2\begin{vmatrix} 2 & 8 \\ 6 & 7 \\ \end{vmatrix}-2\begin{vmatrix} 6 & 8 \\ 2 & 7 \\ \end{vmatrix}+8\begin{vmatrix} 6 & 2 \\ 2 & 6 \\ \end{vmatrix}\right)=6(-68-52+256)=816,$$ $$C=6\begin{vmatrix} 2 & 2 & 8 \\ 6 & 6 & 8 \\ 2 & 3 & 7 \\ \end{vmatrix}=6\left(2\begin{vmatrix} 6 & 8 \\ 3 & 7 \\ \end{vmatrix}-2\begin{vmatrix} 6 & 8 \\ 2 & 7 \\ \end{vmatrix}+8\begin{vmatrix} 6 & 6 \\ 2 & 3 \\ \end{vmatrix}\right)=6(36-52+48)=192$$ and $$D=8\begin{vmatrix} 2 & 2 & 2 \\ 6 & 6 & 2 \\ 2 & 3 & 6 \\ \end{vmatrix}=8\left(2\begin{vmatrix} 6 & 2 \\ 3 & 6 \\ \end{vmatrix}-2\begin{vmatrix} 6 & 2 \\ 2 & 6 \\ \end{vmatrix}+2\begin{vmatrix} 6 & 6 \\ 2 & 3 \\ \end{vmatrix}\right)=8(60-64+12)=64$$ Thus $$\begin{vmatrix} 5 & 6 & 6 & 8 \\ 2 & 2 & 2 & 8 \\ 6 & 6 & 2 & 8 \\ 2 & 3 & 6 & 7 \\ \end{vmatrix}=680-816+192-64=-8\neq0$$ so your matrix has an inverse.

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We have $\det(A)=-8$ and $$ A^{-1}=\frac{1}{4}\begin{pmatrix} -68 & -36 & 48 & 64\cr 68 & 35& -47 & -64 \cr -16 & -9 & 11 & 16 \cr 4 & 3 & -3 &-4 \end{pmatrix} $$ Testing $AA^{-1}=I_4$ shows that $A$ is indeed invertible. So the question "Is this matrix invertible?" is answered, but of course there are many ways to obtain this result.

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Just go on converting to zero the three first elements of the last row. Removing the first one you lose the zeros and second and third position and get nasty fractions instead, but it doesn't matter. Just go on.

Alternatively, you can compute the determinant this way $$5\begin{vmatrix}-1&-4&1\\0&2&6\\0&0&-12\end{vmatrix}-(-1)\begin{vmatrix}6&6&8\\-1&-4&1\\0&2&6\end{vmatrix}$$

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  • $\begingroup$ I think the OP made a mistake when doing the row operations since the determinant you give is $-16$ not $-8$ (my first deleted post) $\endgroup$
    – TheSimpliFire
    Commented Dec 20, 2017 at 20:48
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You may find interesting to use Hamilton-Caley Formula, for $4x4$ matrices:

$$\text{det}\mathsf{A} = \frac{1}{24}\left\{(\text{tr}\mathsf{A}^4 - 6\text{tr}(\mathsf{A}^2)(\text{tr}\mathsf{A})^2 + 3(\text{tr}\mathsf{A})^2 + 8\text{tr}(\mathsf{A}^3)\text{tr}\mathsf{A} - 6\text{tr}(\mathsf{A}^4)\right\}$$

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