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I have a question regarding a co-factor expansion question. I understand that an easy way to check if a matrix is invertible is to do co-factor expansion and if $A \ne 0$ then it's invertible. I'm familiar with the basic co-factor expansion with $3 \times 3$ matrices although I'm unfamiliar with $4\times 4$ matrices.

Could someone please give me a hint, or tell me how to go about checking if this matrix is invertible?

Q: Use the inversion algorithm to find the inverse of the matrix (if the inverse exists)

$$\begin{bmatrix}0 & 0 & 2 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & -1 & 3 & 0\\ 2 & 1 & 5 & -3 \end{bmatrix}$$

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    $\begingroup$ Exactly the same way with 3 by 3. Just checkerboard with +/- and then expand with 3 by 3, and repeat. $\endgroup$ – user223391 Oct 4 '15 at 3:43
  • $\begingroup$ I may have some idea of what you're saying although I'm quite unsure. Could you please show me? $\endgroup$ – Red Queen10101 Oct 4 '15 at 3:46
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To check if matrices are invertible, you need to check the determinant is non-zero:

To find the determinant of this matrix we look for the row or column with the most zeros and do a Laplace development on that row or column. The first row contains the most zeros so we Laplace develop that row:

$$\det=\begin{vmatrix}0 & 0 & 2 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & -1 & 3 & 0\\ 2 & 1 & 5 & -3 \end{vmatrix}$$

$$=0\times\begin{vmatrix} 0 & 0 & 1 \\ -1 & 3 & 0 \\ 1 & 5 & -3 \end{vmatrix}+0\times\begin{vmatrix} 1 & 0 & 1 \\ 0 & 3 & 0 \\ 2 & 5 & -3 \end{vmatrix}+2\times\begin{vmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 2 & 1 & -3 \end{vmatrix}+0\times\begin{vmatrix} 1 & 0 & 0 \\ 0 & -1 & 3 \\ 2 & 1 & 5 \end{vmatrix}$$$$=2\times\begin{vmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 2 & 1 & -3 \end{vmatrix}$$ $$=2\left( 1\times\begin{vmatrix}-1 & 0 \\ 1 & -3 \\ \end{vmatrix}+0\times\begin{vmatrix}0 & 0 \\ 2 & -3 \\ \end{vmatrix}+ 1\times\begin{vmatrix}0 & -1 \\ 2 & 1 \\ \end{vmatrix}\right)$$$$=2\times \left(1\times(3-0)+1\times(0--2)\right)= 2(3+2)=10 \ne 0$$ So your matrix is invertible (or has an inverse). By the way, when I say 'Laplace development', I think that's what you meant by co-factor expansion. They are not the same thing because the matrix of co-factors in something different. Does this help? Or is there something else I can do?

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    $\begingroup$ yes so this proves that it can be inverted. Could you show me how you got to 10? That's what I'm mainly curious about. $\endgroup$ – Red Queen10101 Oct 4 '15 at 3:58
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    $\begingroup$ Hey have you finished editing? Cause im still confused with your working out. $\endgroup$ – Red Queen10101 Oct 4 '15 at 4:49
  • $\begingroup$ @RedQueen10101 Hi sorry I had real format problems plus the answer I have given is wrong, but I know what I did wrong and how too explain it better sorry to take ages it's the matrix syntax that i'm not used to the answer really is 10 and i will show and explain how to get that $\endgroup$ – BLAZE Oct 4 '15 at 4:55
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    $\begingroup$ Hey your solutions still remain at 12 and not 10. Could you please tell me when you have finalised your solutions. $\endgroup$ – Red Queen10101 Oct 4 '15 at 5:31
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    $\begingroup$ @RedQueen10101 Sorry getting tired $\endgroup$ – BLAZE Oct 4 '15 at 5:34

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