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I'm not sure which method would be the best for finding the determinant of \begin{pmatrix}1&2&0&-1&2\\ 2&1&1&0&7\\ 0&3&0&0&-2\\ -1&-10&1&1&1\\ 0&9&0&0&1\end{pmatrix}

G Muf
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    Row/column operations? – Angina Seng Nov 28 '17 at 20:43
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    By hand? Generally above $4\times 4$ it's fastest to use Gaussian elimination (keeping track of the elementary operations as you go) until it's triangular, then take the product of the diagonal. – Mark Schultz-Wu Nov 28 '17 at 20:44
  • Take advantage that the matrix has a lot of zeros. e.g. subtract 2nd column by 5th column, the last row contains only one non-zero entry $1$ at position $(5,5)$. This reduce the determinant to a $4\times 4$ one. In the new matrix, the 3rd row has only one non-zero entry $21$ at position $(3,2)$, this reduce the determinant to a $3 \times 3$ one. This is essentially Laplace expansion together with elementary row/column operations but making a smart choice on which row/column to expand. – achille hui Nov 28 '17 at 21:02
  • I just found this answer posted a while ago: https://math.stackexchange.com/questions/1848952/finding-the-determinant-of-the-5x5-matrix-but-cant-put-it-in-lower-triangular-f Hope this helps! :) – Laura S Nov 28 '17 at 21:01

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First add column $4$ to column $1$ and expand by the relevant rows or columns: \begin{align} &\begin{vmatrix} 0&2&0&-1&2\\2&1&1&0&7\\0&3&0&0&-2\\0&-10&1&1&1\\0&9&0&0&1 \end{vmatrix}=-2 \begin{vmatrix} 2&0&-1&2\\3&0&0&-2\\-10&1&1&1\\9&0&0&1 \end{vmatrix}&&\text{(expanding by the 1st column)}\\[1ex] &=-2\cdot(-3)\begin{vmatrix} 0&-1&2\\1&1&1\\0&0&\color{red}1 \end{vmatrix}+(-2)(-2)\begin{vmatrix} 2&0&-1\\-10&1&1\\\color{red}9&0&0 \end{vmatrix}&&\text{(expanding by the 2nd row)}\\[1ex] &=6\begin{vmatrix} 0&-1\\1&1 \end{vmatrix}+4\cdot9\begin{vmatrix} 0&-1\\1&1 \end{vmatrix}=42.&&\text{(expanding by the last rows)} \end{align}

Bernard
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