I know that I can find the determinant by working my way from 4x4 to 3x3 to 2x2 determinant. However that is long.

Which other method can I use?

\begin{bmatrix} 2 & 1 & 0 & 6 \\ 2 & 5 & 3 & 0 \\ 1 & 6 & 0 &-2 \\ 0 &-2 & 5 & -1 \end{bmatrix}

  • 1
    $\begingroup$ Hints: Use $0$'s to cut down on the work. Also, you can add a multiple of one row to another row without changing the determinant. For example, here, you could start with $-2R_3+R_1\rightarrow R_1$ and $-2R_3+R_2\rightarrow R_2$ to introduce more zeros in the first column. In general, it takes some work to compute a determinant (practice to speed up the calculation) $\endgroup$ – Michael Burr Mar 6 '16 at 22:10
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    $\begingroup$ You could start by decomposing along column 3 since this already has 2 zeros-less work to do. Alternatively, you could just use elementary row operations as suggested in the above comment. $\endgroup$ – John_dydx Mar 6 '16 at 22:16

Substract twice the third row both from the second and the first rows to get:


Develop by the first column and get:


  • $\begingroup$ why does this work? $\endgroup$ – WiccanKarnak Jun 19 '19 at 9:12
  • $\begingroup$ @WiccanKarnak Basic properties of general multilinear functions: as long as you add any scalar multiple of a row/column to a row/column, the determinant doesn't change. The development by row/column is another thing but also based on the same basic properties. $\endgroup$ – DonAntonio Jun 19 '19 at 16:22

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