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I'm calculating the determinant of the matrix $$\begin{bmatrix}1&3&2\\4&1&3\\1&0&1\end{bmatrix}$$ And I know that $$\text{det} \left(\,\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\,\right)$$ is equal to $$\left(a \,\text{det} \left(\begin{bmatrix} e&f\\h&i\end{bmatrix} \right) \right) - \left( b \, \text{det}\left( \begin{bmatrix} d&f\\h&i\end{bmatrix} \right)\right) +\left( c \, \text{det}\left( \begin{bmatrix} d&e\\g&h\end{bmatrix} \right)\right)$$ So for this one I said it was equal to $$\det\left(\begin{bmatrix} 1&3\\0&1\end{bmatrix} \right) - 3\left( \det\left( \begin{bmatrix} 4&3\\1&1\end{bmatrix}\right)\right) +2\left( \det\left( \begin{bmatrix} 4&1\\1&0\end{bmatrix}\right)\right)$$ or $$(1-3) - (3(4-3)) + (2(0-1))$$ or $$-2-3-2$$ or $$-7$$ But the text of the solution is

We can compute the determinant by expanding about any row or column; the most efficient ones to choose are either the second column or the third row. In any case, the determinant will be −4.

I checked over practically everything, and I can't see anything I did wrong. Any help would be appreciated. Thanks!

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    $\begingroup$ First entry on line 4 should be $(1 - 0)$, not $(1-3)$. $\endgroup$
    – John_dydx
    Commented Sep 12, 2016 at 17:56
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    $\begingroup$ You've made an algebraic slip. The third line up from the bottom should be: $$(1-0) - 3(4-3) + 2(0-1) = -4.$$ $\endgroup$ Commented Sep 12, 2016 at 17:57

1 Answer 1

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$\det\begin{bmatrix} a& b \\ c& d\end{bmatrix}=ad-bc$ so for $\det\begin{bmatrix} 1& 3 \\ 0& 1\end{bmatrix}=(1)(1)-(3)(0)=1-0=1\neq-2$

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  • $\begingroup$ Thanks so much! I need to be more careful. =) $\endgroup$
    – auden
    Commented Sep 12, 2016 at 23:46

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