2
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ First entry on line 4 should be $(1 - 0)$, not $(1-3)$. $\endgroup$ – John_dydx Sep 12 '16 at 17:56
  • 1
    $\begingroup$ You've made an algebraic slip. The third line up from the bottom should be: $$(1-0) - 3(4-3) + 2(0-1) = -4.$$ $\endgroup$ – Zestylemonzi Sep 12 '16 at 17:57
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thanks so much! I need to be more careful. =) $\endgroup$ – heather Sep 12 '16 at 23:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.