1
$\begingroup$

I am trying to compute the determinant of

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

I first did the row operation $R_3 \leftarrow R_1-R_3$ so it doesn't change the determinant. So you want the determinant of:

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

and this simplifies to the $-2$ times the determinant of:

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

And this $3\times 3$ matrix has determinant of $-16$ so the determinant of $C$ is $(-2)\cdot (-16) = 32$. However, the book says that the answer is $-32$ and not $32$.

Any hints where the negative comes from?

$\endgroup$
1
$\begingroup$

Recall that the determinant does not change if you add to a row a linear combination of the others.

You should replace $R_3$ with $R_3-R_1$. Then you obtain $$\begin{bmatrix}-2 & 3 & 2 & 1 \\ 0 & 3 & 1 & -3 \\ 0 & -5 & -3 & 1 \\ 0 & 2 & 3 & 1 \end{bmatrix}$$ and the determinant of $$\begin{bmatrix} 3 & 1 & -3 \\ -5 & -3 & 1 \\ 2 & 3 & 1 \end{bmatrix}$$ is just the opposite of $-16$ (your previous computation).

$\endgroup$
6
  • $\begingroup$ I did replace $R_3 \leftarrow R_1-R_3$. However, I am getting negative the correct answer not the correct answer! $\endgroup$ – AspiringMat Dec 3 '16 at 12:35
  • $\begingroup$ @AspiringMat I say replace it with $R_3-R_1$ $\endgroup$ – Robert Z Dec 3 '16 at 12:37
  • $\begingroup$ Sorry I am a little confused. What's the difference here? Why would it change the determinant if we are allowed to add multiple of any row to another? $\endgroup$ – AspiringMat Dec 3 '16 at 12:38
  • $\begingroup$ @AspiringMat Adding to row $R_3$ a linear combination of the others means that you can replace $R_3$ with $R_3+a_1R_1+a_2R_2+a_4R_4$. $\endgroup$ – Robert Z Dec 3 '16 at 12:42
  • $\begingroup$ I see. So essentially I did $R_3 \leftarrow R_1-R_3$ which is two operations one that is yours and one replacing $R_3$ by $-R_3$ which we should inturn compensate for by multiplying the $3\times 3$ matrix determinant by -1 correct? $\endgroup$ – AspiringMat Dec 3 '16 at 12:45

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.