# Error computing determinant of a $4\times 4$ matrix

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?

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).
• I did replace $R_3 \leftarrow R_1-R_3$. However, I am getting negative the correct answer not the correct answer! – AspiringMat Dec 3 '16 at 12:35
• @AspiringMat I say replace it with $R_3-R_1$ – Robert Z Dec 3 '16 at 12:37
• @AspiringMat Adding to row $R_3$ a linear combination of the others means that you can replace $R_3$ with $R_3+a_1R_1+a_2R_2+a_4R_4$. – Robert Z Dec 3 '16 at 12:42
• 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? – AspiringMat Dec 3 '16 at 12:45