Simplify the determinant of a $4 \times 4$ matrix.

Question

I have to find the determinant of the following 4x4 matrix: $\quad A=\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}$

So I apply the Gaussian elimination to obtain an upper-triangle matrix: $$det\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}=\begin{vmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix}\xrightarrow{3R_3-R_1}\begin{vmatrix}3&0&1&0\\0&2&0&0\\0&0&8&0\\0&0&0&-4\end{vmatrix}$$

Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $\quad 0 \quad 0 \quad 2 \quad 0 \quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?

Answer 1

It should be: $$\begin{vmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix}\rightarrow \begin{vmatrix}3&0&1&0\\0&2&0&0\\0&0&\color{red}{3-{1\over 3}}&0\\0&0&0&-4\end{vmatrix}$$

Answer 2

Rather than applying row operations, expand the minors.

$\det A = -4\begin{vmatrix} 3&0&1\\0&2&0\\1&0&3\end{vmatrix} = (-4)(18-2) = -64$

If you apply row operations, you don't want your row operations to change the determinant.

I think of multiplying by an elementary matrix.

$\begin{vmatrix} 1\\&1\\-\frac 13&&1\\&&&1\end{vmatrix}\begin{vmatrix} 3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix} = \begin{vmatrix} 3&0&1&0\\0&2&0&0\\0&0&\frac {8}{3}&0\\0&0&0&-4\end{vmatrix}$

keeps the determinant unchanged.

while

$\begin{vmatrix} 1\\&1\\-1&&3\\&&&1\end{vmatrix}\begin{vmatrix} 3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix} = \begin{vmatrix} 3&0&1&0\\0&2&0&0\\0&0&8&0\\0&0&0&-4\end{vmatrix}$

will change the determinant by a factor of $3.$

Answer 3

Just expand by the first row: \begin{align} \begin{vmatrix}3&0&1&0 \\ 0&2&0&0 \\ 1&0&3&0 \\ 0&0&0&-4 \end{vmatrix}&= 3\,\begin{vmatrix} 2&0&0 \\ 0&3&0 \\ 0&0&-4 \end{vmatrix}+ 1\,\begin{vmatrix} 0&2&0 \\ 1&0&0 \\ 0&0&-4 \end{vmatrix} =3(2\cdot3\cdot(-4))+1(-1)\begin{vmatrix} 2&0 \\ 0&-4 \end{vmatrix}\\ &=-72+8=-64. \end{align}

Answer 4

The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3\to3R_3$ and $R_3\to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.

Alternatively, you could’ve performed $R_3\to R_3-\frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.

As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.