I'm supposed to "use properties of determinants to evaluate the determinant by inspection" on this matrix: $$\pmatrix{ 0 & 0 & 3 \\ 0 & 4 & 1 \\ 2 & 3 & 1 }$$ I don't see anything (zero rows, ways to transform the matrix) that would make it immediately obvious what the determinant is. What am I missing?
[Math] Find the determinant by inspection
linear algebra
Related Solutions
I normally have two ways of viewing determinants without appealing to higher-level math like multilinear forms.
The first is geometric, and I do think that most vector calculus classes nowadays should teach this interpretation. That is that, given vectors $v_1, \ldots, v_n \in \mathbb{R}^n$ dictating the sides of an $n$-dimensional parallelepiped, the volume of this parallelepiped is given by $\det(A)$, where $A = [v_1 \ldots v_n]$ is the matrix whose columns are given by those vectors. We can then view the determinant of a square matrix as measuring the volume-scaling property of the matrix as a linear map on $\mathbb{R}^n$. From here, it would be clear why $\det(A) = 0$ is equivalent to $A$ not being invertible - if $A$ takes a set with positive volume and sends it to a set with zero volume, then $A$ has some direction along which it "flattens" points, which would precisely be the null space of $A$. Unfortunately, I'm under the impression that this interpretation is at least semi-modern, but I think this is one of the cases where the modern viewpoint might be better to teach new students than the old viewpoint.
The old viewpoint is that the determinant is simply the result of trying to solve the linear system $Ax = b$ when $A$ is square. This is most likely how the determinant was first discovered. To derive the determinant this way, write down the generic matrix and then proceed by Gaussian elimination. This means you have to choose nonzero leading entries in each row (the pivots) and use them to eliminate subsequent entries below. Each time you eliminate the rows, you have to multiply by a common denominator, so after you do this $n$ times, you'll end up with the sum of all the permutations of entries from different rows and columns merely by virtue of having multiplied out to get common denominators. The $(-1)^k$ sign flip comes from the fact that at each stage in Gaussian elimination, you're subtracting. So on the first step you're subtracting, but on the second step you're subtracting a subtraction, and so forth. At the very end, by Gaussian elimination, you'll obtain an echelon form (upper triangular), and one knows that if any of the diagonal entries are zero, then the system is not uniquely solvable; the last diagonal entry will precisely be the determinant times the product of the values of previously used pivots (up to a sign, perhaps). Since the pivots chosen are always nonzero, then it will not affect whether or not the last entry is zero, and so you can divide them out.
EDIT: It isn't as simple as I thought, though it will work out if you keep track of what nonzero values you multiply your rows by in Gaussian elimination. My apologies if I mislead anyone.
This is simple. Note that
$$\det(PA) = \det(P)\det(A).$$
If you want $P$ to swap rows $k$ and $l$, then
$$P = \begin{bmatrix} 1 \\ & \ddots \\ & & 1 \\ & & & 0 & 0 & \dots & 0 & 1 \\ & & & 0 & 1 & \dots & 0 & 0 \\ & & & \vdots & \vdots & \ddots & \vdots & \vdots \\ & & & 0 & 0 & \dots & 1 & 0 \\ & & & 1 & 0 & \dots & 0 & 0 \\ & & & & & & & & 1 \\ & & & & & & & & & \ddots \\ & & & & & & & & & & 1 \end{bmatrix}.$$
In other words, $P$ is constructed by swapping rows (or, equivalently, columns) $k$ and $l$ of the identity matrix.
Now, check that $\det(P) = -1$, and you have what you asked about.
Best Answer
Interchange the first and third rows, which negates the determinant. Perhaps the matrix is now in a form that you recognize.