Determinants

Deciding whether a square matrix is invertible could be done by row reduction every time. The determinant answers it with a single number instead, and that number carries more than a yes or no.

Its magnitude is the factor by which the map stretches volume: a matrix with determinant 3 turns any region into one of three times the volume. Its sign says whether orientation survives. And it vanishing is not a failure but the informative case. It says the map has flattened space into something lower-dimensional, which is exactly the condition an eigenvalue computation searches for.

The arithmetic has two routes. Cofactor expansion is defined at any size and unusable beyond small ones; elimination is what anything larger uses. Choosing between them is part of the skill.

The cofactor sign is positional, not a property of the entry. $(-1{)}^{i+j}$ alternates like a chessboard:

The top-left is always $+$. A negative entry in a $+$ position contributes a negative term; the sign of the entry and the sign of the position are separate factors and both apply. Conflating them is the commonest arithmetic error in a hand computation.

Any row or column gives the same answer. This is a theorem, and it has two practical consequences. First, a row or column containing zeros is the one to expand along, since a zero entry annihilates its whole term regardless of the minor. Second, $det(A^{\mathsf{T}})=detA$ follows immediately: transposing swaps the roles of rows and columns, and both expansions were already available.

What multilinearity actually says. The determinant is linear in each row separately, with the other rows held fixed:

This is not linearity in the matrix. Scaling the whole $n\times n$ matrix by $k$ scales every one of its $n$ rows, so

and $det(A+B)\ne detA+detB$ in general. Both mistakes come from reading "linear" as applying to the matrix rather than to one row at a time.

Why $detA=0$ characterises singularity. If the columns are dependent, some column is a combination of the others; subtracting that combination, an operation that leaves the determinant unchanged, produces a zero column, and expanding along it gives zero. Conversely a nonzero determinant makes $A^{-1}=\frac{1}{detA}\mathrm{adj}(A)$ available, so the matrix is invertible. The equivalence with full rank and with a trivial kernel then follows from the earlier units.

Several matrices give up their determinants on sight. Recognising them saves the arithmetic and, more usefully, says what the value means.

Why equal rows force zero. Swapping two rows negates the determinant. Swapping two equal rows leaves the matrix unchanged, so $detA=-detA$, which over the reals forces $detA=0$. The same argument covers proportional rows, since scaling one row first turns the proportional case into the equal case.

Why a zero row forces zero. Expanding along it, every term carries a factor of zero. Geometrically the figure spanned by the rows has been flattened: one edge has length zero, so the volume is zero.

The permutation matrix. swaps the two coordinates. It preserves area, the unit square maps to the unit square, so the magnitude is 1, and it reverses orientation, so the sign is negative. In general a permutation matrix has determinant $+1$ for an even permutation and $-1$ for an odd one, which is the sign of the permutation.

Each value is read from structure rather than computed from entries: a diagonal, a dependency among rows, a single swap. Looking for that structure before starting an expansion is the cheapest step in any determinant calculation, and on a singular matrix it is the whole calculation.

Route 1: cofactor expansion.

1. Choose the row or column with the most zeros.
2. For each entry $a_{ij}$ in it, form the minor $M_{ij}$ by deleting row $i$ and column $j$.
3. Compute $(-1{)}^{i+j}{a}_{ij}det{M}_{ij}$, recursing until $2\times 2$.
4. Sum the terms.

Skip any entry that is zero: its term vanishes whatever the minor is, and computing that minor is wasted work.

Route 2: elimination to triangular form.

1. Reduce using row operations, recording the effect of each on the value.
2. Stop at upper triangular form.
3. Multiply the diagonal entries.
4. Undo the recorded effects.

The third row of that table is why elimination works at all: the workhorse operation costs nothing, so a whole reduction can be carried out with only swaps needing correction. Preferring it over row scaling keeps the bookkeeping trivial.

Choosing. Expansion costs about $n!$ multiplications, elimination about $n^3/3$. At $n=3$ they are comparable and expansion is often quicker by hand; at $n=5$ expansion needs 120 terms and elimination about 40 operations; at $n=10$ expansion is hopeless. Expand when the matrix is small or has a sparse row or column; eliminate otherwise.

Checks worth running.

• A zero row or a zero column forces $det=0$; so do two equal rows, or one row a multiple of another.
• A triangular matrix needs no work: multiply the diagonal.
• After computing, ask whether the verdict is plausible, if the columns visibly satisfy a dependence, the answer must be zero.

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Route 1: expansion along row 1.

Signs along the first row are $+,-,+$.

$j=1$: entry $2$, minor with determinant $4(1)-(-2)(0)=4$. Term: $+2(4)=8$.

$j=2$: entry $-1$, minor with determinant $1(1)-(-2)(3)=7$. Term: $-(-1)(7)=+7$. Both signs apply, the position's $-$ and the entry's own $-$, and they cancel.

$j=3$: entry $3$, minor with determinant $1(0)-4(3)=-12$. Term: $+3(-12)=-36$.

Route 1b: expansion along column 1, as a check. Signs down column 1 are $+,-,+$.

$i=1$: .

$i=2$: .

$i=3$: .

The agreement is the theorem, not a coincidence. Note that column 2 would have been the shrewd choice here: it contains a zero, so one of its three terms vanishes before any minor is computed.

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Route 2: elimination.

Use only "add a multiple of a row to another", which leaves the value unchanged.

$R_2\to R_2-\frac{1}{2}{R}_1$: $(1,4,-2)-\frac{1}{2}(2,-1,3)=(0,4.5,-3.5)$.

$R_3\to R_3-\frac{3}{2}{R}_1$: $(3,0,1)-\frac{3}{2}(2,-1,3)=(0,1.5,-3.5)$.

$R_3\to R_3-\frac{1}{3}{R}_2$: $(0,1.5,-3.5)-\frac{1}{3}(0,4.5,-3.5)=(0,0,-3.5+1.1\bar{6})=(0,0,-\frac{7}{3})$.

Upper triangular, with no swaps and no scalings, so the value is unchanged from $A$:

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$detA=-21\ne 0$, so $A$ is invertible, its columns are independent, its rank is 3, and the map $x\mapsto Ax$ has trivial kernel and image all of ${\mathbb{R}}^3$.

Geometrically: the unit cube maps to a parallelepiped of volume $21$, and the negative sign says orientation is reversed. The image is a mirrored copy, not a rotated one.

What each route cost. Expansion: three $2\times 2$ determinants and three multiplications. Elimination: three row operations and two multiplications. At $3\times 3$ they are close. At $6\times 6$ expansion would need 720 terms and elimination about 70 operations, which is the whole argument for the second route.

Take $A$ from the worked example, with $detA=-21$.

Swapping rows negates. Exchanging rows 1 and 2 gives

One swap, one sign change. Two swaps return the original value, which is why an even number of exchanges costs nothing.

Scaling a row scales the value. Multiplying row 1 by 5:

One row, one factor. Scaling the whole matrix by 5 would scale all three rows and give $5^3(-21)=-2625$, which is the $det(kA)=k^{n}detA$ rule.

Adding a multiple of a row changes nothing. Replacing row 2 by row 2 plus $3\times$ row 1:

The matrix is visibly different and the determinant is identical. Geometrically the figure has been sheared: the base is unmoved and the height is unchanged, so the volume is too.

Multiplicativity. With and :

and $(-2)(-1)=2$.

Read geometrically: $N$ scales volume by $|-1|=1$ and reverses orientation, $M$ scales by 2 and reverses again, so the composition scales by 2 and preserves orientation. Two reversals make a preservation, which is the product of the signs.

Two consequences. $det(A^{-1})=1/detA$, since $A{A}^{-1}=I$ has determinant 1. And similar matrices have equal determinants: $det(P^{-1}AP)=det(P{)}^{-1}detAdetP=detA$, so the determinant belongs to the transformation rather than to the basis it is written in, which is what makes it usable in eigentheory.

A singular check. has $detS=4-4=0$, and indeed its second column is twice the first. The map collapses the plane onto a line: zero area, dependent columns, nontrivial kernel, three statements of one fact.

Adding determinants. " $det(A+B)=detA+detB$." Take $A=I$ and $B=-I$ in dimension 2: $detA=1$, $detB=1$, and $A+B$ is the zero matrix with determinant 0, not 2. The determinant is linear in each row separately with the others fixed, which is a different statement about a different object.

Scaling the matrix as though it were one row. " $det(3A)=3detA$." Scaling the matrix scales all $n$ rows, so $det(3A)=3^{n}detA$. For the worked example's $3\times 3$ matrix that is $27(-21)=-567$, not $-63$.

Losing a sign in the expansion. Expanding along row 1 with entry $a_{12}=-1$ and minor determinant 7, the learner writes $-1\times 7=-7$, forgetting that the position also contributes $(-1{)}^{1+2}=-1$. Both factors apply: $(-1)(-1)(7)=+7$. The chessboard sign and the entry's own sign are independent, and the error is invisible in the final number unless the computation is checked by a second expansion.

Treating a zero determinant as a failed calculation. " $det=0$, so I must have made a mistake." Zero is a legitimate and informative value: the columns are dependent, the rank is below $n$, the kernel is nontrivial, and the map flattens space. Eigenvalue computations search for the values of $\lambda$ making $det(A-\lambda I)=0$, so the zero case is the one being hunted rather than an error to correct.

What separates these. The first two mistake which object the linearity applies to. The third is arithmetic with a conceptual cause, not knowing that two independent sign factors are in play. The fourth misreads an informative result as a broken one, and it is the most costly, because it leads to redoing correct work rather than acting on what it found.