Determinants, rank and invertibility

A size question, a direction count and a reversibility question

The columns b₁=(2,1) and b₂=(1,−1) formed the basis B in lesson 124. They also form adjacent sides of a parallelogram. The number 2(−1)−1(1)=−3 records its signed area: its ordinary area is 3, and the minus sign records the orientation of the ordered columns. Swapping the columns changes that sign while leaving the parallelogram's area unchanged.

Now compare projection P(x,y)=(x,0). Its two standard matrix columns are (1,0) and (0,0), so the unit square collapses onto a line and the determinant is zero. Its rank is 1, because one independent output direction survives. It cannot be reversed on R²: all inputs (x,y) with the same x have the same image.

These observations connect three questions, but the quantities are different. A determinant is one scalar attached to a square matrix; rank counts independent directions and also applies to rectangular matrices; invertibility asks for a two-sided inverse. Preparation: the reversible elimination of Matrices 122, spans and independence from 123, and bases, kernel, image and rank-nullity from 124. All scalars here are real. The general statements allow any finite positive matrix size; the explorer uses exact small rational examples.

A determinant belongs to a square matrix

For A=[[a,b],[c,d]], define det A=ad−bc. The vertical-bar notation |A| is also used, but it means a determinant here, not an entrywise absolute value. A one-by-one matrix [a] has determinant a. The determinant of Iₙ is 1. For example, [[2,1],[1,−1]] has determinant −3, while the shear [[1,2],[0,1]] has determinant 1.

The usual determinant is defined for an n-by-n matrix. A two-by-three matrix has three input coordinates and two output coordinates; it has a rank, but no ordinary square determinant. Neither deleting an arbitrary column nor multiplying the available diagonal entries defines its determinant. We will treat rectangular rank directly.

Already in two dimensions, det A can be negative, zero or positive. Its sign and magnitude will have geometric meanings when the columns are expressed in a fixed ordinary coordinate system. The scalar itself is an exact algebraic quantity; it is not a count of pivots or the number of solutions of Ax=b.

What makes different determinant calculations agree

The determinant is characterized by three column properties. It is linear in each one column when the other columns stay fixed; it is zero when two columns are equal; and it takes value 1 on the ordered standard basis, the columns of Iₙ. Linear in one column means that replacing that column by u+v adds the two resulting determinants, and replacing it by ku multiplies that determinant by k. This separate linearity is called multilinearity.

The equal-column property and multilinearity imply that exchanging two distinct columns negates the determinant. To see the sign, put u+v in both of those positions. The determinant is zero because the columns agree. Expanding those two positions gives terms with u,u and v,v, which vanish, and the two remaining terms with u,v and v,u. Their sum is zero, so the latter terms have opposite signs.

These properties determine the value on any matrix. Write column j as Σᵢaᵢⱼeᵢ and expand one column at a time. Any term with a repeated standard vector vanishes. The surviving terms choose every index exactly once: they correspond to permutations σ of 1,…,n. The value is Σσ sign(σ) aσ(1),1⋯aσ(n),n, where sign(σ) is −1 for an odd number of index inversions and +1 for an even number. An inversion is a pair of positions j<k with σ(j)>σ(k).

This finite sum also supplies an existence formula. Each entry from a given column appears linearly, terms cancel in pairs when two columns coincide, and only the identity selection survives for Iₙ. Thus it has the required properties. The expansion argument shows uniqueness. Different valid computation methods agree because they calculate this same function; agreement is not an assumption about a particular elimination path.

Multilinearity does not say det(A+B)=det A+det B. With A=B=I₂, the left side is det(2I₂)=4, while the right side is 2. Scaling every column by k gives det(kA)=kⁿdet A, since there are n separate column factors. Scaling just one column gives one factor k.

Minors and cofactors organize a computation

Delete row i and column j of a square A to obtain its minor matrix Mᵢⱼ. The cofactor is Cᵢⱼ=(−1)ⁱ⁺ʲdet Mᵢⱼ, with row and column numbers starting at 1. The signs form a checkerboard: +,−,+ in the first row, −,+,− in the second, and so on. A minor is a smaller matrix; a cofactor is a signed scalar computed from it.

Expansion along any fixed row i gives det A=ΣⱼaᵢⱼCᵢⱼ. Expansion along any fixed column j gives det A=ΣᵢaᵢⱼCᵢⱼ. Grouping the permutation terms according to the chosen entry in that row or column gives these formulas. They calculate the same determinant, so choose a row or column with zeros when possible. The cofactor sign belongs to the entry's position in the original matrix, before its row and column are deleted.

For A=[[1,2,0],[0,3,1],[2,1,4]], expand along row 2. Its cofactors are −8, 4 and 3, so det A=0(−8)+3(4)+1(3)=15. Expanding row 1 instead gives 1(12−1)−2(0−2)+0=15. The zero entry removes a term; it does not remove a row or shift the signs of the remaining terms.

The recursion ends at a one-by-one determinant. If the cofactor formula is applied to that case too, the empty zero-by-zero determinant is assigned value 1, the empty product. The explorer's square inputs have sizes 2 or 3. Cofactors are useful for exact small calculations and algebraic arguments; repeatedly expanding dense large matrices creates many subproblems, for which elimination is usually a more practical method.

The checkerboard starts +, −, + in row 1; row 2 starts −, +, −. Here 11 + 4 + 0 = 15. The minor is a matrix; its determinant is a number.

Keep an account of row operations

The permutation formula also gives det(Aᵀ)=det A: a term indexed by a permutation corresponds to the term indexed by its inverse, with the same sign and factors in a different order. Consequently, the column rules apply to rows as well. Swapping two different rows negates the determinant. Multiplying one row by k multiplies it by k. Adding a multiple of a different row to a row leaves it unchanged.

For the replacement rule, expand the changed row Rᵢ+kRⱼ using separate linearity. The first term is the original determinant. The second has row j twice and is zero. A row of zeros also forces a zero determinant, by linearity in that row. Two equal rows force zero because exchanging them both negates the determinant and leaves the matrix unchanged.

For an upper- or lower-triangular square matrix, the determinant is the product of the diagonal entries. In the permutation formula every other choice encounters a required zero. For a diagonal example diag(2,−3,1/2), the determinant is −3. Arbitrary entries on the permitted side of the diagonal do not change that product.

Suppose reversible row operations turn A into a triangular matrix U. If there were s swaps and the product of all nonzero row-scaling factors was q, then det U=(−1)ˢq det A. Therefore det A=(−1)ˢ(det U)/q. Keep replacements, swaps and scales distinct. The diagonal product belongs to the final U; forgetting to undo a row scale or a swap changes the answer for the original A.

The scaling identity also holds for k=0, giving a zero row and a zero determinant. Multiplication by zero is not a reversible elementary row operation. It can destroy rank and solution information, so it cannot be used to infer the original determinant by dividing by its scale factor. The explorer labels this separate erasure explicitly.

Rank counts independent directions

For an m-by-n real matrix A, rank A is the dimension of its column space, equivalently the dimension of the image of x↦Ax from Rⁿ to Rᵐ. It is the number r of pivots in a row-echelon form. Thus 0≤r≤min(m,n), and rank-nullity gives dim ker A=n−r. Original pivot columns, rather than their reduced replacements, supply a basis of the original image.

Why do pivot positions identify independent original columns? A reversible row operation is left multiplication by an invertible matrix E. For any coefficient list c, Ac=0 if and only if EAc=0. Thus all linear relations among the columns are preserved. The pivot columns of a reduced row-echelon matrix are independent, and its other columns are their combinations. Their positions identify an independent spanning selection from the original columns too.

Row rank equals the same r. Reversible row operations replace rows by invertible combinations, preserving their row span. The r nonzero rows of reduced row-echelon form are independent: in a zero combination, the pivot columns force each row coefficient to vanish. They also span its row space. The row and column spaces usually lie in different spaces, Rⁿ and Rᵐ, even though their dimensions agree.

The matrix [[1,0,1],[0,1,1]] has rank 2. It reaches every vector of R² but loses the input direction (−1,−1,1), so it is onto and not injective. The matrix [[1,0],[0,1],[1,1]] also has rank 2. It is injective, but its image is only the plane z=x+y inside R³. Full rank for a rectangle means rank min(m,n); it does not give a two-sided inverse or a square determinant.

For a square matrix, the tests meet

An n-by-n matrix A is invertible if there is a matrix A⁻¹ with A⁻¹A=AA⁻¹=Iₙ. The inverse undoes the map in both orders. For a square matrix the following conditions are equivalent: det A≠0; rank A=n; the columns are a basis of Rⁿ; ker A={0}; the map is both injective and onto Rⁿ; and Ax=b has exactly one solution for every b∈Rⁿ.

Elimination connects the determinant to rank. Reversible row operations have nonzero determinant factors, so they preserve whether the determinant vanishes. If rank A=n, reduced row-echelon form is Iₙ, whose determinant is 1. If rank A<n, that form has a zero row and determinant zero. Therefore det A is nonzero exactly when the rank is n.

The remaining links have a vector-space explanation. With n independent columns in Rⁿ, the columns are a basis, so every b has a unique coefficient list x. This defines a bijective linear map and a linear inverse. Conversely, an inverse makes Ax=0 imply x=0 and gives a solution x=A⁻¹b for every b. Rank-nullity then gives full rank. This proves the equivalences without treating the determinant as a separate count of solutions.

If a square determinant is zero, Ax=b can have no solutions or infinitely many; b decides which. For A=[[1,2],[2,4]], b=(3,6) gives infinitely many solutions, while b=(3,5) is inconsistent. Neither case has a unique solution. The kernel contains (−2,1), so adding any multiple of that vector to one solution produces another when a solution exists.

A zero determinant does not say the matrix is zero or that its rank is zero. The three-by-three matrix with rows (1,2,3),(2,4,6),(1,1,1) has rank 2 and determinant zero. It retains a plane of outputs while losing one input direction. The determinant detects the loss of full n-dimensional reversibility; rank describes how many dimensions remain.

Products and inverses give useful checks

For square matrices A and B of the same size, det(AB)=det A det B, including when a factor is singular. One way to see this fixes A and considers det(Av₁,…,Avₙ) as a function of the columns v₁,…,vₙ. It is multilinear and alternating. The standard-basis expansion from earlier therefore makes it det A times det(v₁,…,vₙ). Taking the v's to be B's columns gives the product formula.

Although matrix multiplication depends on order, determinants are scalars, so det(AB)=det(BA) for square A and B of the same size. The products themselves can still differ. For the shear T=[[1,2],[0,1]] and swap R=[[0,1],[1,0]], both TR and RT have determinant −1, although their actions and matrices differ as in lessons 122 and 124.

If A is invertible, 1=det I=det(AA⁻¹)=det A det(A⁻¹), hence det(A⁻¹)=1/det A. Also det(Aᵀ)=det A, as shown from the permutation formula. For the two-by-two A=[[a,b],[c,d]] with ad−bc≠0, direct multiplication checks A⁻¹=(1/(ad−bc))[[d,−b],[−c,a]]. The zero-determinant case cannot use that division. The explorer computes an inverse only after square full rank is established, using the existing exact system solver.

Keep the coordinate conventions in view

The determinant of a fixed square array is unambiguous. Relating it to a map requires naming the input and output coordinates. For an operator V→V, using the same basis on both sides gives a determinant independent of that common basis. A change to another common basis changes its matrix from A to S⁻¹AS. The product formula gives det(S⁻¹AS)=(1/det S)(det A)(det S)=det A.

Different input and output bases need an extra conversion. In ordinary coordinates, T(x,y)=(x+2y,y) has determinant 1. Lesson 124 used input basis B=((2,1),(1,−1)) and output basis C=((1,1),(0,1)), producing [T]ᶜ←ᴮ=[[4,−1],[−3,0]], whose determinant is −3. These are consistent: the basis-column matrices satisfy det Pᴮ=−3 and det Pᶜ=1, so det T=(det Pᶜ)(det[T]ᶜ←ᴮ)/(det Pᴮ)=1.

Thus the determinant of that mixed-coordinate array is not directly the signed Euclidean area factor in ordinary coordinates. It compares the input B coordinate unit box with the output C coordinate unit box. In the geometric diagrams below, the displayed columns use the same ordinary x,y coordinate axes and unit spacing. The coordinate convention is part of the interpretation, not an optional label.

Using the same basis on both sides instead gives a similarity matrix P⁻¹TP, whose determinant stays 1. Here B and C differ, so that cancellation is unavailable.

Signed size and ordinary volume

In ordinary coordinates, a square A sends the unit cube to the parallelepiped whose edge vectors are its columns. Its nonnegative n-dimensional volume is |det A|. In the plane this is parallelogram area; in three dimensions it is parallelepiped volume. For a triangle determined by two adjacent edge vectors in the plane, the area is half the parallelogram's area.

The sign records orientation when the determinant is nonzero. In the plane, a positive determinant means the ordered first and second columns have the usual counterclockwise orientation; a negative determinant means the opposite. Interchanging two columns reverses this order and the sign, without changing ordinary area. A zero determinant gives zero n-dimensional volume, even if the collapsed image still has positive length or lower-dimensional area.

For an invertible linear map in the same ordinary coordinates, volumes of measurable regions are multiplied by |det A|. The parallelepiped statement gives the basic picture; extending it to general regions uses approximation or integration. The shear with determinant 1 preserves area, but it need not preserve lengths or angles: it sends the unit vector (0,1) to (2,1). Preserving area is a weaker condition than preserving every geometric measurement.

A parameter can change the rank

Consider A(t)=[[1,1],[1,1+t]]. Replacing row 2 by row 2 minus row 1 gives [[1,1],[0,t]], without changing the determinant. Therefore det A(t)=t. For every nonzero t the rank is 2 and the matrix is invertible. At t=0 the rows coincide, the rank is 1 and ker A(0)=span((1,−1)).

For t≠0, A(t)⁻¹=(1/t)[[1+t,−1],[−1,1]]. At t=1/4 this is [[5,−4],[−4,4]]. Multiplying that inverse by (1,0) gives (5,−4), whose image under A(1/4) is (1,0). At t=0 there is no inverse formula to evaluate by continuity or division; the map has changed its rank.

The explorer selects six exact rational t values rather than using a floating-point near-zero test. The nonzero examples illustrate the same algebraic condition, and the zero example illustrates its failure. The formula det A(t)=t establishes the claim for all real t; the finite collection of displayed examples is not its proof.

Exact reversibility and numerical reliability

In exact algebra, any nonzero determinant makes a square matrix invertible. Floating-point computations must also consider roundoff, input error, scaling and the sensitivity of the problem. Testing whether a computed determinant is simply less than an arbitrary fixed decimal threshold can confuse a change of units with a loss of information.

For example, (1/1000)I₂ has determinant 1/1000000. It scales every direction uniformly and is exactly invertible. Conversely, diag(1000000,1/1000000) has determinant 1, just like I₂, but stretches one coordinate direction enormously and shrinks the other. Equal products do not imply equal directional behavior. A small determinant alone does not show that one direction is nearly lost relative to the others.

Numerical conditioning asks how perturbations affect the answer under chosen measures of size; a determinant is not a condition number. The explorer uses reduced rational fractions, not approximate zeros, and rejects arithmetic outside its safe integer range. Its purpose is to expose algebraic identities in bounded examples. Choosing stable methods for large numerical systems is a later numerical-linear-algebra topic.

The first map shrinks both directions equally. The second treats different directions very differently. Neither determinant value alone records that contrast.

Predict a sign, scale or rank change

Choose a matrix and, for the parameter family, a value of t. Predict its rank and whether a square determinant exists. Then select a row action and predict how its determinant and rank will change. For a square matrix, compare an exact cofactor expansion with the determinant accounting along its reversible elimination trace.

Inspect the two rectangular cases rather than assigning them a determinant. Compare the reversible row actions with erasing a row. An inverse appears only for a square full-rank matrix, and the two matrix products with it check both inverse identities. All displayed matrix entries, cofactors and checks come from the selected exact matrix; the fixed examples illustrate the general statements established above.

Worked examples

Try it yourself

1. Find det[[3,1],[2,4]]. Is this square matrix invertible?

   Show the answer and reasoning

   The determinant is 10; the matrix is invertible.

   The determinant is 3·4−1·2=10. It is nonzero, so the matrix is invertible and has rank 2.

2. A square matrix has determinant −5. What happens to the determinant after swapping two different rows?

   Show the answer and reasoning

   It becomes 5.

   It becomes 5. A single row exchange negates the determinant; it does not change rank.

3. For a three-by-three A with det A=5, compare scaling one row by 2 with replacing A by 2A.

   Show the answer and reasoning

   The determinants become 10 and 40, respectively.

   Scaling one row gives determinant 10. Replacing A by 2A scales all three rows, so det(2A)=2³·5=40.

4. Does R2←R2+7R1 change a determinant? Does R2←0R2 preserve rank?

   Show the answer and reasoning

   The row addition preserves both determinant and rank. Erasing a row need not preserve rank.

   The first operation leaves the determinant and rank unchanged because it adds a multiple of a different row and is reversible. The second creates a zero row and is not reversible. For a square matrix its determinant becomes zero; rank may decrease, or may stay the same if the erased row was already redundant.

5. Find the determinant of the triangular matrix [[2,1,0],[0,−3,4],[0,0,1/2]].

   Show the answer and reasoning

   The determinant is −3.

   Multiply the diagonal entries: 2·(−3)·(1/2)=−3. The matrix is invertible and has rank 3.

6. What sign multiplies the minor determinant for the entry in row 2, column 3?

   Show the answer and reasoning

   The sign is −1.

   It is (−1)²⁺³=−1. The cofactor is the negative of that minor determinant; its contribution also includes the original entry a₂₃.

7. A three-by-three matrix has determinant zero. Must its rank be zero?

   Show the answer and reasoning

   No. The rank can be 0, 1 or 2.

   No. Its rank is less than 3, and may be 0, 1 or 2. The dependent-row example in the article has rank 2.

8. For A=[[1,2],[2,4]], compare Ax=(3,6)ᵀ and Ax=(3,5)ᵀ.

   Show the answer and reasoning

   The first system has infinitely many solutions; the second has none.

   The first system has infinitely many solutions because the second equation repeats twice the first. The second is inconsistent because twice 3 is 6 rather than 5. The same zero determinant permits either outcome, depending on b.

9. A two-by-three matrix has rank 2. Does it have a nonzero determinant and a two-sided inverse?

   Show the answer and reasoning

   It has no ordinary determinant and no two-sided inverse.

   It has no ordinary square determinant and no two-sided inverse. As a map R³→R² it is onto, and rank-nullity gives a one-dimensional kernel, so it is not injective.

10. For same-size square matrices with det A=−3 and det B=2, find det(AB), det(Aᵀ) and det(A⁻¹).

    Show the answer and reasoning

    The values are −6, −3 and −1/3, respectively.

    They are −6, −3 and −1/3 respectively. The nonzero determinant of A justifies using its inverse.

11. Use I₂ to test whether det(A+B)=det A+det B always holds.

    Show the answer and reasoning

    No. Taking A=B=I₂ gives a counterexample.

    Take A=B=I₂. Then det(A+B)=det(2I₂)=4, whereas det A+det B=2. Separate linearity in one row or column does not imply additivity in the whole matrix.

12. Can matrices for the same operator have different determinants when different input and output bases are used?

    Show the answer and reasoning

    Yes, when the input and output coordinate bases differ.

    Yes. The standard shear has determinant 1, but the B-to-C matrix in lesson 124 has determinant −3. Converting the input and output basis unit boxes recovers the ordinary determinant. A common basis change on both sides instead produces S⁻¹AS and preserves the determinant.

13. For A(t)=[[1,1],[1,1+t]], which real t gives rank 1? What is its kernel then?

    Show the answer and reasoning

    Only t=0; the kernel then is span((1,−1)).

    Only t=0, since det A(t)=t. At that value both rows are (1,1), and the kernel is span((1,−1)). All nonzero t give rank 2.

14. Does det((1/1000)I₂)=1/1000000 by itself prove that one direction is nearly lost relative to the others?

    Show the answer and reasoning

    No.

    No. Both directions have the same nonzero scale factor. Exact invertibility and numerical sensitivity are different questions, and determinant magnitude alone is not a condition number.

What to carry forward

• A determinant is a scalar for a square matrix; rank counts independent directions and also applies to rectangular matrices.
• The determinant is separately linear in each row or column and alternating; it is not linear in the whole matrix.
• A cofactor includes its original position's sign and the determinant of its minor matrix. Any complete row or column expansion gives the same result.
• Swaps negate a determinant, one-row scales multiply it, and adding a multiple of a different row to a row leaves it unchanged. Undo all scale and sign factors after elimination.
• Rank is the pivot count and equals both row-space and column-space dimension. Original pivot columns identify an image basis.
• For square matrices, nonzero determinant, full rank, a trivial kernel and a two-sided inverse are equivalent. Every right-hand side then has a unique solution.
• det(AB)=det A det B and det(Aᵀ)=det A. An invertible matrix has inverse determinant 1/det A.
• Ordinary volume uses |det A| in stated ordinary coordinates. A common basis change preserves an operator's determinant; mixed input and output bases require their conversion factors.
• A zero determinant detects loss of full-dimensional reversibility, not necessarily loss of every direction. Determinant magnitude alone does not measure numerical conditioning.