Every square matrix hides a single number: its determinant. It tells you whether a system of equations has exactly one solution, how much a matrix stretches areas, whether three points lie on a line, and whether a matrix can be undone.
Every square matrix has one special number attached to it: its determinant.
Down-diagonal product minus up-diagonal product. For order 1 it is even simpler: |[a]| = a.
To every square matrix A we attach a number, its determinant, written in any of these ways. Only square matrices have determinants.
The bars do not mean “modulus” (absolute value) here.
Evaluate |2 4; −1 2|.
Evaluate |x x + 1; x − 1 x|.
A 3 × 3 determinant breaks into three 2 × 2 determinants, one for each entry of a row or column you choose.
Every place has a sign, in a checkerboard: + − +. The sign of row i, column j is (−1)i + j.
All six choices (three rows, three columns) give the same value, so choose the one with the most zeros.
How to expand
Every 0 in the chosen line wipes out a whole term.
Instead of (−1)^(i+j), use + when i + j is even and − when it is odd.
A = [2 2; 4 0] = 2B with B = [1 1; 2 0]: |A| = −8 = 2²(−2) = 4|B|. Scaling every entry by k scales each of the n factors in every product.
Evaluate |1 2 4; −1 3 0; 4 1 0|.
Evaluate |0 sin α −cos α; −sin α 0 sin β; cos α −sin β 0|.
Find x if |3 x; x 1| = |3 2; 4 1|.
Put the three corners of a triangle into a 3 × 3 determinant. Half of it is the area.
The area of the triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃).
This is the textbook formula. The determinant is the same thing in compact form.
Area is positive, so take the absolute value. If the area is given, use both signs. Three collinear points (on one line) give area 0.
Area of the triangle with vertices (3, 8), (−4, 2), (5, 1).
Line through A(1, 3) and B(0, 0); find k if D(k, 0) makes area ABD = 3.
Cross out an entry’s row and column: the small determinant left is its minor. Give it its sign and you have its cofactor.
The minor Mᵢⱼ of an entry is the determinant left after deleting its row i and column j. Here: the minor of 6 in Example 8.
The cofactor is the minor with its checkerboard sign.
Entries of any row or column times their own cofactors, added up, give Δ. Shown here for row 1.
Entries of one row times the cofactors of a different row always add to 0. Example 11 checks this one.
Minor of 6 in |1 2 3; 4 5 6; 7 8 9|.
Minors and cofactors of |1 −2; 4 3|.
M₁₁, A₁₁, M₂₁, A₂₁ of a general 3 × 3 determinant.
All minors and cofactors of |2 −3 5; 6 0 4; 1 5 −7|; show a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃ = 0.
The adjoint is built from cofactors. Put each cofactor in its entry’s place, then turn rows into columns.
adj A is the transpose of the matrix of cofactors.
Swap the two diagonal entries and change the sign of the other two (Example 12).
Each row meets its own cofactors on the diagonal (giving |A|) and other rows’ cofactors elsewhere (giving 0).
For example [1 2; 4 8]: its rows are proportional. It squashes the plane onto a line.
For example [1 2; 3 4], with |A| = −2. Theorem 2: products of non-singular matrices are non-singular.
adj A for A = [2 3; 1 4].
Divide the adjoint by the determinant and you have the inverse. It works exactly when |A| ≠ 0.
Divide Theorem 1 by the number |A|: A · (adj A / |A|) = I.
Theorem 4: A is invertible if and only if it is non-singular.
The first is Theorem 3. The second holds for an n × n matrix.
This continues the idea of inverses from Chapter 3 and inverse functions from Chapter 1.
A = [1 3 3; 1 4 3; 1 3 4]: verify A adj A = |A| I and find A⁻¹.
A = [2 3; 1 −4], B = [1 −2; −1 3]: verify (AB)⁻¹ = B⁻¹A⁻¹.
A = [2 3; 1 2] satisfies A² − 4A + I = O; find A⁻¹.
Write the equations as one matrix equation. The determinant of A tells you whether there is exactly one solution.
A system is consistent if it has a solution and inconsistent if it has none.
A⁻¹ exists, so this is the unique solution (the matrix method).
Then there is no solution.
Then there may be infinitely many solutions, or none.
Solve 2x + 5y = 1, 3x + 2y = 7.
Solve 3x − 2y + 3z = 8, 2x + y − z = 1, 4x − 3y + 2z = 4.
Three numbers: sum 6; second + 3 × third = 11; first + third = 2 × second.
Use the product [1 −1 2; 0 2 −3; 3 −2 4][−2 0 1; 9 2 −3; 6 1 −2] to solve a system.
Every multiple-choice question from the exercises, in one place. Tap an option to check it and see the reasoning.
The questions come from Exercises 4.1–4.4 and the miscellaneous exercise.
|[a]| = a; |a₁₁ a₁₂; a₂₁ a₂₂| = a₁₁a₂₂ − a₁₂a₂₁; a 3 × 3 determinant expands along any row or column with signs (−1)^(i+j).
Area of a triangle = ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1| (absolute value). Zero area ⇔ collinear.
Minor Mᵢⱼ: delete row i and column j. Cofactor Aᵢⱼ = (−1)^(i+j) Mᵢⱼ. |A| = Σ (entries of a row) × (their cofactors); with another row’s cofactors the sum is 0.
adj A = (cofactor matrix)′; A(adj A) = (adj A)A = |A| I. Singular: |A| = 0; non-singular: |A| ≠ 0.
A⁻¹ exists ⇔ |A| ≠ 0, and A⁻¹ = (1/|A|) adj A. |AB| = |A||B|, |adj A| = |A|ⁿ⁻¹.
AX = B: |A| ≠ 0 ⇒ unique solution X = A⁻¹B; |A| = 0 and (adj A)B ≠ O ⇒ no solution; |A| = 0 and (adj A)B = O ⇒ infinitely many or none.
Coefficients of equations were laid out on a counting board in exactly the pattern of a determinant, and rows and columns were subtracted to eliminate unknowns.
In Japan, his work on eliminating a quantity from equations contains the idea of determinants and their expansion.
Vandermonde treated determinants as functions in their own right; Laplace gave expansion by minors; Gauss used them in number theory.
Both proved the product rule |AB| = |A||B|; Cauchy gave the word "determinant" its present meaning. Jacobi then made it standard.
Each example is a small animated story. Diagonals are drawn in, rows and columns are crossed out, cofactors fly across the diagonal, and triangles and lines appear on a grid. Press Next step (or Play). Under the controls, each step is explained in detail, with links to the exact basics it uses; after reading a basic, press ← Back to return to the same step. Most examples have a Try it panel where you type your own numbers.
2 × 2 by the cross rule, 3 × 3 by expanding along a row or column, and an equation between determinants.
Half a determinant gives the area; zero area gives a straight line.
Every entry has a minor (the leftover determinant) and a cofactor (the minor with its checkerboard sign).
Cofactors, transpose, divide by |A|; and two shortcuts.
Write AX = B, check |A| ≠ 0, and read off X = A⁻¹B.
A product that turns out to be I solves a system in two lines.
Surveyors use them to find the area of a plot from its corner coordinates, engineers to check that things line up, phones to know how much an image was stretched or flipped, and anyone solving several conditions at once to know whether a single answer exists.
A surveyor walks round a plot with GPS and records each corner. Cut the plot into triangles from one corner; each triangle’s area is ½|x y 1|. Add them up.
A microwave link needs three towers in a straight line. Slide the middle tower. The determinant of the three positions is 0 exactly when they line up, and only then does the beam get through.
Photo and game apps move pictures with 2 × 2 matrices. The determinant says how much every area is multiplied, and a negative sign means the picture was mirrored.
Three shopping trips, three receipts, three unknown prices. The receipts are a system AX = B. If |A| ≠ 0 the prices are fixed; if one receipt is just a multiple of another, |A| = 0 and the prices cannot be pinned down.
When the sun is overhead, every point (x, y) casts its shadow at (x, 0). That is the matrix [1 0; 0 0], with determinant 0. A kite, a bird and a balloon above the same spot give the same shadow, so no matrix can turn shadows back into objects.
Decide first, then tap each card.
Every question that asks you to evaluate, show, find or solve has an animated walkthrough here; the multiple-choice questions are in the Lesson’s self-check. Pick a question from the chips above each player. Press Play to watch it solved, or step through with Next. Each step has a short "why" under it and links to the basics it needs.
2 × 2: down-diagonal minus up-diagonal. 3 × 3: pick the row or column with the most zeros, cross out, and add the signed terms. (Q8 is in the quiz.)
Plot the points, write ½|x y 1|, and remember that area is never negative. Zero area means the points lie on one line. (Q5 is in the quiz.)
Cross out the row and the column, take the leftover determinant, then apply the + − + checkerboard. (Q5 is in the quiz.)
Cofactors, transpose, divide by the determinant, or use a matrix equation to find the inverse. (Q17–Q18 are in the quiz.)
Write AX = B. If |A| ≠ 0 there is exactly one solution, X = A⁻¹B. If |A| = 0, test (adj A)B.
Trigonometric and algebraic determinants that simplify completely, inverse identities, and a system in 1/x, 1/y, 1/z. (Q8–Q9 are in the quiz.)
Determinants look like a recipe of crosses and signs. Here each part of the recipe is explained from its roots: where the number comes from, why it measures area, why the signs alternate, and why you divide by it to undo a matrix. Every example and exercise links back here exactly where you need it, and a button takes you straight back to the problem.
Nobody invented the formula a₁b₂ − a₂b₁ out of thin air. It falls out of the most ordinary job in algebra: solving two equations in two unknowns.
Multiply the first by b₂ and the second by b₁, so both have b₁b₂y. Subtract: the y terms vanish.
The second line comes from doing the same to eliminate x. The same number, a₁b₂ − a₂b₁, multiplies both unknowns.
Because this number determines whether the system has exactly one solution, it is called the determinant of the coefficient array [a₁ b₁; a₂ b₂]. Read it as down-diagonal product minus up-diagonal product. Bigger square arrays have determinants built the same way (step 3).
A determinant is the "can I divide by it?" number of a system of equations. Not zero: one answer. Zero: either no answer or endless answers.
Place two arrows (a, c) and (b, d) at the origin. They are two sides of a parallelogram, and its area is |ad − bc|: the 2 × 2 determinant again. A triangle with the same two sides is half of it.
Draw the smallest rectangle around a triangle. The triangle is the rectangle minus three right-angled corner triangles, whose areas are ½ × base × height. Doing this with letters gives ½[x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)], which is exactly ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|.
The determinant is positive when the corners are listed anticlockwise and negative when clockwise, so the area takes its absolute value. When the area is given, the determinant could be +2 × area or −2 × area, so solve both. If the three points are on one line, the triangle is flat and the determinant is 0.
A 3 × 3 determinant must keep the two properties that make the 2 × 2 one useful: it is 0 when two rows are equal, and it changes sign when two rows are swapped. The expansion a₁₁M₁₁ − a₁₂M₁₂ + a₁₃M₁₃ is the only way to build it from 2 × 2 minors so that both stay true.
The minor of an entry is the determinant of what remains after deleting the entry’s row and column. It is one size smaller, so a 3 × 3 determinant breaks into three 2 × 2 ones.
Moving an entry to the top-left corner takes i − 1 row swaps and j − 1 column swaps, and each swap flips the sign. So the sign is (−1)^(i+j−2) = (−1)^(i+j): + when i + j is even, − when it is odd. That makes a chessboard pattern, starting with + at the top-left.
Two equal (or proportional) rows or columns give 0. Adding a multiple of one row to another changes nothing. Using one row’s entries with another row’s cofactors gives 0, because that is the expansion of a determinant with a repeated row.
Multiply every term of one bracket by every term of the other, then collect: x² + x − x − 1. This "difference of squares" is used in Example 2.
A minus sign in front of a bracket changes the sign of everything inside it.
Squares hide the sign, so x² = k has two answers. √8 = √(4 × 2) = 2√2.
This is why "area 3" gives two positions for an unknown vertex.
If a polynomial becomes 0 when x = y, then (x − y) is a factor. Symmetry then suggests the others.
The point (cos θ, sin θ) is 1 unit from the centre, so cos²θ + sin²θ = 1². It makes every θ vanish in Exercise 4.1 Q2 and Miscellaneous Q1–Q2.
These come from Chapter 3’s Basics; here is the short version.
A matrix is a table: aᵢⱼ sits in row i, column j. Only square matrices (as many rows as columns) have determinants.
The entry in row i, column j of AB is row i of A times column j of B: multiply the pairs and add. A system of equations is one such product: AX = B, where row i of A times the column of unknowns gives the left side of equation i.
kA multiplies every entry by k; A + B adds entries in the same place. That is why |kA| = kⁿ|A| for an n × n matrix: each of the n factors in every product gets a k.
The transpose A′ swaps rows and columns (entries fly across the diagonal). The identity I has 1s on the diagonal and 0s elsewhere, and IA = AI = A. The adjoint is a transpose, and A(adj A) = |A| I.
For numbers, multiplying by a is undone by multiplying by 1/a, and 0 has no reciprocal because it sends everything to 0. Matrices behave the same way: a 2 × 2 matrix multiplies every area by |A|. To undo it we need a matrix that multiplies areas by 1/|A|, and that is impossible when |A| = 0 because the plane has been squashed flat.
Row i of A times column i of adj A is the expansion of |A| along row i, so the diagonal of A(adj A) is |A|. Every other entry uses one row with another row’s cofactors, which is 0 (step 3). So A(adj A) = |A| I, and dividing by the number |A| gives A · (adj A / |A|) = I.
(AB)⁻¹ = B⁻¹A⁻¹: to undo "B, then A" you undo A first. Socks, then shoes; shoes off, then socks.
Each linear equation in x and y is a straight line; a solution of the system is a point on all of them. Two lines either cross once (one solution), never meet (none: inconsistent) or lie on top of each other (infinitely many). The determinant tells the first case apart from the other two.
With three unknowns each equation is a flat plane in space, and the same three cases happen. |A| ≠ 0 is the "they meet in exactly one point" case.
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