Determinants

A determinant is a unique number (a scalar) associated with every square matrix. If A is a square matrix, its determinant is written as |A| or det(A) or Δ. Determinants let us test whether a matrix has an inverse, find the area of a triangle, check collinearity of points, and solve systems of linear equations.

For a matrix of order 1, A = [a], the determinant is simply |A| = a.

For a square matrix of order 2, A = [[a₁, b₁], [a₂, b₂]], the determinant is defined as |A| = a₁b₂ − a₂b₁.

For a square matrix of order 3, the determinant is evaluated by expansion along any row or column. Expanding along the first row of A = [[a₁,b₁,c₁],[a₂,b₂,c₂],[a₃,b₃,c₃]] gives:

|A| = a₁(b₂c₃ − b₃c₂) − b₁(a₂c₃ − a₃c₂) + c₁(a₂b₃ − a₃b₂)

Each term is the entry multiplied by the 2×2 determinant left after deleting the row and column of that entry, with alternating signs (+, −, +) across the row. Expanding along any other row or column always gives the same value — this is proved formally in the next section.

Worked Example: Evaluate the determinant of A = [[2, 3, 5], [6, 0, 4], [1, 5, 7]].

Expanding along the first row:

|A| = 2(0×7 − 4×5) − 3(6×7 − 4×1) + 5(6×5 − 0×1)

|A| = 2(0 − 20) − 3(42 − 4) + 5(30 − 0)

|A| = 2(−20) − 3(38) + 5(30) = −40 − 114 + 150 = −4

So |A| = −4.

Evaluating large determinants directly can be tedious, so several properties let us simplify a determinant using row and column operations before expanding — very similar in spirit to row-reducing a matrix.

• P1 (Transpose): The value of a determinant is unchanged if its rows and columns are interchanged, i.e., |A| = |A′| (A′ is the transpose of A).
• P2 (Interchange): If any two rows (or two columns) of a determinant are interchanged, the sign of the determinant changes, though its magnitude stays the same.
• P3 (Repetition): If any two rows (or columns) of a determinant are identical or proportional, the value of the determinant is zero.
• P4 (Scalar multiple): If every element of one row (or column) is multiplied by a constant k, the value of the determinant is multiplied by k. Consequently, for an n×n matrix A, |kA| = kⁿ|A|.
• P5 (Sum splitting): If some elements of a row (or column) are each written as a sum of two terms, the determinant can be expressed as the sum of two determinants.
• P6 (Invariance under row/column operations): The value of a determinant is unchanged if a scalar multiple of one row (or column) is added to another row (or column), e.g. R_i → R_i + kR_j.

Worked Example 1: Without full expansion, show that |3 6 9; 1 2 3; 4 5 6| = 0.

Row R₁ = (3, 6, 9) = 3 × (1, 2, 3) = 3R₂. Since R₁ is a scalar multiple of R₂, the two rows are proportional, so by property P3 the determinant is zero.

Worked Example 2: Evaluate |1 2 3; 4 5 6; 7 8 9| using row operations.

Apply R₂ → R₂ − 4R₁ and R₃ → R₃ − 7R₁ (value unchanged by P6): new R₂ = (0, −3, −6), new R₃ = (0, −6, −12). Now R₃ = 2R₂, so by P3 the determinant equals 0.

Determinants give a compact formula for the area of a triangle when the coordinates of its three vertices are known, without needing to separately find the base and height.

If the vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃), the area of the triangle is:

Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

which is the same as ½ times the absolute value of the determinant |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|. The modulus sign is essential because area is always non-negative, while the determinant itself can come out negative depending on the order in which the vertices are taken.

Because the same expression appears whichever way the points are arranged, three points (x₁,y₁), (x₂,y₂) and (x₃,y₃) are collinear exactly when this determinant equals zero — that is, when the “triangle” they form has zero area.

Worked Example: Find the area of the triangle whose vertices are (1, 0), (6, 0) and (4, 3).

Area = ½ |1(0 − 3) + 6(3 − 0) + 4(0 − 0)| = ½ |−3 + 18 + 0| = ½ × 15 = 7.5 square units.

To expand determinants systematically (and to build the adjoint of a matrix later), we define the minor and cofactor of each element.

The minor M_ij of an element a_ij in a determinant is the determinant obtained by deleting the i-th row and the j-th column that contain that element.

The cofactor A_ij of the element a_ij is the minor with a sign attached depending on position: A_ij = (−1)^i+j M_ij.

Using cofactors, a determinant can be evaluated by multiplying each element of any one row (or column) by its own cofactor and adding the results — this is exactly the expansion used earlier, now written formally. For expansion along row i:

|A| = a_i1A_i1 + a_i2A_i2 + a_i3A_i3

An important related fact: if the elements of a row are multiplied by the cofactors of a different row, the sum is always zero, i.e. a_i1A_k1 + a_i2A_k2 + a_i3A_k3 = 0 for i ≠ k.

Worked Example: For A = [[2, −3, 5], [6, 0, 4], [1, 5, −7]], find the minor and cofactor of a₁₂ = −3, and verify |A| by expanding along the first row.

M₁₂ is obtained by deleting row 1 and column 2: M₁₂ = |6 4; 1 −7| = 6(−7) − 4(1) = −42 − 4 = −46. Since i+j = 1+2 = 3 (odd), A₁₂ = (−1)³(−46) = 46.

Similarly, M₁₁ = |0 4; 5 −7| = 0(−7) − 4(5) = −20, so A₁₁ = (−1)²(−20) = −20. And M₁₃ = |6 0; 1 5| = 30 − 0 = 30, so A₁₃ = (−1)⁴(30) = 30.

Expanding along the first row: |A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ = 2(−20) + (−3)(46) + 5(30) = −40 − 138 + 150 = −28.

A square matrix A is called non-singular if |A| ≠ 0, and singular if |A| = 0. Only a non-singular matrix has a multiplicative inverse.

The adjoint of a square matrix A, written adj(A), is the transpose of the matrix formed by the cofactors of every element of A. That is, replace each element of A by its cofactor to get the cofactor matrix, then transpose it.

Every square matrix satisfies the identity:

A · adj(A) = adj(A) · A = |A| I

where I is the identity matrix of the same order. This identity is the key that leads to the formula for the inverse. If |A| ≠ 0, dividing both sides by |A| gives:

A⁻¹ = (1/|A|) adj(A)

If |A| = 0, then A · adj(A) is the zero matrix, and A has no inverse.

Worked Example: Find the inverse of A = [[2, 3], [1, 4]] using the adjoint method.

|A| = 2(4) − 3(1) = 8 − 3 = 5, which is non-zero, so A⁻¹ exists.

For a 2×2 matrix [[a, b], [c, d]], adj(A) = [[d, −b], [−c, a]]. Here, adj(A) = [[4, −3], [−1, 2]].

Therefore A⁻¹ = (1/5) [[4, −3], [−1, 2]].

Check: A · adj(A) = [[2,3],[1,4]] × [[4,−3],[−1,2]] = [[8−3, −6+6],[4−4, −3+8]] = [[5, 0],[0, 5]] = 5I = |A|I, confirming the result.

A system of linear equations can be written compactly in matrix form AX = B, where A is the matrix of coefficients, X is the column matrix of unknowns, and B is the column matrix of constants. For three equations in x, y, z:

a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂, a₃x + b₃y + c₃z = d₃

can be written as A = [[a₁,b₁,c₁],[a₂,b₂,c₂],[a₃,b₃,c₃]], X = [[x],[y],[z]], B = [[d₁],[d₂],[d₃]], giving AX = B.

If A is non-singular (|A| ≠ 0), A⁻¹ exists, and multiplying both sides on the left by A⁻¹ gives the unique solution X = A⁻¹B. This is called the matrix method. When |A| = 0, the system does not have a unique solution and must be checked further:

• If |A| = 0 and (adj A)B ≠ O (a non-zero column), the system is inconsistent — no solution exists.
• If |A| = 0 and (adj A)B = O, the system is consistent with infinitely many solutions.

Worked Example: Solve using the matrix method: x − y + z = 4, 2x + y − 3z = 0, x + y + z = 2.

Here A = [[1,−1,1],[2,1,−3],[1,1,1]] and B = [[4],[0],[2]]. Expanding along the first row, |A| = 1(1+3) − (−1)(2+3) + 1(2−1) = 4 + 5 + 1 = 10, which is non-zero, so a unique solution exists.

Computing all nine cofactors and transposing gives adj(A) = [[4, 2, 2], [−5, 0, 5], [1, −2, 3]], so A⁻¹ = (1/10)[[4, 2, 2], [−5, 0, 5], [1, −2, 3]].

Then X = A⁻¹B gives: x = (1/10)(4×4 + 2×0 + 2×2) = (1/10)(20) = 2; y = (1/10)(−5×4 + 0×0 + 5×2) = (1/10)(−10) = −1; z = (1/10)(1×4 + (−2)×0 + 3×2) = (1/10)(10) = 1.

So x = 2, y = −1, z = 1. Checking in the original equations confirms all three are satisfied.