Determinants

In this chapter: determinants of order 1, 2 and 3, properties used to simplify them, area of a triangle and collinearity, minors and cofactors, adjoint and inverse of a matrix, and solving a system of linear equations with the matrix method.

Evaluating a determinant

The determinant is defined only for a square matrix. For A = [a] of order 1, |A| = a. For order 2:

For order 3, expand along any row or column. Each element is multiplied by the 2 × 2 determinant left after deleting its row and column, with the sign pattern below. Along the first row:

Expanding along any row or column gives the same value, so choose the row or column with the most zeros. That single habit saves a lot of time in the exam.

Properties that make evaluation faster

• |A′| = |A|, so anything true for rows is true for columns.
• Interchanging two rows (or columns) changes the sign of the determinant.
• If two rows (or columns) are identical or proportional, the determinant is 0.
• Multiplying one row by k multiplies the determinant by k. Hence, for a square matrix of order n, |kA| = kⁿ|A|.
• The operation R_i → R_i + kR_j (or the same for columns) does not change the determinant. Use it to create zeros before expanding.
• If each element of a row is a sum of two terms, the determinant splits into a sum of two determinants.
• |AB| = |A| |B| for square matrices of the same order.
• The determinant of a diagonal or triangular matrix is the product of its diagonal elements.

Area of a triangle

For a triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃):

If the determinant is zero, the "triangle" has no area, so the three points are collinear. This gives the equation of the line through two points (x₁, y₁) and (x₂, y₂): set the determinant with third row (x, y, 1) equal to zero.

Worked example: Find k if the area of the triangle with vertices (k, 0), (4, 0) and (0, 2) is 4 square units.
Solution: ½ |k(0 − 2) − 0(4 − 0) + 1(8 − 0)| = 4, so |8 − 2k| = 8. Then 8 − 2k = 8 gives k = 0, and 8 − 2k = −8 gives k = 8. Both k = 0 and k = 8 are answers. Dropping the modulus and giving only one value is the usual mistake.

Minors and cofactors

The minor M_ij of the element a_ij is the determinant left after deleting row i and column j. The cofactor is A_ij = (−1)^{i + j} M_ij. In this language,

• |A| = sum of the products of the elements of any row (or column) with their own cofactors, for example |A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃.
• If the elements of one row are multiplied by the cofactors of a different row, the sum is 0. For example, a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0.

Adjoint and inverse

The adjoint of A is the transpose of the matrix of cofactors: adj A = [A_ij]′. The two facts in the previous list combine into one identity:

A is singular if |A| = 0 and non-singular if |A| ≠ 0. A square matrix is invertible if and only if it is non-singular. For a 2 × 2 matrix, the adjoint is quick: swap the diagonal elements and change the sign of the other two, so abcd has adjoint d−b−ca.

Worked example: Find the inverse of A = 123014560.
Solution: |A| = 1(0 − 24) − 2(0 − 20) + 3(0 − 5) = −24 + 40 − 15 = 1, so A is invertible. Cofactors: A₁₁ = −24, A₁₂ = 20, A₁₃ = −5; A₂₁ = 18, A₂₂ = −15, A₂₃ = 4; A₃₁ = 5, A₃₂ = −4, A₃₃ = 1. Writing these as columns (the transpose) gives adj A, and since |A| = 1, A⁻¹ = adj A = −2418520−15−4−541. Check one entry of AA⁻¹: row 1 of A times column 1 is −24 + 40 − 15 = 1.

Solving linear equations: the matrix method

Write the system as AX = B, with A the coefficient matrix, X the column of unknowns and B the column of constants. A system is consistent if it has at least one solution and inconsistent if it has none.

Worked example: Solve 2x + 5y = 1 and 3x + 2y = 7.
Solution: A = 2532, B = 17. |A| = 4 − 15 = −11 ≠ 0, so there is a unique solution. adj A = 2−5−32, so X = A⁻¹B = (−1/11) 2 − 35−3 + 14 = (−1/11) −3311 = 3−1. Hence x = 3, y = −1. Check: 2(3) + 5(−1) = 1 and 3(3) + 2(−1) = 7.

JEE also uses Cramer's rule in the form D = |A| and D₁, D₂, D₃ (replace one column by B). If D ≠ 0, the solution is unique with x = D₁/D and so on. If D = 0 and at least one of D₁, D₂, D₃ is non-zero, there is no solution. For a homogeneous system (B = O), x = y = z = 0 is always a solution, and a non-zero solution exists only when D = 0.

Common mistakes: (1) Wrong signs on cofactors: use (−1)^{i + j}, not the sign of the element. (2) Forgetting to transpose the cofactor matrix when forming adj A. (3) Writing |kA| = k|A| for a 3 × 3 matrix; it is k³|A|. (4) Dropping the modulus in area questions and so losing the second value of k. (5) Concluding "infinitely many solutions" whenever |A| = 0; you must still check (adj A)B or the individual equations.

JEE and MHT‑CET focus

• Expanding 3 × 3 determinants quickly, using row and column operations to create zeros first.
• Results on |kA|, |adj A|, |A⁻¹|, adj(adj A) and |AB|, which appear as one-line numerical questions.
• Area of a triangle, collinearity and the equation of a line through two points.
• Finding A⁻¹ by the adjoint method and verifying AA⁻¹ = I.
• Consistency of a system, unique solution by X = A⁻¹B, and parameter values for which a system has no solution or infinitely many.

Practice questions