MathematicsClass 12Determinants

Determinants | Class 12 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 4 min read

Determinants | Class 12 Mathematics Notes

Determinants – this guide gives you a concise, exam-ready overview of Determinants from Class 12 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

4.5 Adjoint and Inverse of a Matrix

This section discusses the adjoint and inverse of a square matrix and their relation to determinants. The adjoint of a square matrix A = [a_ij] of order n is defined as the transpose of the cofactor matrix [A_ij], where A_ij is the cofactor of element a_ij. Denoted as adj A, it is constructed by taking cofactors of all elements and then transposing the resulting matrix. For a 2×2 matrix, adjoint can be found by swapping diagonal elements and changing signs of off-diagonal elements. The key theorem states that A × adj A = adj A × A = |A| × I, where I is the identity matrix. A matrix is singular if its determinant is zero and nonsingular otherwise. Only nonsingular matrices have inverses. The inverse of a nonsingular matrix A is given by A^{-1} = (1/|A|) × adj A. The section includes proofs and examples verifying these results, including matrix multiplication to confirm the theorem and finding inverses of specific matrices. It also states that the determinant of a product of matrices equals the product of their determinants and that the inverse of a product is the product of inverses in reverse order. The section concludes with examples applying these concepts and solving matrix equations.

📊 Diagram: See figure_2: We state the following theorem without proof.

🧪 Activity: No specific activity in this section.

🔗 Connection: Prepares for applications of determinants and matrices in solving linear systems.

Frequently asked questions

If the area of triangle is 4 square unit and vertices are (k, 0), (4, 0), (0, 2) then the value of k is..

0

If (6, 0), (4, 3), (2, 1) is the vertices of a triangle then the area of the triangle is

5 square unit

If any two rows or any two columns are identical or proportional, then the value of determinant is

0

Evaluate the determinants in Exercises 1 and 2. 1. \(\left| \begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array} \right|\) 2. (i) \(\left| \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right|\) (ii) \(\left| \begin{array}{cc} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} \right|\) 3. If \(A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\), then show that \(|2A| = 4 |A|\) 4. If \(A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}\), then show that \(|3A| = 27 |A|\) 5. Evaluate the determinants (i) \(\left| \begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array} \right|\) (ii) \(\left| \begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array} \right|\) (iii) \(\left| \begin{array}{lll}0 & 1 & 2\\ -1 & 0 & -3\\ -2 & 3 & 0 \end{array} \right|\) (iv) \(\left| \begin{array}{rrr}2 & -1 & -2\\ 0 & 2 & -1\\ 3 & -5 & 0 \end{array} \right|\) 6. If \(\mathrm{A} = \left[ \begin{array}{rrr}1 & 1 & -2\\ 2 & 1 & -3\\ 5 & 4 & -9 \end{array} \right]\) , find \(|\mathrm{A}|\) 7. Find values of \(x\) , if (i) \(\left| \begin{array}{lll}2 & 4\\ 5 & 1 \end{array} \right| = \left| \begin{array}{lll}2x & 4\\ 6 & x \end{array} \right|\) (ii) \(\left| \begin{array}{ll}2 & 3\\ 4 & 5 \end{array} \right| = \left| \begin{array}{ll}x & 3\\ 2x & 5 \end{array} \right|\) 8. If \(\left| \begin{array}{ll}x & 2\\ 18 & x \end{array} \right| = \left| \begin{array}{ll}6 & 2\\ 18 & 6 \end{array} \right|\) , then \(x\) is equal to (A) 6 (B) \(\pm 6\) (C) -6 (D) 0

1. Evaluate \(\left| \begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array} \right| = (2)(-1) - (4)(-5) = -2 + 20 = 18.\n 2.(i) \(\left| \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right| = (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta) = \cos^2 \theta + \sin^2 \theta = 1.\n (ii) \(\left| \begin{array}{cc} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} \right| = (x^{2} - x + 1)(x + 1) - (x - 1)(x + 1) = (x^{3} + x^{2} - x^{2} - x + x + 1) - (x^{2}

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