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CBSE • Class XII • Mathematics • Ch 4
Estimated Time: 45 Mins
Study Progress: In Progress

Determinants

In Class 12 Mathematics, "Determinants" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

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📐 Have You Ever Wondered?

How can a single scalar number calculated from a square matrix immediately tell you whether a system of three simultaneous equations has a unique solu...

How can a single scalar number calculated from a square matrix immediately tell you whether a system of three simultaneous equations has a unique solution, or calculate the exact area of a triangle in coordinate space? Determinants quantify multidimensional scaling volume.

Why This Chapter Matters

In Class 12 Mathematics, "Determinants" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • Matrix definition from Chapter 2.
  • Cramer's rule.
  • Area of triangles in coordinate geometry.

What You Will Learn (Core Objectives)

  • Compute Determinants of order 1, 2, and 3 using cofactor expansion.
  • Calculate Minors ($M_{ij}$) and Cofactors ($A_{ij} = (-1)^{i+j} M_{ij}$).
  • Evaluate Area of a Triangle using determinants: $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
  • Find Adjoint of a Matrix ($\text{adj } A = [A_{ij}]^T$) and verify $A(\text{adj } A) = |A| I$.
  • Find Matrix Inverse ($A^{-1} = \frac{1}{|A|}\text{adj } A$) and solve systems of linear equations using the Matrix Method ($X = A^{-1}B$).

Chapter Roadmap & Progression

1 1. Minors, Cofactors & Adjoint
2 2. Matrix Inversion & Singular Matr...
3 3. Matrix Method for Linear Systems

Complete Concept Guide (100% Curriculum Coverage)

1. Minors, Cofactors & Adjoint

For square matrix $A = [a_{ij}]$:
• Minor ($M_{ij}$): Determinant of submatrix deleting row $i$ and column $j$.
• Cofactor ($A_{ij}$): $$\mathbf{A_{ij} = (-1)^{i+j} M_{ij}}$$
• Adjoint ($\text{adj } A$): Transpose of cofactor matrix: $\mathbf{\text{adj } A = [A_{ij}]^T}$. Key Theorem: $$\mathbf{A \cdot (\text{adj } A) = (\text{adj } A) \cdot A = |A| \cdot I_n}$$

2. Matrix Inversion & Singular Matrices

A matrix is Singular if $|A| = 0$; Non-Singular if $|A| \ne 0$. A square matrix possesses an inverse $A^{-1}$ if and only if it is non-singular: $$\mathbf{A^{-1} = \frac{1}{|A|} \text{adj } A} \quad (|A| \ne 0)$$ Determinant of adjoint: $|\text{adj } A| = |A|^{n-1}$.

3. Matrix Method for Linear Systems

System $AX = B$: If $|A| \ne 0$, the system is consistent with a unique solution: $$\mathbf{X = A^{-1} B}$$

Visual Learning & Conceptual Map

Determinants Master Matrix

Conceptual framework, core mechanisms, and analytical relationships
Academic Architecture

1. Minors, Cofactors & Adjoint • 2. Matrix Inversion & Singular Matrices

Chapter Summary & 10 Key Takeaways

Takeaway 1
Adjoint Theorem: $A(\text{adj } A) = |A| I$ establishing foundational inverse formula.
Takeaway 2
Invertibility Condition: Matrix inverse exists if and only if determinant $|A| \ne 0$.
Takeaway 3
Adjoint Determinant: $|\text{adj } A| = |A|^{n-1}$ for an $n \times n$ matrix.
Takeaway 4
Matrix Method: Solving linear systems via $X = A^{-1}B$.
Takeaway 5
Collinearity Test: Three points are collinear if the triangle determinant equals zero.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Evaluate the determinant: $\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix}$.
Reveal Answer & Explanation
Answer: $\Delta = (2)(2) - (4)(-1) = 4 + 4 = 8$.
8.
2
If $A$ is a square matrix of order 3 and $|A| = 5$, find the value of $|\text{adj } A|$.
Reveal Answer & Explanation
Answer: $|\text{adj } A| = |A|^{n-1} = |A|^{3-1} = |A|^2 = 5^2 = 25$.
25.
3
Find the inverse of $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.
Reveal Answer & Explanation
Answer: $|A| = 2(4) - 3(1) = 8 - 3 = 5 \ne 0$. Adjoint of a $2\times 2$ matrix swaps diagonal and negates off-diagonal: $\text{adj } A = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$. $A^{-1} = \frac{1}{5}\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$.
1/5 [[4, -3], [-1, 2]].
4
Using determinants, show that points $A(a, b+c)$, $B(b, c+a)$, and $C(c, a+b)$ are collinear.
Reveal Answer & Explanation
Answer: $\Delta = \frac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix}$. Apply $C_2 \to C_2 + C_1$: $\Delta = \frac{1}{2} \begin{vmatrix} a & a+b+c & 1 \\ b & a+b+c & 1 \\ c & a+b+c & 1 \end{vmatrix} = \frac{a+b+c}{2} \begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{vmatrix} = 0$ (since $C_2 = C_3$). Area is 0, so points are collinear.
Area is 0, points are collinear.
5
Solve the system of equations using matrix method: $2x + 5y = 1$ and $3x + 2y = 7$.
Reveal Answer & Explanation
Answer: $A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}$. $|A| = 4 - 15 = -11$. $A^{-1} = -\frac{1}{11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$. $X = A^{-1}B = -\frac{1}{11}\begin{bmatrix} 2(1) - 5(7) \\ -3(1) + 2(7) \end{bmatrix} = -\frac{1}{11}\begin{bmatrix} -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$. Solution: $x = 3, y = -1$.
x = 3, y = -1.
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