Formula Sheet

Matrices & Determinants Formulas

All key formulas grouped by subtopic. Each one has a quick reminder and common mistakes to watch for.

10 formulas · 7 subtopics

Determinant of 2×2 Matrix

#1
abcd=adbc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

💡 Product of main diagonal minus product of off-diagonal.

Determinant of 3×3 Matrix

#2
A=a(eifh)b(difg)+c(dheg)|A| = a(ei-fh) - b(di-fg) + c(dh-eg)

💡 Expand along the row/column with most zeros to simplify calculation.

Determinant of Product

#3
AB=AB|AB| = |A| \cdot |B|

💡 Works for square matrices of same order. Also |A^n| = |A|^n.

Assuming |A+B| = |A| + |B|. This is FALSE in general.

Adjoint and Inverse

#4
A1=adj(A)A,Aadj(A)=AIA^{-1} = \frac{\text{adj}(A)}{|A|}, \quad A \cdot \text{adj}(A) = |A| \cdot I

💡 adj(A) = transpose of cofactor matrix. |adj(A)| = |A|^{n-1} for n×n matrix.

Forgetting that adj(adj(A)) = |A|^{n-2} · A for n×n matrix.

Scalar Multiple of Determinant

#5
kA=knA for n×n matrix|kA| = k^n |A| \text{ for } n \times n \text{ matrix}

💡 Each of the n rows gets multiplied by k, so the determinant picks up k^n.

Writing |2A| = 2|A| instead of |2A| = 2^n|A|. The exponent n is crucial.

Cramer's Rule

#6
x=DxD,y=DyD,z=DzDx = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}

💡 D ≠ 0: unique solution. D = 0 and all Dᵢ = 0: infinite or no solutions. D = 0 and some Dᵢ ≠ 0: no solution.

Characteristic Equation (2×2)

#7
AλI=0    λ2tr(A)λ+A=0|A - \lambda I| = 0 \implies \lambda^2 - \text{tr}(A)\lambda + |A| = 0

💡 tr(A) = sum of diagonal elements = sum of eigenvalues. |A| = product of eigenvalues.

Cayley-Hamilton Theorem

#8
Every matrix satisfies its own characteristic equation\text{Every matrix satisfies its own characteristic equation}

💡 For 2×2: A² - tr(A)·A + |A|·I = O. Use this to express A⁻¹ in terms of A and I.

Orthogonal Matrix

#9
AAT=ATA=I    A1=AT,A=±1AA^T = A^TA = I \implies A^{-1} = A^T, \quad |A| = \pm 1

💡 Rotation matrices are orthogonal. If det = 1, it's a proper rotation.

Nested Adjoint

#10
adj(adj(A))=An2A,adj(A)=An1\text{adj}(\text{adj}(A)) = |A|^{n-2} A, \quad |\text{adj}(A)| = |A|^{n-1}

💡 For adj applied k times: det = |A|^{(n-1)^k}. Very common in JEE numerical problems.