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Determinants - Adjoint of a matrix

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The minor MijM_{ij} of an element aija_{ij} of a determinant is the determinant obtained by deleting its ithi^{th} row and jthj^{th} column in which element aija_{ij} lies.

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The cofactor AijA_{ij} of an element aija_{ij} is defined by Aij=(−1)i+jMijA_{ij} = (-1)^{i+j} M_{ij}.

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The adjoint of a square matrix A=[aij]n×nA = [a_{ij}]_{n \times n} is defined as the transpose of the matrix [Aij]n×n[A_{ij}]_{n \times n}, where AijA_{ij} is the cofactor of the element aija_{ij}. It is denoted by adjAadj A.

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For a square matrix AA of order nn, A(adjA)=(adjA)A=∣A∣IA(adj A) = (adj A) A = |A|I, where II is the identity matrix of order nn.

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A square matrix AA is said to be singular if ∣A∣=0|A| = 0 and non-singular if ∣A∣≠0|A| \neq 0.

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If AA is a non-singular matrix of order nn, then ∣adjA∣=∣A∣n−1|adj A| = |A|^{n-1}.

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If AA and BB are non-singular matrices of the same order, then adj(AB)=(adjB)(adjA)adj(AB) = (adj B)(adj A) (Reversal Law).

📐Formulae

adjA=adj A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \ A_{12} & A_{22} & A_{32} \ A_{13} & A_{23} & A_{33} \end{bmatrix} (Transpose of the cofactor matrix) \text{ (Transpose of the cofactor matrix)}

A(adjA)=(adjA)A=∣A∣IA(adj A) = (adj A)A = |A|I

∣adjA∣=∣A∣n−1|adj A| = |A|^{n-1}

adj(AB)=(adjB)(adjA)adj(AB) = (adj B)(adj A)

adj(kA)=kn−1adjA (where n is the order of matrix A)adj(kA) = k^{n-1} adj A \text{ (where } n \text{ is the order of matrix } A \text{)}

adj(adjA)=∣A∣n−2Aadj(adj A) = |A|^{n-2} A

💡Examples

Problem 1:

Find the adjoint of the matrix A=[2314]A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}.

Solution:

  1. Find the cofactors of the elements: A11=4A_{11} = 4 A12=−1A_{12} = -1 A21=−3A_{21} = -3 A22=2A_{22} = 2

  2. Form the cofactor matrix: C=[4−1−32]C = \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}

  3. Take the transpose: adjA=CT=[4−3−12]adj A = C^T = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}

Explanation:

For a 2×22 \times 2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, the adjoint can be quickly found by swapping the diagonal elements (aa and dd) and changing the signs of the off-diagonal elements (bb and cc).

Problem 2:

If AA is a square matrix of order 33 and ∣A∣=5|A| = 5, find the value of ∣adjA∣|adj A|.

Solution:

We use the property ∣adjA∣=∣A∣n−1|adj A| = |A|^{n-1}. Given n=3n = 3 and ∣A∣=5|A| = 5. ∣adjA∣=53−1=52=25|adj A| = 5^{3-1} = 5^2 = 25

Explanation:

This property is frequently asked in CBSE 1-mark questions. The power of the determinant is always one less than the order of the square matrix.

Problem 3:

Compute adjAadj A for A=[1−12235−201]A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}.

Solution:

Calculate cofactors: A11=(3−0)=3A_{11} = (3-0) = 3 A12=−(2−(−10))=−12A_{12} = -(2 - (-10)) = -12 A13=(0−(−6))=6A_{13} = (0 - (-6)) = 6 A21=−(−1−0)=1A_{21} = -(-1-0) = 1 A22=(1−(−4))=5A_{22} = (1 - (-4)) = 5 A23=−(0−2)=2A_{23} = -(0 - 2) = 2 A31=(−5−6)=−11A_{31} = (-5-6) = -11 A32=−(5−4)=−1A_{32} = -(5-4) = -1 A33=(3−(−2))=5A_{33} = (3 - (-2)) = 5

adjA=[31−11−125−1625]adj A = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}

Explanation:

The adjoint is the transpose of the cofactor matrix. Note how A12A_{12} (row 1, col 2) becomes the element at row 2, col 1 in the adjoint matrix.

Adjoint of a matrix Class 12 Notes & Examples | CBSE Maths