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Determinants - Adjoint and inverse of a square matrix

Grade 12CBSE

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

πŸ”‘Concepts

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Minors and Cofactors: The minor MijM_{ij} of an element aija_{ij} of a determinant is the determinant of the sub-matrix left after deleting its ii-th row and jj-th column. The cofactor CijC_{ij} is defined as (βˆ’1)i+jMij(-1)^{i+j} M_{ij}. Visually, you can imagine 'cross-out' lines passing through the element aija_{ij} horizontally and vertically; the remaining numbers form the minor.

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Adjoint of a Square Matrix: The adjoint of a square matrix A=[aij]A = [a_{ij}] is the transpose of the matrix of its cofactors. It is denoted by adjAadj A. Visually, if you arrange all cofactors CijC_{ij} into a matrix, the adjoint is obtained by 'reflecting' that matrix across its main diagonal (switching rows and columns).

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Singular and Non-singular Matrices: A square matrix AA is called singular if its determinant ∣A∣=0|A| = 0. It is called non-singular if ∣Aβˆ£β‰ 0|A| \neq 0. Geometrically, a singular matrix collapses the area or volume it represents to zero, meaning it cannot be 'undone' or inverted.

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Existence of Inverse: A square matrix AA has an inverse Aβˆ’1A^{-1} if and only if AA is a non-singular matrix (i.e., ∣Aβˆ£β‰ 0|A| \neq 0). The inverse is the unique matrix such that AAβˆ’1=Aβˆ’1A=IAA^{-1} = A^{-1}A = I, where II is the identity matrix.

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Product of Matrix and Adjoint: For any square matrix AA of order nn, the product A(adjA)A(adj A) or (adjA)A(adj A)A results in a scalar matrix ∣A∣I|A|I. Visually, this product yields a diagonal matrix where every entry on the main diagonal is the value of the determinant ∣A∣|A|, and all other entries are zero.

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Properties of Adjoint Determinants: The determinant of the adjoint of a matrix AA of order nn is given by ∣adjA∣=∣A∣nβˆ’1|adj A| = |A|^{n-1}. This means the 'scaling factor' of the adjoint is the determinant raised to a power one less than the matrix dimension.

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Reversal Law for Inverses: For two non-singular matrices AA and BB of the same order, (AB)βˆ’1=Bβˆ’1Aβˆ’1(AB)^{-1} = B^{-1}A^{-1}. This 'socks-and-shoes' principle implies that to undo a sequence of transformations, you must undo the last one first.

πŸ“Formulae

Cij=(βˆ’1)i+jMijC_{ij} = (-1)^{i+j} M_{ij}

adjA=[Cij]Tadj A = [C_{ij}]^T

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

Aβˆ’1=1∣A∣adjAΒ where ∣Aβˆ£β‰ 0A^{-1} = \frac{1}{|A|} adj A \text{ where } |A| \neq 0

∣adjA∣=∣A∣nβˆ’1|adj A| = |A|^{n-1}

(AB)βˆ’1=Bβˆ’1Aβˆ’1(AB)^{-1} = B^{-1}A^{-1}

(Aβˆ’1)βˆ’1=A(A^{-1})^{-1} = A

(AT)βˆ’1=(Aβˆ’1)T(A^T)^{-1} = (A^{-1})^T

πŸ’‘Examples

Problem 1:

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

Solution:

Step 1: Calculate the determinant ∣A∣|A|. ∣A∣=(2Γ—4)βˆ’(3Γ—1)=8βˆ’3=5|A| = (2 \times 4) - (3 \times 1) = 8 - 3 = 5 Since ∣Aβˆ£β‰ 0|A| \neq 0, the inverse exists.

Step 2: Find the cofactors of elements. C11=(βˆ’1)1+1(4)=4C_{11} = (-1)^{1+1}(4) = 4 C12=(βˆ’1)1+2(1)=βˆ’1C_{12} = (-1)^{1+2}(1) = -1 C21=(βˆ’1)2+1(3)=βˆ’3C_{21} = (-1)^{2+1}(3) = -3 C22=(βˆ’1)2+2(2)=2C_{22} = (-1)^{2+2}(2) = 2

Step 3: Form the cofactor matrix and transpose it to find adjAadj A. adjA=[4βˆ’1βˆ’32]T=[4βˆ’3βˆ’12]adj A = \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}^T = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}

Step 4: Use the inverse formula Aβˆ’1=1∣A∣adjAA^{-1} = \frac{1}{|A|} adj A. Aβˆ’1=15[4βˆ’3βˆ’12]=[4/5βˆ’3/5βˆ’1/52/5]A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{bmatrix}

Explanation:

To find the inverse, we first check if the matrix is non-singular. Then we find the cofactors, transpose them to get the adjoint, and finally divide the adjoint by the determinant.

Problem 2:

If AA is a square matrix of order 3 and ∣A∣=4|A| = 4, find the value of ∣adjA∣|adj A|.

Solution:

Step 1: Identify the given values. Order n=3n = 3, Determinant ∣A∣=4|A| = 4.

Step 2: Use the formula for the determinant of an adjoint. ∣adjA∣=∣A∣nβˆ’1|adj A| = |A|^{n-1}

Step 3: Substitute the values into the formula. ∣adjA∣=43βˆ’1=42=16|adj A| = 4^{3-1} = 4^2 = 16

Therefore, ∣adjA∣=16|adj A| = 16.

Explanation:

This problem uses the property relating the determinant of the adjoint to the determinant of the original matrix and its order.