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Algebra - Adjoint and Inverse of a Matrix

Grade 12ICSE

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

🔑Concepts

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Determinant of a Square Matrix: Every square matrix AA is associated with a scalar called its determinant, denoted by ∣A∣|A| or det⁡(A)\det(A). Only square matrices have determinants. Visually, the determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is calculated by subtracting the product of the secondary diagonal from the product of the main diagonal: ad−bcad - bc.

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Minors and Cofactors: For an element aija_{ij} of a matrix, the minor MijM_{ij} is the determinant of the sub-matrix formed by deleting the ithi^{th} row and jthj^{th} column. The cofactor CijC_{ij} is given by (−1)i+jMij(-1)^{i+j} M_{ij}. Visually, cofactors follow a 'checkerboard' sign pattern: (+−+−+−+−+)\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix} for a 3×33 \times 3 matrix.

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Adjoint of a Matrix (adjAadj A): The adjoint (or adjugate) of a square matrix AA is the transpose of its cofactor matrix. Visually, for a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the adjoint is easily found by swapping the positions of aa and dd, and changing the signs of bb and cc, resulting in (d−b−ca)\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.

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Singular and Non-Singular Matrices: A square matrix AA is called singular if its determinant is zero (∣A∣=0|A| = 0). It is non-singular if ∣A∣≠0|A| \neq 0. An inverse exists only for non-singular matrices.

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Inverse of a Matrix (A−1A^{-1}): The inverse is a unique matrix such that A⋅A−1=A−1⋅A=IA \cdot A^{-1} = A^{-1} \cdot A = I, where II is the identity matrix. Visually, multiplying a matrix by its inverse 'undoes' the transformation, returning the system to its original state (the identity).

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Existence Condition: A square matrix AA has an inverse if and only if it is non-singular (∣A∣≠0|A| \neq 0). If ∣A∣=0|A| = 0, the matrix is non-invertible or singular.

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Properties of Inverse and Adjoint: Important relationships include the 'reversal law' (AB)−1=B−1A−1(AB)^{-1} = B^{-1} A^{-1} and the property that the determinant of the adjoint of AA of order nn is ∣adjA∣=∣A∣n−1|adj A| = |A|^{n-1}.

📐Formulae

Cij=(−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

A−1=frac1∣A∣adjAA^{-1} = \\frac{1}{|A|} adj A, provided ∣A∣≠0|A| \neq 0

(AB)−1=B−1A−1(AB)^{-1} = B^{-1} A^{-1}

(AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T

∣A−1∣=frac1∣A∣|A^{-1}| = \\frac{1}{|A|}

∣adjA∣=∣A∣n−1|adj A| = |A|^{n-1}, where nn is the order of the matrix

💡Examples

Problem 1:

Find the inverse of the matrix A=(2358)A = \begin{pmatrix} 2 & 3 \\ 5 & 8 \end{pmatrix}.

Solution:

Step 1: Calculate the determinant of AA. ∣A∣=(2times8)−(3times5)=16−15=1|A| = (2 \\times 8) - (3 \\times 5) = 16 - 15 = 1 Since ∣A∣neq0|A| \\neq 0, A−1A^{-1} exists.

Step 2: Find the cofactors of the elements. C11=8C_{11} = 8, C12=−5C_{12} = -5, C21=−3C_{21} = -3, C22=2C_{22} = 2

Step 3: Write the cofactor matrix and find its transpose to get adjAadj A. CofactorMatrix=(8−5−32)Cofactor Matrix = \begin{pmatrix} 8 & -5 \\ -3 & 2 \end{pmatrix} adjA=(8−3−52)adj A = \begin{pmatrix} 8 & -3 \\ -5 & 2 \end{pmatrix}

Step 4: Use the formula A−1=frac1∣A∣adjAA^{-1} = \\frac{1}{|A|} adj A. A−1=frac11(8−3−52)=(8−3−52)A^{-1} = \\frac{1}{1} \begin{pmatrix} 8 & -3 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 8 & -3 \\ -5 & 2 \end{pmatrix}

Explanation:

To find the inverse of a 2×22 \times 2 matrix, we first ensure it is non-singular (∣A∣≠0|A| \neq 0). Then we swap the main diagonal elements, negate the off-diagonal elements to find the adjoint, and divide by the determinant.

Problem 2:

If AA is a square matrix of order 3 such that ∣A∣=5|A| = 5, find the value of ∣adjA∣|adj A|.

Solution:

Step 1: Identify the order of the matrix nn and the value of ∣A∣|A|. Here, n=3n = 3 and ∣A∣=5|A| = 5.

Step 2: Apply the formula for the determinant of an adjoint matrix. ∣adjA∣=∣A∣n−1|adj A| = |A|^{n-1}

Step 3: Substitute the values into the formula. ∣adjA∣=53−1=52=25|adj A| = 5^{3-1} = 5^2 = 25

Therefore, ∣adjA∣=25|adj A| = 25.

Explanation:

This problem uses the property ∣adjA∣=∣A∣n−1|adj A| = |A|^{n-1}, which relates the determinant of the adjoint matrix to the determinant of the original matrix and its order.