The Adjugate Matrix and the Inverse Matrix This Is a Version of Part of Section 8.5

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The adjugate matrix and the inverse matrix This is a version of part of Section 8.5. The adjugate of a square matrix Let A be a square matrix. By deﬁnition, the adjugate of A is a matrix B, often denoted by adj(A), with the property that AB = det(A)I = BA where I is the identity matrix the same size as A. We shall show how to construct the adjugate of any square matrix. The 2 × 2 case Let a b A = : c d We construct the adjugate as follows. • Replace each entry aij of A by the element remaining when the ith row and jth column are crossed out d c : b a • Use the following matrix of signs + − ; − + where the entry in row i and column j is the sign of (−1)i+j, to get d −c : −b a • Take the transpose of this matrix to get the adjugate of A d −b adj(A) = : −c a • Observe that A adj(A) = det(A)I = adj(A) A: Example Let 1 2 A = : 3 1 Then 1 −2 adj(A) = : −3 1 1 2 The general case Let A be an n × n matrix with entries aij. We deﬁne its adjugate as the result of the following sequence of operations. • Choose an entry aij in the matrix A. • Crossing out the entries in row i and column j, an (n − 1) × (n − 1) matrix is constructed, denoted by M(A)ij, and called a submatrix. • The determinant det(M(A)ij) is called the minor of the element aij. • If det(M(A)ij) is multiplied by the corresponding sign, we get the cofactor i+j cij = (−1) det(M(A)ij) of the element aij. • Replace each element aij by its cofactor to obtain the matrix C(A) of cofactors of A. • The transpose of the matrix of cofactors C(A) is the adjugate. Example Let 0 1 2 3 1 A = @ 2 0 1 A : −1 1 2 The matrix of minors is 0 −1 5 2 1 @ 1 5 3 A : 2 −5 −4 The matrix of cofactors is 0 −1 −5 2 1 @ −1 5 −3 A : 2 5 −4 The adjugate is the transpose of the matrix of cofactors 0 −1 −1 2 1 @ −5 5 5 A : 2 −3 −4 Theorem A square matrix A is invertible if and only if det(A) 6= 0. If det(A) 6= 0 then the inverse of A is given by 1 A−1 = adj(A): det(A) Example The matrix 0 1 2 3 1 A = @ 2 0 1 A −1 1 2 has det(A) = −5. The inverse of A therefore exists and is equal to 0 −1 −1 2 1 1 A−1 = − −5 5 5 : 5 @ A 2 −3 −4.

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