1. LAPLACIAN EXPANSION 1.1. Laplacian Expansion, Cofactor, Adjugate, Inverse Matrix Formula. In this section, We generalize the method that we used to compute the formula of 2 by 2 and 3 by 3 determinants to n by n matirx, the goal is not to write down the explicite formula, but to make use of it to analyses a method to find the inverse. Definition 1.1.1 The block diagonal matrix is the block matrix with the same partition on rows and columns, such that all non-zero blocks lies on the diagonal cells. Like: 0 1 A1 B .. C @ . A An The following could be viewed as block diagonal matrix with approperate partition 0 1 1 0 1 1 @ 2 5 A = @ 2 5 A 6 7 6 7 The above matrix could be seen as a block diagonal matrix with partition (1; 2) 0 2 3 1 0 2 3 1 B 4 6 C B 4 6 C B C B C B 2 4 7 C B 2 4 7 C B C = B C B 1 2 0 C B 1 2 0 C B C B C @ 3 5 6 A @ 3 5 6 A 6 6 The above matrix could be seen as block diagonal matrix with partition (1; 3; 2) 0 1 3 1 B 2 4 C B C B 5 6 C B C @ 5 A 7 The above matrix is not a block diagonal matrix, as the partition on its columns and rows are not the same Proposition 1.1 The determinant of block diagonal matrix is the product of the determinant of all diagonal blocks. 1 Compute the following determinant: 1 2 3 2 2 3 1 5 Answer: The determinant is just the product of the determinant of diagonal matrices. That is 1 2 3 2 2 3 1 5 1 2 2 3 = 3 2 1 5 = (−4) × 7 = −28 Definition 1.1.2 The block upper triangular matrix is the block matrix with the same partition on rows and columns, such that all non-zero blocks lies on or above the diagonal cells. Like: 0 1 A11 ··· A1n B .. C @ . A Ann Definition 1.1.3 The block lower triangular matrix is the block matrix with the same partition on rows and columns, such that all non-zero blocks lies on or below the diagonal cells. Like: 0 1 A11 B . .. C @ . A An1 ··· An Proposition 1.2 The determinant of block upper triangular matrix or block lower triangular matrix is the product of determinant of blocks of diagonals. 2 Definition 1.1.4 Let M be a n × n square matrix over F , and 1 ≤ i ≤ n, 1 ≤ j ≤ n, then the cofactor of M at i,j is defined to be the determinant of (n − 1) × (n − 1) matrix which is obtained by deleting the ith row and jth column. If we multiply this determinant by (−1)i+j, then the value is called the algebraic cofactor, denoted by Mij, where the M is the notation for original matrix. 0 1 2 2 1 Let M=@ 8 2 8 A, What is M12? −1 2 0 Answer: M12 is got by computing the determinant of cancelling the 1st row and 2nd column, and times (−1)i+j i.e. 1+2 8 8 M12 = (−1) = −8 −1 0 Now we trying to use our method to find the determinant for n × n matirx inductively, and look what we could find. Trying to simplify the determinant of 5×5 matrix into the computation of determinant 4×4 matrix 1 2 2 1 1 2 4 5 4 4 4 6 7 8 6 5 7 1 1 2 6 1 0 0 7 We use our standard method: 3 1 2 2 1 1 2 4 5 4 4 4 6 7 8 6 5 7 1 1 2 6 1 0 0 7 1 2 2 2 4 5 4 4 2 4 5 4 4 2 4 5 4 4 = 4 6 7 8 6 + 4 6 7 8 6 + 4 6 7 8 6 5 7 1 1 2 5 7 1 1 2 5 7 1 1 2 6 1 0 0 7 6 1 0 0 7 6 1 0 0 7 1 1 2 4 5 4 4 2 4 5 4 4 + 4 6 7 8 6 + 4 6 7 8 6 5 7 1 1 2 5 7 1 1 2 6 1 0 0 7 6 1 0 0 7 1 2 2 2 4 5 4 4 4 2 5 4 4 5 2 4 4 4 = 4 6 7 8 6 − 6 4 7 8 6 + 7 4 6 8 6 5 7 1 1 2 7 5 1 1 2 1 5 7 1 2 6 1 0 0 7 1 6 0 0 7 0 6 1 0 7 1 1 4 2 4 5 4 4 2 4 5 4 − 8 4 6 7 6 + 6 4 6 7 8 1 5 7 1 2 2 5 7 1 1 0 6 1 0 7 7 6 1 0 0 4 5 4 4 2 5 4 4 2 4 4 4 6 7 8 6 4 7 8 6 4 6 8 6 = 1 × − 2 × + 2 × 7 1 1 2 5 1 1 2 5 7 1 2 1 0 0 7 6 0 0 7 6 1 0 7 2 4 5 4 2 4 5 4 4 6 7 6 4 6 7 8 −1 × + 1 × 5 7 1 2 5 7 1 1 6 1 0 7 6 1 0 0 As you can see, in the expansion, we expanded out the all the entry of the first row, and left with the determinant of numbers that in the different row and column with the entry, with the sign alternating, that exactly the first row times the cofactor corresponds to first row. To conclude, we now have the Laplacian Expansion: 4 Definition 1.1.5 Suppose A is a square matrix, then we can expand A by any row or any column. namely Row expansion: det(A) = ai1Ai1 + ai2Ai2 + ··· + ainAin Column expansion: det(A) = a1iA1i + a2iA2i + ··· + aniAni (Please keep in mind to change the sign, as the term appeared on the formula means algebraic cofac- tors) Expand the following matrix by the second row 6 8 9 2 4 1 1 2 5 8 9 6 9 6 8 = −2 × + 4 × − 1 × 2 5 1 5 1 2 Expand the following matrix by the second column 6 8 9 2 4 1 1 2 5 6 9 2 1 2 6 = −4 × + 8 × − 2 × 1 5 1 5 1 9 As you can see, the method we used above can inductively give the formula, but what is more than that is the following observation: What if I change the coefficient of each corfactor by the entry of other row? Then in the previous example it would be the laplacian expansion of the determinant: In the expansion of 6 8 9 2 4 1 1 2 5 8 9 6 9 6 8 = −2 × + 4 × − 1 × 2 5 1 5 1 2 Now if we change the coefficient by the corresponding entries of other lines, that is 5 8 9 6 9 6 8 −1 × + 2 × − 5 × 2 5 1 5 1 2 What the value would be? as you can see, the line that we expanded only determined the coefficient in front of each algebraic cofacter, not appear in any of the cofacter . if we change the coefficient, the only change is the line that we expand by. 6 8 9 8 9 6 9 6 8 −1 × + 2 × − 5 × = −1 −2 −5 = 0 2 5 1 5 1 2 1 2 5 we get the last equation because the second row is scalar multiple of the third row Proposition 1.3 The othogonal property of cofactor: let A = mataijnn be a matirx, r = 0 1 k Al1 B Al2 C a a ··· a cl = B C k1 k2 kn be any row of A, and B . C be the cofactor of some row. then @ . A Aln l rkc = 0 if k 6= l l rkc = detA if k = l At this point, we can make the propersition: Now we construct a matrix, namely, Adjugate Matrix of A, it is constructed by putting Aij in the j’th row and i’th column(remember the order). Then with the adjugate matrix, we have the following property: Proposition 1.4 ∗ ∗ let A = (aij)1≤i≤n be a matirx, we define A = (Aji)1≤j≤n to be the adjunct matrix, then A A = 1≤j≤n 1≤i≤n det(A)In Then if the determinant is not 0, simply divide the adjugate matrix by determinant, we can get the inverse Proposition 1.5 ∗ let A = (aij)1≤i≤n be a matirx, we define A = (Aji)1≤j≤n to be the adjunct matrix, then 1≤j≤n 1≤i≤n −1 A = (Aji)1≤j≤n 1≤i≤n 0 2 4 6 1 Find the inverse of @ 1 3 4 A 6 7 2 Answer: 6 0 2 4 6 1 @ 1 3 4 A 6 7 2 0 1 3 4 4 6 4 6 − B 7 2 7 2 3 4 C B C 1 B 1 4 2 6 2 6 C = B − − C 6 2 6 2 1 4 2 4 6 B C B C 1 3 4 1 3 2 4 2 4 @ A − 6 7 2 6 7 6 7 1 3 0 −22 34 −2 1 1 = − 64 @ 22 −32 −2 A −11 10 2 7.