Question

Given the square matrix: A = 0:3 a) Find the cofactors A11, A12, and A13- b) Find the determinant of the matrix A. c) Do you think that the matrix A is nonsingular? If yes find A-' using Elementary Row Operations. (Justify your answer) Remark: You may type your solution in the box below, or you can upload your solution as a pdf file.

linear algebra

Answer 1

Solution. The given square meet six is --/ (a) now, which is A,, - Cofactor of the elemento first element of the first row =(-1) (0)*( 7 ) = 6-152 (1x1–0x2) = 1x (1-0) = 1 HE A12 = cofactor of the element (-1) which is the second element of the first row = 6-10 (1X1 – 1X2) (-1,542 L ?! =(-1) (1-2) =(-1)*(-1) = 1 A2 = 1

A13 cofactor of the element o which is the third element of the first row. (-1)145 | o/= (-1)" (140–1X1) = 1x (0-1) A 13 = (6) The determinant of the of the given matrise A is IAI = 2 - Oy 16?! – (-1)*], 2/+ox liol =ox (1X1 – 042) – (-1) (1x141x2)+0 (140-11 - Ox (1-0) - (-1 (1-2)+ ox(0-1) = 0 + 1 X (-1) + 0 TAI = -11

(c) since we have from (b) that IAI=-170 So we Therefore the meetix A is nonsingulast Now, the cofactor of elements of the first row of can find A IAI cute 1 The cofactor of the element of the second dow of (Al are -/- % 10 % t/ l.e. 1, 0,-1 The cofactor of the element of third row 1727-12%.107 of IAI але die -2, o 1 Therefore Adj. (A) = The transpose of the matrix 0 2

2 Mody.( A ) Now, A ' Adjila) (-1) IAI - -1 2 .. . T' 0